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2 years ago

Explain why a potassium atom would not bond with a calcium atom?

Chemistry

1 answer:

docker41 [41]2 years ago

6 0

Answer:

Calcium atoms have smaller radii than potassium atoms since calcium atoms have a greater nuclear charge. Each calcium atom will be closer to the delocalized electrons. Thus, the bonds in calcium will be stronger than that in potassium.

Explanation:

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What volume of 0.194 MNa3PO4 solution is necessary to completely react with 85.5 mL of 0.109 MCuCl2 ?

ira [324]

Answer:

35.9 ml

Explanation:

Start with the balanced equation:

3CuCl2(aq)+2Na3PO4(aq)→Cu3(PO4)2(s)+6NaCl(aq)

This tells us that 3 moles of CuCI2 react with 2 moles Na3PO4-

∴ 1 mole CuCl2 will react with 2/3 moles Na3PO4

We know that concentration = moles/volume i.e:

c= n/v

∴n=c×v

∴nCuCl2=0.107×91.01000=9.737×10−3

I divided by 1000 to convert ml to L

∴nNa3PO4=9.737×10−3×23=6.491×10−3

v=nc=6.491×10−30.181=35.86×10−3L

∴v=35.86ml

4 0

2 years ago

A substance has a boiling point of 78 °C. Which of the following is true about the substance? (5 points) a It will also change f

vovangra [49]

Answer: If a substance has a boiling point of $78^{o}C$ then it is true that it will also change from a gas to a liquid at 78 °C while the gas loses energy.

Explanation:

The temperature at which vapor pressure of a liquid substance becomes equal to the atmospheric pressure is called boiling point of substance.

At the boiling point, liquid phase and vapor phase remains in equilibrium.

This means that as liquid phase changes into vapor phase and also vapor phase changes into liquid phase at the boiling point.

Thus, we can conclude that if a substance has a boiling point of $78^{o}C$ then it is true that it will also change from a gas to a liquid at 78 °C while the gas loses energy.

6 0

3 years ago

Please work out all the problems on this page, no links, no i don't knows please do not take advantage of me or my points please

sp2606 [1]

The answer is an atom.

4 0

3 years ago

Read 2 more answers

For this question, the "entropy term" refers to "-TΔS". Addition reactions are generally favorable at low temperatures because _

nata0808 [166]

Answer:

Lowering the temperature typically reduces the significance of the decrease in entropy. That makes the Gibbs Free energy of the reaction more negative. As a result, the reaction becomes more favorable overall.

Explanation:

In an addition reaction there's a decrease in the number of particles. Consider the hydrogenation of ethene as an example.

$\rm H_2C\text{=}CH_2\; (g) + H_2\; (g) \stackrel{\text{Ni}^\ast}{\to} H_3C\text{-}CH_3\; (g)$ .

When $\rm H_2$ is added to $\rm H_2C\text{=}CH_2$ (ethene) under heat and with the presence of a catalyst, $\rm H_3C\text{-}CH3$ (ethane) would be produced.

Note that on the left-hand side of the equation, there are two gaseous molecules. However, on the right-hand side there's only one gaseous molecule. That's a significant decrease in entropy. In other words, $\Delta S < 0$ .

The equation for the change in Gibbs Free Energy for a particular reaction is:

$\Delta G = \Delta H + (\underbrace{- T \, \Delta S}_{\text{entropy}\atop \text{term}})$ .

For a particular reaction, the more negative $\Delta G$ is, the more spontaneous ("favorable") the reaction would be.

Since typically $\Delta S < 0$ for addition reactions, the "entropy term" of it would be positive. That's not very helpful if the reaction needs to be favorable.

$T$ (absolute temperature) is always nonnegative. However, lowering the temperature could help bring the value of

8 0

3 years ago

Fossil fuels are:

Scorpion4ik [409]

Answer:

Explanation:

7 0

3 years ago