All categories

3 years ago

#1: In an icy glass of lemonade, which of the following is the solvent?

1 answer:

4 0

In a solution, there are two components that is solvent and solute. Solute is one that is dissolved and solvent is one in which is used for dissolving solute or it can be said that it is the which is used for dissolving. So in the icy glass of lemonade, there is a substance that is used for dissolving solute and it will be water as it is used for dissolving solute.

You might be interested in

How many atoms are in a sample of 68.7 g copper (Cu)?

dlinn [17]

<span>Divide the number of grams present in the sample by copper's gram atomic weight to find the number of gram atomic weights present. Then multiply that result by Avogadro's Number: 6.022137 x 10^23 atoms/gram atomic weight.1,200 g/(63.54 g/gram atomic weight) ? 18.885741 gram-atomic weights. Hope this helps. </span>

8 0

3 years ago

Read 2 more answers

An interpenetrating primitive cubic structure like that of CsCl with anions in the corners has an edge length of 664 pm. If the

san4es73 [151]

Answer:

the ionic radius of the anion $r^- = 312.52 \ pm$

Explanation:

From the diagram shown below :

The anion $Cl^-$ is located at the corners

The cation $Cs^+$ is located at the body center

The Body diagonal length = $\sqrt{3 \ a }$

∴ $2 \ r^+ \ + 2r^- \ = \sqrt{3 \ a} \\ \\ r^+ +r^- = \frac{\sqrt{3}}{2} a$

Given that :

$\frac{r^+}{r^-} =0.84$ (i.e the ratio of the ionic radius of the cation to the ionic radius of

the anion )

$0.84r^- \ + r^- \ = \frac{\sqrt{3}}{2}a \\ \\ 1.84 r^- = \frac{3}{2}a \\ \\ r^- = \frac{\sqrt{3}}{2*1.84}a$

Also ; a = 664 pm

Then :

$r^- = \frac{\sqrt{3} }{2*1.84}*664 \ pm\\ \\ r^- = 312.52 \ pm$

Therefore, the ionic radius of the anion $r^- = 312.52 \ pm$

4 0

3 years ago

If a temperature increase from 10.0 ∘C to 22.0 ∘C doubles the rate constant for a reaction, what is the value of the activation

maria [59]

The activation energy barrier is 40.1 kJ·mol⁻¹

Use the Arrhenius equation

$\ln( \frac{k_2 }{k_1 }) = (\frac{E_{a} }{R })(\frac{ 1}{T_1} - \frac{1 }{T_2 })\\$

$\ln( \frac{2k }{k}) = (\frac{E_{a} }{\text{8.314 J} \cdot \text{K}^{-1} \text{mol}^{-1} })(\frac{ 1}{\text{283.15 K}} - \frac{1 }{\text{295.15 K }})\\$

$\ln2 = (\frac{ E_{a} }{\text{8.314 J} \cdot \text{K}^{-1} \text{mol}^{-1}}) \times 1.436 \times10^{-4}\\$

$\ln2 = E_{a} \times 1.727 \times 10^{-5} \text{ mol} \cdot \text{J}^{-1}$

$E_{a} = \frac{\ln2 }{ 1.727 \times10^{-5}\text{ mol} \cdot \text{J}^{-1}}\\$

$E_{a} = \text{40 100 J}\cdot\text{mol}^{-1} = \textbf{40.1 kJ}\cdot \textbf{mol}^{-1}$

3 0

3 years ago

14. As the moles of salt added are increased, what<br> happens to the melting point of the water?

Vika [28.1K]

Salt lowers the freezing/melting point of water, so in both cases the idea is to take advantage of the lower melting point. Ice forms when the temperature of water reaches 32 degrees Fahrenheit (0 degrees Celsius).

3 0

2 years ago

Read 2 more answers

cylinder contains 250 g of Helium at 200 K. The external pressure is constant at 1 atm. The temperature of the surroundings is l

brilliants [131]

Answer:

The system gains 126100 J

Explanation:

The heat can be calculated by the equation:

Q = nxCxΔT, where Q is the heat, C is the heat capacity,n is the number of moles and ΔT is the variation of temperature (final - initial). The number of moles is the mass divided by the molar mass, so:

n = 250/4 = 62.5 mol.

The system must be in thermal equilibrium with the surroundings, so if the temperature of the surroundings decreased 97 K, the temperature of the system increased by 97 K, so ΔT = 97 K

Q = 62.5x20.8x97

Q = 126100 J

6 0

3 years ago