Answer:
a) The temperature of the beaker rises as this transfer of heat goes on.
b) Check Explanation.
Explanation:
a) The heat lost by the piece of metal is normally gained by the all the components that it comes in contact with after the heating procedure.
(Heat lost by piece of metal) = (Heat gained by the cold water) + (Heat gained by the beaker).
So, since heat is also gained by the Beaker, its temperature should rise under normal conditions.
That is essentially what the zeroth law of thermodynamics about thermal equilibrium talks about.
If two bodies are at thermal equilibrium with reach other and body 2 is in thermal equilibrium with a third body, then body 1 and body 3 are also in thermal equilibrium
Temperature of the piece of metal decreases, temperature of water rises and the temperature of the beaker rises as they all try to attain thermal equilibrium.
b) In calorimetry, the aim is usually for the water (in this case) to take up all of the heat supplied by the piece of metal. Hence, the calorimeter is usually heavily insulated (or properly called lagged). Thereby, reducing the amount of heat that the calorimeter would gain.
But in cases where the heat lost to the insulated calorimeter isn't negligible, the heat capacity of the calorimeter is usually obtained and included it is included in the heat transfer calculations.
Hope this Helps!!!
The temperature of the beaker of cold water will increase due to the transfer of heat from the hot metal. In calorimetry, the heat gained by the water and the beaker is equal to the heat lost by the metal.
When the hot metal is dropped into the beaker of cold water, a heat transfer process occurs. According to the principle of conservation of energy, energy cannot be created or destroyed, only transferred or transformed. In this case, heat energy is transferred from the higher temperature metal to the lower temperature water and beaker.
The specific heat capacity of a substance is the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius. The heat lost by the metal can be calculated using the formula:
[tex]\[ q_{\text{metal}} = m_{\text{metal}} \cdot c_{\text{metal}} \cdot \Delta T_{\text{metal}} \][/tex]
where [tex]\( q_{\text{metal}} \)[/tex]is the heat lost by the metal, [tex]\( m_{\text{metal}} \)[/tex] is the mass of the metal, [tex]\( c_{\text{metal}} \)[/tex] is the specific heat capacity of the metal, and [tex]\( \Delta T_{\text{metal}} \)[/tex] is the change in temperature of the metal.
Similarly, the heat gained by the water can be calculated using the formula:
[tex]\[ q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}} \][/tex]
where [tex]\( q_{\text{water}} \)[/tex] is the heat gained by the water, [tex]\( m_{\text{water}} \)[/tex] is the mass of the water,[tex]\( c_{\text{water}} \)[/tex] is the specific heat capacity of water, and [tex]\( \Delta T_{\text{water}} \)[/tex] is the change in temperature of the water.
In calorimetry, assuming no heat is lost to the surroundings, the heat lost by the metal is equal to the heat gained by the water (and the beaker, if its heat capacity is considered):
[tex]\[ q_{\text{metal}} = q_{\text{water}} + q_{\text{beaker}} \][/tex][tex]\[ m_{\text{metal}} \cdot c_{\text{metal}} \cdot \Delta T_{\text{metal}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}} + m_{\text{beaker}} \cdot c_{\text{beaker}} \cdot \Delta T_{\text{beaker}} \][/tex]
Here, [tex]\( m_{\text{beaker}} \)[/tex] is the mass of the beaker, [tex]\( c_{\text{beaker}} \)[/tex] is the specific heat capacity of the beaker material, and [tex]\( \Delta T_{\text{beaker}} \)[/tex] is the change in temperature of the beaker. The temperature change of the beaker and the water will be the same if they reach thermal equilibrium.
The temperature of the beaker will rise until thermal equilibrium is reached, at which point the temperature of the metal, water, and beaker will be the same. The heat that causes the temperature of the glass beaker to rise is accounted for in calorimetry by including the beaker's mass and specific heat capacity in the calculations.
The size of a population increases if the number of individuals added to the population is equal to the number of individuals leaving the population, is this true
Answer:
The answer to the question
The size of a population increases if the number of individuals added to the population is equal to the number of individuals leaving the population, is this true?
is No
Explanation:
The size of a population increases if the number of individuals added to the population is more than the number of individuals leaving the population
In biology, a population is the total number of organism of a particular species or group living together at a given geographical location. It is the number of inhabitants of a place.
Population growth is the increase in the number count of the specie or group in a population.
a 125g bar of aluminum at 22 degrees celsius. determine the final temperature of the aluminum, if the amount of energy applied is equal to 3600 calories. the specific heat of aluminum is .90 j/gc
Answer : The final temperature of the aluminum is, [tex]155.9^oC[/tex]
Explanation :
Formula used :
[tex]q=m\times c\times (T_{final}-T_{initial})[/tex]
where,
q = heat = 3600 cal = 15062.4 J (1 cal = 4.184 J)
m = mass of aluminum = 125 g
c = specific heat of aluminum = [tex]0.90J/g^oC[/tex]
[tex]T_{final}[/tex] = final temperature = ?
[tex]T_{initial}[/tex] = initial temperature = [tex]22^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]15062.4J=125g\times 0.90J/g^oC\times (T_{final}-22)^oC[/tex]
[tex]T_{final}=155.9^oC[/tex]
Thus, the final temperature of the aluminum is, [tex]155.9^oC[/tex]
A sample of carbon dioxide occupies 2.77 L at a pressure of 18.1 psi and a temperature of 56ºC. What pressure will the carbon dioxide exert if the volume is increased to 4.81 L while the temperature is held constant?
Answer:
10.4psi
Explanation:
Since the problem involves pressure and volume at constant temperature, Boyle's law is the appropriate formula to use.
Boyle's law States that at a constant temperature, the volume of a given mass of a bass is inversely proportional to its pressure.
Using Boyle's law
P1v1 = p2v2
P1 is 18.1psi
V1 is 2.77L
P2 =?
V2 is 4.81
Temperature is constant
P2 =P1v1/ v2
= 18.1 x 2.77 / 4.82
= 10.4psi
For the reaction 2 H2O(g) equilibrium reaction arrow 2 H2(g) + O2(g), K = 2.4 ✕ 10−3 at a given temperature. At equilibrium in a 2.7 L container, it is found that [H2O(g)] = 2.0 ✕ 10−1 M and [H2(g)] = 2.5 ✕ 10−2 M. Calculate the moles of O2(g) present under these conditions.
To calculate the moles of O2(g) present at equilibrium, use the equilibrium constant expression and the given concentrations of H2O(g) and H2(g). The moles of O2(g) present at equilibrium is approximately 7.23 × 10-3 mol.
Explanation:To calculate the moles of O2(g) present at equilibrium, we can use the equilibrium constant expression and the given concentrations of H2O(g) and H2(g). The equilibrium constant expression for the given reaction is K = [H2]^2/[H2O]. Since we have the values for [H2] and [H2O], we can rearrange the expression to solve for [O2], which is the unknown.
By substituting the known values into the expression and solving for [O2], we get [O2] ≈ 2.67 × 10-3 M. This represents the concentration of O2(g), which we can then convert to moles of O2 using the volume of the container.
Given that the container has a volume of 2.7 L, we can multiply the concentration of O2 by the volume to get the moles of O2(g). In this case, the moles of O2(g) present at equilibrium would be approximately 7.23 × 10-3 mol.
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To calculate the moles of O2(g) present under the given conditions, divide the concentration of H2(g) by 2 and use the ideal gas law to find the number of moles in the 2.7 L container. The number of moles of O2(g) in the 2.7 L container is equal to the number of moles in any other volume multiplied by the ratio of the volumes.
Explanation:To calculate the moles of O2(g) present under the given conditions, we can first start by calculating the concentration of O2(g) using the given equilibrium concentrations of H2O(g) and H2(g). Since the reaction is 2 H2O(g) <=> 2 H2(g) + O2(g), the concentration of O2(g) is equal to half the concentration of H2(g). Therefore, the concentration of O2(g) is 2.5 x 10^-2 M / 2 = 1.25 x 10^-2 M.
Next, we can use the ideal gas law to calculate the number of moles of O2(g) in the 2.7 L container. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin. In this case, we don't have the pressure or the temperature, so we can't directly calculate the number of moles. However, since the pressure and temperature are not changing, we can assume that the pressure and temperature are constant, and therefore the number of moles is directly proportional to the volume. So, if we know the number of moles of O2(g) in the 2.7 L container, we can calculate the number of moles in any other volume using the equation: n1V1 = n2V2.
Therefore, if we let n1 be the number of moles of O2(g) in the 2.7 L container and n2 be the number of moles of O2(g) in another volume, we can set up the equation: n1 x 2.7 = n2 x V2. Substituting the values, we get: n1 x 2.7 = 1.25 x 10^-2 x V2. Solving for n2, we get: n2 = n1 x 2.7 / (1.25 x 10^-2). Plugging in the value of n1 as 1.25 x 10^-2 M x 2.7 L, we can calculate n2.
Therefore, the moles of O2(g) present under the given conditions can be calculated as follows: n2 = (1.25 x 10^-2 M) x (2.7 L) / (1.25 x 10^-2) = 2.7 moles.
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When powdered zinc is heated with sulfur, a violent reaction occurs, and zinc sulfide forms: Some of the reactants also combine with oxygen in air to form zinc oxide and sulfur dioxide. When 83.2 g of Zn reacts with 52.4 g of S8, 104.4 g of ZnS forms. (a) What is the percent yield of ZnS? (b) If all the remaining reactants combine with oxygen, how many grams of each of the two oxides form?
Answer:
Explanation:
The detailed steps and step by step calculations is as shown in the attached file.
To calculate the percent yield of ZnS, use the given masses of Zn and S8 and compare to the theoretical yield. For the oxides formed, determine the moles of remaining Zn and S8 using the balanced chemical equation.
Explanation:To calculate the percent yield of ZnS, we need to compare the actual yield (104.4 g) to the theoretical yield. The theoretical yield can be calculated using the stoichiometry of the reaction. The balanced chemical equation is: Zn + S8 → ZnS. The molar ratio between Zn and ZnS is 1:1, which means that the moles of ZnS formed will be equal to the moles of Zn used. Convert the given masses of Zn and S8 to moles using their molar masses, then compare the moles to calculate the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction. Once you have determined the limiting reactant and calculated the theoretical yield, you can calculate the percent yield using the formula: (actual yield / theoretical yield) x 100.
For part (b), if all the remaining reactants combine with oxygen, we need to determine the remaining moles of Zn and S8 after the reaction with ZnS is complete. Convert these moles to masses using their molar masses, and then use the balanced chemical equation to determine the moles of each oxide formed. The balanced chemical equation for the reaction of zinc with oxygen is: 2Zn + O2 → 2ZnO. The molar ratio between Zn and ZnO is 2:2, so the moles of ZnO formed will be equal to the moles of Zn remaining. The balanced chemical equation for the reaction of sulfur with oxygen is: S8 + 8O2 → 8SO2. The molar ratio between S8 and SO2 is 1:8, so the moles of SO2 formed will be 8 times the moles of S8 remaining. Convert the moles of each oxide formed to masses using their molar masses.
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How many grams of NaCl would need to be added to 1001 g of water to increase the boiling temperature of the solution by 1.500 °C? (Kb for water is 0.5100 °C/m)
To find the amount of NaCl needed to increase the boiling temperature of water by 1.500°C, one must understand molal boiling point elevation and van't Hoff factor. Using the formula and given constants, we calculate the grams of NaCl required through the steps of determining molality, calculating the moles of NaCl, and then converting to grams.
Explanation:To calculate the amount of NaCl needed to raise the boiling temperature of water by 1.500°C, we first understand the concept of molal boiling point elevation and van't Hoff factor. Given that the Kb for water is 0.5100°C/m, we use the formula ΔTb = i*Kb*m, where ΔTb is the boiling point elevation, i is the van't Hoff factor for NaCl (which is 2 because NaCl disassociates into Na+ and Cl- ions), Kb is the ebullioscopic constant for the solvent (water), and m is the molality of the solution.
To find the molality (m), we rearrange the equation as m = ΔTb / (i*Kb). Substituting the known values, we get m = 1.500 / (2*0.5100) = 1.4706 mol/kg. Knowing the molality and the mass of the solvent (water), we can then calculate the moles of NaCl required, which is molality * mass of solvent in kg. Finally, converting moles of NaCl to grams using its molar mass (58.44 g/mol), gives us the total grams of NaCl needed. This step-by-step process provides a clear link between the theoretical concepts and their practical application.
Approximately 85.78 grams of [tex]NaCl[/tex] would need to be added to 1001 grams of water to increase the boiling temperature of the solution by [tex]1.500 \°C[/tex].
To solve this problem, we will use the concept of boiling point elevation, which states that the boiling point of a solvent is increased by an amount proportional to the molal concentration of the solute. The proportionality constant is known as the ebullioscopic constant (Kb). The formula to calculate the boiling point elevation is:
[tex]\[ \Delta T_b = i \cdot K_b \cdot m \][/tex]
where:
- [tex]\( \Delta T_b \)[/tex] is the increase in boiling point temperature,
- [tex]\( i \)[/tex] is the van 't Hoff factor (the number of moles of particles in solution per mole of solute; for [tex]NaCl[/tex], [tex]\( i = 2 \)[/tex] because it dissociates into two ions, [tex]Na^+ and Cl^-)[/tex],
- [tex]\( K_b \)[/tex] is the ebullioscopic constant for the solvent (given as [tex]0.5100 \°C[/tex]/m for water),
- [tex]\( m \)[/tex] is the molality of the solution (moles of solute per kilogram of solvent).
Given:
- [tex]\( \Delta T_b = 1.5000 °C \)[/tex]
- [tex]\( K_b = 0.5100 °C/m \)[/tex]
- [tex]\( i = 2 \)[/tex] for [tex]NaCl[/tex]
- Mass of water (solvent) = 1001 g (which is approximately 1 kg, since 1000 g = 1 kg)
First, we will rearrange the formula to solve for the molality (m):
[tex]\[ m = \frac{\Delta T_b}{i \cdot K_b} \][/tex]
Substitute the given values:
[tex]\[ m = \frac{1.5000 °C}{2 \cdot 0.5100 °C/m} \][/tex]
[tex]\[ m = \frac{1.5000 °C}{1.0200 °C/m} \][/tex]
[tex]\[ m \approx 1.4695 m \][/tex]
Now that we have the molality, we can calculate the number of moles of [tex]NaCl[/tex] needed to achieve this molality in 1 kg of water:
[tex]\[ \text{moles of NaCl} = m \cdot \text{mass of solvent in kg} \][/tex]
[tex]\[ \text{moles of NaCl} = 1.4695 \text{ m} \cdot 1 \text{ kg} \][/tex]
[tex]\[ \text{moles of NaCl} \approx 1.4695 \][/tex]
The molar mass of [tex]NaCl[/tex] is approximately 58.44 g/mol. To find the mass of [tex]NaCl[/tex] needed, we multiply the number of moles by the molar mass:
[tex]\[ \text{mass of NaCl} = \text{moles of NaCl} \cdot \text{molar mass of NaCl} \][/tex]
[tex]\[ \text{mass of NaCl} \approx 1.4695 \text{ mol} \cdot 58.44 \text{ g/mol} \][/tex]
[tex]\[ \text{mass of NaCl} \approx 85.7797 \text{ g} \][/tex]
Describe a simple chemical test that could be used to distinguish between members of each of the following pairs of compounds (a) 4-Chlorophenol and 4-chloro-1-methylbenzene (c) 4-Methylphenol and 2,4,6-trinitrophenol (b) 4-Methylphenol and 4-methylbenzoic acid (d) Ethyl phenyl ether and 4-methylphenol
Answer:
Explanation:
(A) Phenol gives violet color by complexation with Fe3+ solution . It is best identification...
(B) For b , same phenolic test can be done ...But other esterification is also possibility...
(C) Picric acid has a particular identification test ,
(D) Here , like the first one , phenolic test with FeCl3 gives violet color for 4-ethylphenol and no color for ethyl phenyl ether...
Copper(I) ions in aqueous solution react with NH 3 ( aq ) according to Cu + ( aq ) + 2 NH 3 ( aq ) ⟶ Cu ( NH 3 ) + 2 ( aq ) K f = 6.3 × 10 10 Calculate the solubility (in g·L−1) of CuBr ( s ) ( K sp = 6.3 × 10 − 9 ) in 0.76 M NH 3 ( aq ) .
Answer:
53.18 gL⁻¹
Explanation:
Given that:
[tex]Cu^{2+}_{(aq)}[/tex] [tex]+[/tex] [tex]2NH_{3(aq)}[/tex] [tex]------>[/tex] [tex][Cu(NH_3)_2]^+_{(aq)}[/tex] ------equation (1)
where;
Formation Constant [tex](k_f) =[/tex] [tex]6.3*10^{10}[/tex]
However, the Dissociation of [tex]CuBr_{(s)[/tex] yields:
[tex]CuBr_{(s)}[/tex] ⇄ [tex]Cu^{+}_{(aq)}[/tex] [tex]+[/tex] [tex]Br^-_{(aq)}[/tex] -------------- equation (2)
where;
the Solubility Constant [tex](k_{sp})[/tex] [tex]= 6.3 *10^{-9[/tex]
From equation (1);
[tex](k_f) =[/tex] [tex]\frac{[[Cu(NH_3)_2]^+]}{[Cu^{2+}][NH_3]^{2}}[/tex] --------- equation (3)
From equation (2)
[tex](k_{sp})[/tex] [tex]= [Cu^+][Br^-][/tex] --------- equation (4)
In [tex]NH_3[/tex], the net reaction for [tex]CuBr_{(s)[/tex] can be illustrated as:
[tex]CuBr_{(s)[/tex] [tex]+[/tex] [tex]2NH_{3(aq)}[/tex] ⇄ [tex][Cu(NH_3)_2]^+_{(aq)}[/tex] [tex]+ Br^-_{(aq)}[/tex]
The equilibrium constant (K) can be written as :
[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}[/tex]
If we multiply both the numerator and the denominator with [tex][Cu^+][/tex] ; we have:
[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}*\frac{[Cu^+]}{[Cu^+]}[/tex]
[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+}{[NH_3]^2[Cu^+]}*{[Cu^+][Br^-]}[/tex]
[tex]K = k_f *k_{sp}[/tex]
[tex]K= (6.3*10^{10})*(6.3*10^{-9})[/tex]
[tex]K= 3.97*10^2[/tex]
[tex]K[/tex] ≅ [tex]4.0*10^2[/tex]
Now; we can re-write our equilibrium constant again as:
[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}[/tex]
[tex]4.0*10^2 = \frac{(x)(x)}{(0.76-2x)^2}[/tex]
[tex]4.0*10^2 = \frac{(x)^2}{(0.76-2x)^2}[/tex]
[tex]4.0*10^2 = (\frac{(x)}{(0.76-2x)})^2[/tex]
By finding the square of both sides, we have
[tex]\sqrt {4.0*10^2} = \sqrt {(\frac{(x)}{(0.76-2x)})^2[/tex]
[tex]2.0*10 = \frac{x}{(0.76-2x)}[/tex]
[tex]20(0.76-2x) =x[/tex]
[tex]15.2 -40x=x[/tex]
[tex]15.2 = 40x +x[/tex]
[tex]15.2 = 41x[/tex]
[tex]x = \frac{15.2}{41}[/tex]
[tex]x = 0.3707 M[/tex]
In gL⁻¹; the solubility of [tex]CuBr_{(s)[/tex] in 0.76 M [tex]NH_3[/tex] solution will be:
[tex]= \frac{0.3707 mole of CuBr}{1L}*\frac{143.45 g}{mole of CuBr}[/tex]
= 53.18 gL⁻¹
The problem involves finding the solubility of copper(I) bromide in a solution of ammonia. The solution entails expressing the formation of complex ions and dissolution of CuBr in terms of equilibrium expressions. Solving the system of expressions for solubility 's' can eventually give the solubility in g·L-1.
Explanation:In this problem, we are looking for the solubility of copper(I) bromide (CuBr) in a solution with 0.76 M NH3. Since CuBr does not have high solubility in water, we are using ammonia which forms complex ions with copper, enhancing its solubility. We will utilize the given equilibrium constant (Kf) which helps us to understand how far the forward reaction (complex ion formation) proceeds.
To start, we'll write the reaction of Cu+ ion with NH3 and the corresponding Kf expression:
Cu+ (aq) + 2NH3 (aq) <=> Cu(NH3)2+ (aq); Kf = [Cu(NH3)2+] / [Cu+] [NH3]^2
Next, we need to find an expression for [Cu+] in terms of solubility and solve for solubility. Let's denote the solubility of CuBr as 's'. The dissolution of CuBr can be represented as:
CuBr (s) <=> Cu+ (aq) + Br- (aq)
We can see that for every mole of CuBr that dissolves, one mole of Cu+ ion is produced. Thus, [Cu+] = s.
Now using the Kf expression, we can solve the system of equations to find s, which effectively gives us the solubility of CuBr in 0.76 M NH3. After that, by taking the molecular weight of CuBr into account, we can finally express the solubility in g·L−1.
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an aqeous solution of oxalic acid h2c2o4 was prepared by dissolving a 0.5842g of solute in enough water to make a 100 ml solution a 10 ml aliquot of this solution was then transferred to a volumetric flask and diluted to a final volume of 250 ml?propanoic
The question is incomplete, here is the complete question:
An aqeous solution of oxalic acid was prepared by dissolving a 0.5842 g of solute in enough water to make a 100 ml solution a 10 ml aliquot of this solution was then transferred to a volumetric flask and diluted to a final volume of 250 ml. How many grams of oxalic acid are in 100. mL of the final solution?
Answer: The mass of oxalic acid in final solution is 0.0234 grams
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex] ......(1)
Given mass of oxalic acid = 0.5842 g
Molar mass of oxalic acid = 90 g/mol
Volume of solution = 100 mL
Putting values in equation 1, we get:
[tex]\text{Molarity of oxalic acid solution}=\frac{0.5842\times 1000}{90\times 100}\\\\\text{Molarity of oxalic acid solution}=0.0649M[/tex]
To calculate the molarity of the diluted solution, we use the equation:
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of the concentrated oxalic acid solution
[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted oxalic acid solution
We are given:
[tex]M_1=0.0649M\\V_1=10mL\\M_2=?M\\V_2=250.0mL[/tex]
Putting values in above equation, we get:
[tex]0.0649\times 10=M_2\times 250.0\\\\M_2=\frac{0.0649\times 10}{250}=0.0026M[/tex]
Now, calculating the mass of glucose by using equation 1, we get:
Molarity of oxalic acid solution = 0.0026 M
Molar mass of oxalic acid = 90 g/mol
Volume of solution = 100 mL
Putting values in equation 1, we get:
[tex]0.0026=\frac{\text{Mass of oxalic acid solution}\times 1000}{90\times 100}\\\\\text{Mass of oxalic acid solution}=\frac{0.0026\times 90\times 100}{1000}=0.0234g[/tex]
Hence, the mass of oxalic acid in final solution is 0.0234 grams
The question involves dilution of an oxalic acid solution and subsequent calculation of the molarity for titration, which is a typical procedure in a college-level chemistry course.
Explanation:The student's question pertains to the dilution of an aqueous solution of oxalic acid and the calculation of molarity after dilution and titration. Starting with a mass of 0.5842g of oxalic acid dissolved to make a 100 ml solution, a 10 ml aliquot is further diluted to 250 ml. The goal is to understand the change in concentration as a result of the dilution process and to apply this understanding to various problems presented, such as titration against potassium permanganate (KMnO4).
How many grams of potassium fluoride can form if 4.00 grams of potassium are reacted with 3.00 grams of fluorine gas according to the reaction: 2K (s) + F2 (g) → 2KF (s)
Answer:
We can for 5.93 grams potassium fluoride
Explanation:
Step 1: Data given
Mass of potassium = 4.00 grams
Mass of fluorine = 3.00 grams
Molar mass potassium = 39.10 g/mol
Molar mass fluorine gas =38.00 g/mol
Step 2: The balanced equation
2K (s) + F2 (g) → 2KF (s)
Step 3: Calculate moles potassium
Moles potassium = 4.00 grams / 39.10 g/mol
Moles potassium = 0.102 moles
Step 4: Calculate moles F2
Moles F2 = 3.00 grams / 38.00 g/mol
Moles F2 = 0.0789 moles
Step 5: Calculate limiting reactant
Potassium is the limiting reactant. There will react 0.102 moles
Fluorine gas is in excess. There will react 0.102/ 2 = 0.051 moles
There will remain 0.0789 - 0.051 = 0.0279 moles
Step 6: Calculate moles potassium fluoride
For 2 moles potassium we need 1 mol fluorine to produce 2 moles potassium fluoride
For 0.102 moles K we need 0.102 moles KF
Step 7: Calculate mass KF
Mass KF = moles KF * molar mass KF
Mass KF = 0.102 moles * 58.10 g/mol
Mass KF = 5.93 grams
Final answer:
By calculating the moles of potassium and fluorine used in the reaction from their given masses, and considering the stoichiometry of the balanced equation, it's determined that 4.59 grams of potassium fluoride can be produced.
Explanation:
The question asks how many grams of potassium fluoride can form when 4.00 grams of potassium react with 3.00 grams of fluorine gas according to the balanced chemical equation 2K (s) + F2 (g) → 2KF (s). To solve this, we first find the molar mass of potassium (K) and fluorine (F2), which are approximately 39.10 g/mol and 38.00 g/mol, respectively. Then, we calculate the moles of K and F2 available by dividing the given masses by their respective molar masses. With potassium: 4.00 g / 39.10 g/mol = 0.102 moles of K; with fluorine: 3.00 g / 38.00 g/mol = 0.079 moles of F2. As the equation shows a 2:1 ratio, fluorine limits the reaction. Therefore, we can form 0.079 moles of KF. The molar mass of KF is approximately 58.10 g/mol (39.10 g/mol for K + 19.00 g/mol for F). Thus, the mass of KF that can form is 0.079 moles × 58.10 g/mol = 4.59 grams.
Choose the correct statement about melting points. * Melting point can tell us if we have a mixture of compounds (for example, compound of interest impurity). Melting point can tell us the identity of the components of a mixture. If we have a sample with a lowered melting range as compared to our standards, we can do mixed melting point determinations, combining our sample with each standard, to determine the identity of our sample. All of these are true. None of these are true.
Answer: All of these statements are true
Explanation:
Melting point help us to determine if a mixture is pure or has impurities by the virtues of it melting range..
The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochloric acid (HCl) is 3.32 × 10−10 s−1 at 25°C. Calculate the rate constant at 48.5°C if the activation energy is 116 kJ/mol.
Answer:
K(48.5°C) = 1.017 E-8 s-1
Explanation:
CH3Cl + H2O → CH3OH + HClat T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1
at T2 = 48.5°C (321.5 K) ⇒ K2 = ?
Arrhenius eq:
K(T) = A e∧(-Ea/RT)Ln K = Ln(A) - [(Ea/R)(1/T)]∴ A: frecuency factor
∴ R = 8.314 E-3 KJ/K.mol
⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)
⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)
(1)/(2):
⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)
⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)
⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)
⇒ Ln (K1/K2) = - 3.422
⇒ K1/K2 = e∧(-3.422)
⇒ (3.32 E-10 s-1)/K2 = 0.0326
⇒ K2 = (3.32 E-10 s-1)/0.0326
⇒ K2 = 1.017 E-8 s-1
The rate constant at 48.5°C if the activation energy is 116 kJ/mol is [tex]1.017 E-8 s^{-1}[/tex]
Chemical reaction:CH₃Cl + H₂O → CH₃OH + HCl
At T₁ = 25°C (298 K) ⇒ K₁ = 3.32 E-10 s-1
At T₂ = 48.5°C (321.5 K) ⇒ K₂ = ?
According to Arrhenius equation:
[tex]K(T) = A*e^{(-Ea/RT)}\\\\ln K = ln(A) - [(Ea/R)(1/T)][/tex]
where,
A = frequency factor
R = 8.314 E-3 KJ/K.mol
[tex]ln K_1 = ln(A) - [Ea/R)*(1/T_1)][/tex]..........(1)
[tex]ln K_2 = ln(A) - [(Ea/R)*(1/T_2)][/tex]...........(2)
On dividing equation 1 by 2:
ln (K₁/K₂) = (Ea/R)* (1/T₂ - 1/T₁)
ln (K₁/K₂) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)
ln (K₁/K₂) = (13952.37 K)*(- 2.453 E-4 K-1)
ln (K₁/K₂) = - 3.422
(K₁/K₂)= [tex]e^{(-3.422)}[/tex]
[tex](3.32 E-10 s^{-1})[/tex] / K₂ = 0.0326
K₂ = [tex](3.32 E-10 s^{-1})[/tex] /0.0326
K₂ = [tex]1.017 E-8 s^{-1}[/tex]
Thus, the rate constant at 48.5°C if the activation energy is 116 kJ/mol is 1[tex]0.017 E-8 s^{-1}[/tex]
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Describes the difference between the formulas for nitrogen monoxide and nitrogen dioxide?
Answer:
nitrogen monoxide: NO
nitrogen dioxide: NO₂
Explanation:
Nitrogen monoxide is composed by 1 atom of O (prefix "mono-") and 1 atom of N. Nitrogen dioxide is composed by 2 atoms of O (prefix "di-") and 1 atom of N. As the oxigen atom in oxides has the valency -2 (it shares 2 electrons), the nitrogen has valency +2 in NO and +4 in NO₂.
PLEASE HELP WILL GIVE BRAINLIEST
C8H18 + O2à CO2 + H2O is the unbalanced combustion reaction between octane and oxygen. What is the coefficient for carbon dioxide in the balanced equation?
A.2
B.8
C.16
D.18
Answer:
16. Option c.
Explanation:
The reaction is:
C₈H₁₈ + O₂ → CO₂ + H₂O
Let's count, we have 8 C, 18 H and 2 O in reactant side
We have 1 C, 2 H and 3 O in product side
In order to balance the C, we add 8 to CO₂ and to balance the H, we add a 9 to H₂O
Now we have 8 C on both sides and 18 H On both sides. Finally we have 25 O on product side.
C₈H₁₈ + O₂ → 8CO₂ + 9H₂O
To balance the O in product side we must add 25/2, but as it is a rational number, we must multiply x2 to get an integer number (x2 in all the stoichiometry). The balanced reaction is :
2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O
with 16 C, 36 H and 50 O, on both sides
Hydrogenotrophy is Choose one: A. an award given to fuel companies that have the most fuel-efficient vehicles. B. the oxidation of water during photosynthesis to liberate electrons, protons, and oxygen gas. C. the use of hydrogen gas as an electron donor. D. the generation of hydrogen gas by methanogens in syntrophy.
Answer:
C. the use of hydrogen gas as an electron donor.
Explanation:
Hydrogenotrophy is the convertion of hydrogen gas to other compounds as part of its metabolism.
What is true about the chemiosmotic synthesis of atpt?
Explanation:
The chemiosmotic theory
The theory suggests that the actual production of adenosine triphosphate ATP in cellular respiration of respiring cells takes place through the process of chemiosmosis that makes the electrochemical gradient across the inner and outer membranes of mitochondria.Breaking down of energy-rich molecules such as glucose the energy is released in the form of NADH and FADH2.It is found that for each glucose molecule 36 molecules of ATP can be produced during cellular respiration. These molecules are generated by the Krebs cycle, the electron transport system, and chemiosmosis.Two ATP molecules are produced through glycolysis, so the total number of molecules formed is 38 molecules of ATP.20 mL of an approximately 10% aqueous solution of ethylamine, CH3CH2NH2, is titrated with 0.3000 M aqueous HCl. Which indicator would be most suitable for this titration
20 mL of an approximately 10% aqueous solution of ethylamine, CH3CH2NH2, is titrated with 0.3000 M aqueous HCl. Which indicator would be most suitable for this titration? The pKa of CH3CH2NH3+ is 10.75.
Answer:
Bromocresol green, color change from pH = 4.0 to 5.6
Explanation:
The equation for the reaction is as follows:
[tex]C_2H_5NH_{2(aq)[/tex] + [tex]H^+_{(aq)[/tex] ⇄ [tex]C_2H_5NH_{3(aq)}^+[/tex]
Given that concentration of [tex]C_2H_5NH_{2(aq)[/tex] = 10%
i.e 10 g of [tex]C_2H_5NH_{2(aq)[/tex] in 100 ml solution
molar mass = 45.08 g/mol
number of moles = [tex]\frac{10}{45.08}[/tex]
= 0.222 mol
Molarity of [tex]C_2H_5NH_{2(aq)[/tex] = 0.222 × [tex]\frac{1000}{100}mL[/tex]
= 2.22 M
However, number of moles of [tex]C_2H_5NH_{2(aq)[/tex] in 20 mL can be determined as:
number of moles of [tex]C_2H_5NH_{2(aq)[/tex] = 20 mL × 2.22 M
= [tex]44*10^{-3} mole[/tex]
Concentration of [tex]C_2H_5NH_{2(aq)[/tex] = [tex]\frac{44*10^{-3}*1000}{20}[/tex]
= 2.22 M
Similarly, The pKa Value of [tex]C_2H_5NH_{2(aq)[/tex] is given as 10.75
pKb value will be: 14 - pKa
= 14 - 10.75
= 3.25
Finally, the pH value at equivalence point is:
pH= [tex]\frac{1}{2}pKa - \frac{1}{2}pKb-\frac{1}{2}log[C][/tex]
pH = [tex]\frac{14}{2}-\frac{3.25}{2}-\frac{1}{2}log [2.22][/tex]
pH = 5.21
∴
The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6.
For titration of the weak base ethylamine with a strong acid like HCl, the pH at the equivalence point will be less than 7, so an indicator that changes color in an acidic pH range such as methyl orange would be most suitable.
Explanation:The appropriate indicator in this case would depend on the equivalence point of the titration. Ethylamine, CH3CH2NH2, is a weak base, and it's being titrated with HCl, a strong acid. The reaction between them generates water, and ethylammonium chloride which functions as a weak acid. This setup indicates that the pH at the equivalence point will be less than 7.
Because the pH at the equivalence point will be in the acidic range (below 7), we should select an indicator which changes color around this pH. A good choice would be methyl orange, which changes color from red to yellow across a pH range of approximately 3.1 to 4.4.
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The limiting reactant can be described as: Entry field with incorrect answer the amount actually obtained from a reaction the substance that is depleted first and stops a reaction the maximum amount that can be produced in a reaction the substance left over at the end of the reaction the ratio of the amount produced to the maximum possible
Answer:
The limiting reactant can be described as the substance that is depleted first and stops
Explanation:
Imagine you have hydrogen and oxygen to produce water.
The reaction is: 2H₂(g) + O₂(g) → 2H₂O(g)
You have 1 mol of each reactant. As you see ratio is 2:1, so the limiting reactant is the hydrogen.
You know by stoichiometry, that 2 moles of H₂ need 1 mol of O₂ to react
If I have 1 mol of H₂, I will need the half of moles of O₂, so 0.5 moles. It is ok because I have 1 mole, as I need the half, then half a mole will remain unreacted. This is what is called excess reagent,
If I make to react 1 mol of oxgen I need 2 moles of H₂. As I have 1 mol, of course I will need 2 moles but the thing is I have 1 mol.
This is the limiting reactant. I do not have enough of reactant so the reaction will happen until I complete to use it, that's why we can say that is depleted first and stops.
In a chemical reaction, when you have data of both reactants you can determine the limiting. Otherwise the excersise must tell that one ractant is in excess, to work with the limiting. Limiting reactant is the first step to work with the reaction, all the operations must be done by it. You do not use the reagent in excess
Sodium chloride as a compound does not truly exist in the ocean. True or False
. The molar heat of vaporization of acetone, C3H6O, is 30.3 kJ/mol at its boiling point. How many kilojoules of heat would be liberated by the condensation of 5.00 g of acetone?
2.61 kilojoules of heat would be liberated by the condensation of 5.00 g of acetone
Explanation:
To convert grams into moles
[tex]moles = grams \times \frac{1 mole}{grams}[/tex]
We have 5.00 g acetone
[tex]moles = 5 \times \frac{1}{58.1}[/tex]
[tex]moles = 0.0861[/tex]
Heat liberated = moles [tex]\times[/tex] heat of vapourization
=0.0861 mol x 30.3 kJ/mol
= 2.61 kJ
Therefore, 2.61 kilojoules of heat would be liberated by the condensation of 5.00 g of acetone
if 12 molecules of methane reacted with plenty of oxygen, we should expect to produce____ molecules of car on dioxide and _____ molecules of water
Answer:
12 molecules of carbon dioxide
24 molecules of water
Explanation:
Given parameters:
Number of molecules of methane = 12 molecules
Unknown:
Number of molecules of carbon dioxide = ?
Number of molecules of water = ?
Solution:
The given amount of methane is the limiting reagent in this reaction. By this, we can ascertain the amount of products that will be formed and the extent of the reaction.
We first write the balanced chemical equation for this reaction;
CH₄ + 2O₂ → CO₂ + 2H₂O
From balanced equation;
1 mole of methane produced 1 mole of carbon dioxide
12 molecules of methane will produce 12 molecules of carbon dioxide
Also;
1 mole of methane produced 2 moles of water;
12 molecules of methane will produce (12 x 2)molecules = 24 molecules of water
When 12 molecules of methane react with excess oxygen, 12 molecules of carbon dioxide and 24 molecules of water are produced, following a 1:2:1:2 ratio of methane to oxygen to carbon dioxide to water.
Explanation:If 12 molecules of methane (CH4) react with plenty of oxygen (O2), the balanced chemical equation for the combustion of methane shows a 1:2:1:2 ratio. This means for every one methane molecule, two oxygen molecules are consumed, and one carbon dioxide (CO2) molecule and two water (H2O) molecules are produced. Therefore, 12 molecules of methane reacting with excess oxygen will produce 12 molecules of carbon dioxide and 24 molecules of water.
This equation states that for every 1 molecule of methane and 2 molecules of oxygen that react, we will produce 1 molecule of carbon dioxide and 2 molecules of water.
When a nonvolatile solute is added to a solvent, it tends to ___________________ the freezing point and ___________________ the boiling point, compared to the pure solvent.
Answer:
decreases and increases
Explanation:
the addition of solutes can alter the freezing point and the boiling point .the changes in solute content where there is an addition of solutes results in the dropping of freezing point and increase in boiling point.this is known as freezing point depression and boiling point depression.
Carbon, nitrogen, oxygen and hydrogen are considered _______ because they are required in relatively large quantities, whereas manganese, zinc, and copper are considered _____ because they are required in very small quantities.
Answer:
macro-nutrients , micro-nutrients
Explanation:
For plant growth soil needs some basic elements which is basic need for plant growth and metabolic synthesis.
plant need approx 17 essential nutrients for growth and for completing its life cycle among which some are found in water and air. Nitrogen, carbon, hydrogen and oxygen needed the most and in large quantity as nitrogen helps in rapid growth and green color and is mobile and its deficiency lead to reduced growth and yellow foliage.
Of the reactions below, only ________ is not spontaneous. Mg (s) 2HCl (aq) MgCl2 (aq) H2(g) 2Ag (s) 2HNO3 (aq) 2AgNO3 (aq) H2 (g) 2Ni (s) H2SO4 (aq) Ni2SO4 (aq) H2 (g) 2Al (s) 6HBr (aq) 2AlBr3 (aq) 3H2 (g) Zn (s) 2HI (aq) ZnI2(aq) H2 (g)
Answer:
2Ag (s) + 2HNO₃ (aq) 2AgNO₃ (aq) + H₂ (g)
Explanation:
2Ag (s) + 2HNO₃ (aq) 2AgNO₃ (aq) + H₂ (g)
Ag is below H₂ in reactivity series. Therefore Ag does not spontaneously replace H₂ from any compound.
Answer:
The answer to the question is
2Ag (s) +2HNO₃ (aq) →2AgNO₃ (aq)+ H₂ (g)
Explanation:
The position of a cell on the reduction potential table determines if a reduction reaction will be spontaneous
The higher the positive reduction potential the higher the spontaneity of the reduction reaction. that is if the E⁰ cell is positive the cell is a spontaneous voltaic cell, however if the E⁰ cell is negative the cell is a electrolytic non-spontaneous cell
On the standard potential table silver has a low positive potential of
Ag⁺ + e⁻ → Ag = 0.799
It is an oxidizing agent tending to scarcely release electrons, therefore for the reaction to take place, there has to be some external supply of electromotive force.
A small toy car draws a 0.50-mA current from a 3.0-V NiCd (nickel-cadmium) battery. In 10 min of operation, (a) how much charge flows through the toy car, and (b) how much energy is lost by the battery? 4. (Resistance and Ohm’s law, Prob. 17.16, 1.0 point) How
Answer:
0.3 Coulomb charge flows through the toy car.
0.9 Joules of energy is lost by the battery
Explanation:
[tex]Current(I)=\frac{charge(Q)}{Time (T)}[/tex]
a)
Current drawn from the battery = 0.50 mA = (0.50 × 0.001 A)
milli Ampere = 0.001 Ampere
Duration of time current drawn = T = 10 min = 10 × 60 s = 600 s
1 min = 60 seconds
Charge flows through the toy car be Q
[tex]0.50\times 0.001 A=\frac{Q}{600 s}[/tex]
[tex]Q=0.50 \times 0.001 A\times 600 s=0.3 C[/tex]
0.3 Coulomb charge flows through the toy car.
b)
[tex]Heat(H)=Voltage(V)\times Current(I)\times Time(T)[/tex]
Voltage of the battery = V = 3.0 V
Current drawn from the battery = 0.50 mA = (0.50 × 0.001 A)
Duration of time current drawn = T = 10 min = 10 × 60 s = 600 s
[tex]H=V\times I\time T=3.0 V\times 0.50 \times 0.001 A\times 600 s[/tex]
H = 0.9 Joules
0.9 Joules of energy is lost by the battery
When solid (NH4)(NH2CO2) is introduced into an evacuated flask at 25 ∘C, the total pressure of gas at equilibrium is 0.116 atm. What is the value of Kp at 25 ∘C? Kp =
Answer : The value of [tex]K_p[/tex] is, [tex]2.32\times 10^{-4}[/tex]
Explanation :
For the given chemical reaction:
[tex](NH_4)(NH_2CO_2)(s)\rightleftharpoons 2NH_3(g)+CO_2(g)[/tex]
The expression of [tex]K_p[/tex] for above reaction follows:
[tex]K_p=(P_{NH_3})^2\times P_{CO_2}[/tex] ........(1)
[tex](NH_4)(NH_2CO_2)(s)\rightleftharpoons 2NH_3(g)+CO_2(g)[/tex]
Initial: 0 0
At eqm: 2x x
As we are given that:
Total pressure of gas at equilibrium = 0.116 atm
2x + x = 0.116 atm
3x = 0.116 atm
x = 0.0387 atm
Putting values in expression 1, we get:
[tex]K_p=(2x)^2\times (x)[/tex]
[tex]K_p=(2\times 0.0387)^2\times (0.0387)[/tex]
[tex]K_p=2.32\times 10^{-4}[/tex]
Thus, the value of [tex]K_p[/tex] is, [tex]2.32\times 10^{-4}[/tex]
What mass of silver chloride can be produced from 1.30 L of a 0.245 M solution of silver nitrate? The reaction described in Part A required 3.36 L of calcium chloride. What is the concentration of this calcium chloride solution?
Please explain, I don't know where to begin.
Answer:
Explanation:
Given parameters:
Volume of AgNO₃ = 1.3L
Molarity or concentration AgNO₃ = 0.245M
Unknown:
Mass of AgCl produced = ?
Solution:
To solve this problem, we have to work from the known compound to the unknown one using the mole concept.
The known compound is the one that we can obtain the value of the number of moles from. Here it is the given AgNO₃ that can furnish us with this piece of information;
Now let us establish the balanced reaction equation;
2AgNO₃ + CaCl₂ → 2AgCl + Ca(NO₃)₂
Now;
Find the number of moles of AgNO₃;
Number of moles of AgNO₃ = concentration x volume
= 1.3 x 0.245
= 0.32moles
From the balanced reaction equation;
2 mole of AgNO₃ produced 2 moles of AgCl
0.32 moles of AgNO₃ will produce 0.32moles of AgCl
Now that we know the number of moles of the AgCl, we can find the mass;
Mass of AgCl = number of moles x molar mass
Molar mass of AgCl = 107.9 + 35.5 = 143.4g/mol
Mass of AgCl = 0.32 x 143.4 = 45.89g
Cytotoxic t cells can attack target cells with which chemical weapons?
Answer:
secrete cytotoxic substance which triggers apoptosis of target cell.
Explanation:
Cytotoxic T cells have cell surface receptor which recognizes the antigen present on the receptor of target cell. This interaction initiates the process of killing of target cell.
After interaction cytotoxic t cell release cytotoxic substance called granzyme and perforin. Granzyme triggers apoptosis through the activation of caspases or by making the release of cytochrome c and activation of the apoptosome.
Perforin make pores in the cell and its action is similar to complement membrane attack complex. Therefore cytotoxic substances are released by Tc cells which trigger apoptosis of target cell.
Phosphine, PH₃, is a colorless, toxic gas that is used in the production of semiconductors as well as in the farming industry. When heated, phosphine decomposes into phosphorus and hydrogen gases.
4 PH₃(g) ⟶ P₄(g) + 6 H₂(g)
This decomposition is first order with respect to phosphine, and has a half‑life of 35.0 s at 953 K. Calculate the partial pressure of hydrogen gas that is present after 70.5 s if a 2.20 L vessel containing 2.29 atm of phosphine gas is heated to 953 K.
1.72 atm of hydrogen gas is created when 2.29 atm of phosphine breaks down at 953 K for 70.5 seconds.
Explanation:
The first step in this calculation is to use the first-order decay equation to find the remaining moles of phosphine. For a first-order reaction with a half-life of 35.0 s, after 70.5 s (or two half-lives), 25% of the phosphine would remain. To find the remaining moles of phosphine, we multiply 2.29 atm by 25%, yielding 0.57 atm.
From the balanced equation for the decomposition of phosphine (4PH3 -> P4 + 6H2), we see that 1 mole of phosphine yields 3/2 moles of H2. Thus, the number of moles of H2 produced is 75% of the original moles of PH3 (1.72 atm).
The sum of the partial pressures of phosphine and hydrogen is 0.57 + 1.72 = 2.29 atm, which is the final pressure in the 2.20 L vessel after 70.5 s.
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Polar molecules have a partial positive charge on one side and a partial negative charge on the other side, a separation of charge called a dipole. Which of the following molecules has this kind of a dipole?a. IonicB. Covalent
Answer:Covalent
Explanation:
Covalent bonds are formed by sharing of electrons between two atoms. The shared electrons are normally situated between the nuclei of the both atoms. However, atoms of some elements posses an usual ability to attract the shared electrons of a bond towards itself. We say that such atoms have a high electronegativity. High electronegativity leads to the existence of polar covalent bonds since the shared electron pair is now closer to one of the bonding atoms than the other. The atom to which the electron pair is closer becomes partially negative while the other becomes partially positive. This is only possible in a covalent bond where electrons are shared between bonding atoms. The charge separation is known as a dipole.
Note that ionic bonds involve a complete transfer of electrons leading to the formation of ions and is not applicable here.