How to write the formulas

Density of Silver = 10.5 g / cm^3

Density of Lead = 11.34 g / cm^3

So the masses of each given a volume of 1 cm^3 is:

mass of Silver = 10.5 grams

mass of Lead = 11.34 grams

So we see that Lead has greater mass for similar volume.

The difference is:

11.34 – 10.5 = 0.84 grams

To answer this question, we must first look at the reactivity series table. From the table, we can see that those which are located above are the most reactive while those below are less reactive. Which means that elements cannot replace those elements which are above them.

From the series, platinum is below lead, hence there will be no reaction.

Answer:

No reaction

Final answer:

Platinum (Pt) does not react with a lead(II) nitrate solution because of its low reactivity, indicated by its position near the bottom of the activity series, which means it will not displace lead to form a new compound.

Explanation:

Platinum metal (Pt) is a highly inert metal and does not react easily with other substances. In the activity series, which is a list of metals arranged in order of their reactivity with acids and water, platinum is near the bottom, indicating very low reactivity. Considering the context provided about reactions of lead compounds, such as the reaction of lead(II) nitrate with potassium iodide to form a precipitate of lead(II) iodide, we can infer a similar lack of reactivity would apply to platinum in this case.

While lead(II) nitrate reacts with potassium iodide to form a precipitate of lead(II) iodide, such a double-replacement reaction is not likely to occur between platinum and lead(II) nitrate. The reason is that platinum will not displace lead from its salt because lead is above platinum in the activity series. Therefore, no reaction between Pt metal and lead(II) nitrate solution would be expected. Since the question appears to be about the reaction profiles of different metals, we have determined that there is no reactive interaction between these specific substances.

Hey Brainly Student! Your answer us true, SUPRISE FACT ABOUT ATOMS!!: When you think of an atom what do you think of? Problily a small circle with afew rings around it right? Well, actually an atom doesnt look like that at all!! An atom looks sort of like the Solar System, and the little rights are like the rotation the planets spin in, also the little star like balls/dots dont actually go around the atom's center, but go in many diffrent places it can be far from the atom, or can be very close! An atom is unpredictable, so there is no correct answer to where the atom's little balls/dots will end up next. I hope this helped, and also hope you liked/loved the Fun-Fact!! Your fellow Brainly user, GalaxyGamingKitty

Final answer:

In biology, high school students learn about the conversion process of pyruvate into acetyl CoA for the Krebs cycle. The components involved include Pyruvate Dehydrogenase Enzyme, Acetyl CoA, and Oxaloacetate.

Explanation:

In the process of converting three-carbon pyruvate molecules into two-carbon acetyl groups for entry into the Krebs cycle, several components are involved:

Pyruvate Dehydrogenase Enzyme: This enzyme catalyzes the conversion of pyruvate into acetyl Coenzyme A (acetyl CoA) in the mitochondrial matrix.

Acetyl CoA: The two-carbon molecule formed from the conversion of pyruvate, ready to enter the Krebs cycle.

Oxaloacetate: A four-carbon molecule in the Krebs cycle that combines with acetyl CoA to initiate the cycle by forming citrate.

Answer:

415.825 kJ/mol.

Explanation:

Average enthalpy = [tex]\Delta H_f(CH_4)-4\Delta H_f(H)-\Delta H_f(C)[/tex]

Now, [tex]\Delta H_f(CH_4)=-74.6 kJ/mol[/tex]

         [tex]\Delta H_f(H)=218 kJ/mol[/tex]

         [tex]\Delta H_f(C)=716.7 kJ/mol[/tex]

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Putting all these values above.

We get,

Average enthalpy=-1663.3 kJ.

Therefore, averagemolar bond enthalpy= [tex]\dfrac{1663.3 }{4}=415.825\ kJ/mol.[/tex]

Hence, this is the required solution.

When patterning a shotgun, a sufficient percentage of pellets within a 30-inch circle depends on the specific shooting requirements and preferences. However, a commonly recommended standard is to have at least 55% of the load with an even distribution within the 30-inch circle.

Having  55% of the load with an even distribution within the 30-inch circle ensures a dense and consistent pattern, increasing the chances of hitting the target effectively. It is important to note that individual shooting styles and the intended use of the shotgun may vary, so it is advisable to pattern the shotgun and adjust the load distribution accordingly for optimal performance.

Hence, at least 55% of the load, with even distribution, is prescribed.

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Isotopes are the forms of same element with same atomic number but different mass number due to different number of neutrons. The name of isotope of oxygen is Oxygen-17.

Element generally consist of atoms or we can atoms combine to form element. Atoms of an element is always same, means all the properties of all atoms of one type of element is same. Different element have different types of atoms.

Oxygen-16 has atomic number 8, which means number of electron is 8 , number of proton is 8 and number of neutron is also 8. If we increase the number of neutron by 1 that is 9 then the mass number of oxygen will be increased by 1 that is 17. Now the new element is oxygen-17 which is an isotope.

Therefore, the name of isotope of oxygen is Oxygen-17.

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The possible values of all quantum numbers for the electron gained by Br atom to form [tex]{\text{B}}{{\text{r}}^-}[/tex] is [tex]\boxed{n=4,\;l=1,\;{m_l}=1,\;{m_s}=-\frac{1}{2}}[/tex]

Further explanation:

Quantum numbers:

Quantum numbers govern the size, energy, shape, and orientation of an orbital. The four quantum numbers are as follows:

1. Principal Quantum Number (n): It denotes the principle electron shell. The values of n are positive integer (1, 2, 3,…).

2. Angular Momentum Quantum Number (l): It represents the shape of an orbital. The value of l is an integer from 0 to (n-1). (Refer to the table in the attached image)

3. Magnetic Quantum Number[tex]\left( {{m_l}} \right)[/tex]: This quantum number represents the orientation of the orbital in space. The value of [tex]{m_l}[/tex]lies between –l to +l. The formula to calculate the value of [tex]{m_l}[/tex] is as follows:

[tex]{m_l}=-l,(-l+1),.....,0,1,2,.....,(l-1),l[/tex]

Therefore, the total number of [tex]{m_l}[/tex] values for a given value of l is 2l+1.

4. Electron Spin Quantum Number[tex]({m_s})[/tex]: It represents the direction of the electron spin. Its value can be [tex]+ \frac{1}{2}[/tex]or[tex]- \frac{1}{2}[/tex].

The atomic number of bromine (Br) is 35 and its electronic configuration is [tex]\left[ {{\text{Ar}}} \right]\;3{d^{10}}4{s^2}4{p^5}[/tex]. When it gains an electron, it forms [tex]{\text{B}}{{\text{r}}^ - }[/tex] whose electronic configuration becomes [tex]\left[ {{\text{Ar}}} \right]\;3{d^{10}}4{s^2}4{p^6}[/tex].

The principal electron shell in [tex]{\text{B}}{{\text{r}}^ - }[/tex] is 4p. So its principal quantum number for the added electron in [tex]{\text{B}}{{\text{r}}^ - }[/tex] is 4.

The angular momentum quantum number for p orbital is 1. So the value of l for the added electron in [tex]{\text{B}}{{\text{r}}^ - }[/tex] is 1.

The value of magnetic quantum number for the electron in [tex]{\text{B}}{{\text{r}}^ - }[/tex] ranges from -1 to 1, including 0. But by convention this electron is added to [tex]{p_z}[/tex] orbital so the value of [tex]{m_l}[/tex] for the added electron in [tex]{\text{B}}{{\text{r}}^ - }[/tex] is 1.

The spin quantum number has two values, either [tex]+ \frac{1}{2}[/tex] or [tex]- \frac{1}{2}[/tex]. But the electron added in Br is the second electron being added to [tex]{p_z}[/tex] orbital so by convention, the value of [tex]{m_s}[/tex] for the added electron in [tex]{\text{B}}{{\text{r}}^-}[/tex] is [tex]{\mathbf{ - }}\frac{{\mathbf{1}}}{{\mathbf{2}}}[/tex].

The possible set of four quantum numbers for the electron gained by Br atom to form [tex]{\text{B}}{{\text{r}}^ - }[/tex] is n = 4, l = 1, [tex]{m_l} = 1[/tex] and [tex]{m_s} =  - \frac{1}{2}[/tex].

Learn more:

1. Allowed values of [tex]{m_l}[/tex]: https://brainly.com/question/2920448

2. Calculation of volume of gas: https://brainly.com/question/3636135

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Structure of the atom

Keywords: quantum numbers, Br, Br-, electron, 35, electronic configuration, n, l, ml, ms, principal quantum number, angular momentum quantum number, electron spin quantum number, magnetic quantum number, n = 4, ml=1, ms=-1/2, l=1.

The voltage-gated ion channel opens in response to a change in membrane potential and is involved in the generation and spreading of action potentials. As the membrane depolarizes, voltage-gated sodium channels open, allowing sodium ions to enter the muscle fiber, leading to the spread of action potential.

The ion channel that opens in response to a change in membrane potential and participates in the generation and conduction of action potentials is known as a voltage-gated ion channel. As the membrane depolarizes, voltage-gated sodium channels are triggered to open. Sodium ions then enter the muscle fiber initiating a process known as excitation-contraction coupling which results in an action potential, a transmission of an electrical signal, swiftly spreading along the entire membrane. This mechanism is critical to the function of neurons and muscle fibers.

Typically, the voltage of the inner portion of the membrane is negative. When this voltage becomes less negative, the voltage-gated channel allows ions to cross the membrane. It is also worth noting, that voltage-gated channels can become inactivated for a brief period after activation, during which they will not open in response to a signal.

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Final answer:

Overheating in a hydration experiment of copper sulfate can cause decomposition of the substance, leading to an elevated calculation of water content due to additional weight loss incorrectly assumed to be water.

Explanation:

In the experiment involving hydrated copper sulfate, overheating can lead to a high calculated percent value for water because it may cause the copper sulfate to decompose further into copper oxide and sulfur dioxide, or lose additional bound water that is not part of the water of crystallisation. These additional losses are assumed to be water, resulting in an incorrect, elevated calculation of the water content. The precise weighing of starting and ending products is crucial to determine the amount of water in copper sulfate accurately.

The total number of iron (Fe) atoms in 6 L of blood in an adult human can be found by converting the volume of blood to the mass of hemoglobin, converting that to the moles of hemoglobin, and then considering how many Fe atoms that would be. It turns out to be approximately 3.44 x 10^23 Fe atoms.

To approximate the number of iron (Fe) atoms in the 6 L of blood in an adult human, we must first convert the volume of blood to the mass of hemoglobin it contains. From the information provided, we know that blood contains about 15.4 g of hemoglobin per 100 ml. Therefore, in 6 L of blood, we have approximately 9240 g of hemoglobin, obtained by a simple conversion from liters to milliliters, then multiplying by the concentration of hemoglobin (15.4 g/100ml).

Next, we need to understand how many moles of hemoglobin that would be. The molar mass of hemoglobin is around 64500 g/mol. So, 9240 g of hemoglobin is about 0.143 mol.

Because there are four Fe atoms in each hemoglobin molecule, this is equivalent to 0.572 mol of Fe atoms. To calculate the actual number of Fe atoms, we multiply this by Avogadro's number (6.022 x 10^23 mol^-1), resulting in about 3.44 x 10^23 Fe atoms.

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Answer:

Amount of water that can be vaporized = 369 g

Explanation:

Heat content of the cookie (q) = 835 kJ

Latent heat of vaporization of water (ΔHvap) = 40.7 kJ/mol

Density of water = 1 g/ml

Calculation:

Heat content of cookie = Heat absorbed by water

q = n(ΔHvap)

n = moles of water

n = q/ΔH = 835 kJ/ 40.7 kJ/mol = 20.516 moles

Molar mass of water = 18 g/mol

Mass of water = moles * molar mass = 20.516 moles * 18 g/mol = 369.3 g

Data:

Hooke represented mathematically his theory with the equation:

F = K * x  

On what:

F (elastic force) = ?

K (elastic constant) = 3 N/cm

x (deformation or elongation of the elastic medium) = 2cm

Solving:

[tex]F = K * x[/tex]

[tex]F = 3\:\dfrac{N}{\diagup\!\!\!\!\!\!cm}*2\:\diagup\!\!\!\!\!\!cm[/tex]

[tex]\boxed{\boxed{F = 6N}}\end{array}}\qquad\checkmark[/tex]

Answer:

C. 6 N

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The force needed to stretch the spring 2 cm is 6 N.

Hooke's Law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position, and this force is expressed as [tex]\( F = k \Delta x \),[/tex]  where [tex]\( F \)[/tex] is the force exerted by the spring, [tex]\( k \)[/tex]  is the spring constant (elastic constant), and [tex]\( \Delta x \)[/tex]  is the displacement from the spring's equilibrium position.

Given:

- The displacement [tex]\( \Delta x \)[/tex] is 2 cm.

- The spring constant [tex]\( k \)[/tex] is 3 N/cm.

Using Hooke's Law, we can calculate the force [tex]\( F \)[/tex] as follows:

[tex]\[ F = k \Delta x \] \[ F = (3 \, \text{N/cm}) \times (2 \, \text{cm}) \] \[ F = 6 \, \text{N} \][/tex]

Therefore, the force needed to stretch the spring 2 cm is 6 N.

I do believe is Mechanical Energy

Answer:

The answer is D) on USA Test prep

Explanation:

Answer:That would be 16.1 degrees Celsius 

Actually each molecule of Hb can bind four oxygen molecules. So given 1 x 10^6 Hb, so there are 4 x 10^6 binding sites available.

If only 50% of the Hb are bound, so this means that 2 x 10^6 binding sites are occupied by O2. However we must remember that hemoglobin binding is an all-or-nothing phenomenon. Meaning that each molecule of Hb binds four molecules of oxygen or zero at all.

If 23% is saturated, so the number of binding sites occupied is 0.92 x 10^6 binding sites or 9.2 x 10^5 binding sites.