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2 years ago

How should students prepare to use chemicals in the lab?.

Chemistry

1 answer:

Serggg [28]2 years ago

6 0

Answer:

Become familiar with the chemicals to be used, including exposure or spill hazards.

Locate the spill kits and understand how they are used.

Explanation:

Hope this helped! :)

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A gas has a pressure of 853.0 millibars at a temperature of 29.0 °C. If the volume is unchanged but the temperature is increased

Nitella [24]

The new pressure of the gas that initially have a pressure of 853.0 millibars at a temperature of 29.0 °C is 1011.17 millibars. Details about pressure can be found below.

<h3>How to calculate pressure?</h3>

The pressure of a given gas can be calculated using the following formula:

P1/T1 = P2/T2

Where;

P1 = initial pressure = 853.0 millibars
P2 = final pressure = ?
T1 = initial temperature = 29°C + 273 = 302K
T2 = final temperature = 85°C + 273 = 358K

853/302 = P2/358

358 × 853 = 302P2

305374 = 302P2

P2 = 305374 ÷ 302

P2 = 1011.17 millibars

Therefore, the new pressure of the gas that initially have a pressure of 853.0 millibars at a temperature of 29.0 °C is 1011.17 millibars.

Learn more about pressure at: brainly.com/question/15175692

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8 0

2 years ago

. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are

olga nikolaevna [1]

<u>Answer:</u> The standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

<u>Explanation:</u>

We are given:

$K_{sp}\text{ of }Ag_2S=8\times 10^{-51}$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

K = equilibrium constant or solubility product = $8\times 10^{-51}$

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ$

For the given chemical equation:

$Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)$

The equation used to calculate Gibbs free change is of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]$

We are given:

$\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ$

Putting values in above equation, we get:

$285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol$

Hence, the standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

8 0

3 years ago

Which scientific law states the relationship between an object's mass, acceleration, and amount of force acting on it? *

jenyasd209 [6]

Newton's second law of motion

5 0

2 years ago

You and your roommates have already decided to spend a week volunteering in the relief effort over the winter break, but in orde

Vlada [557]

Answer:

I dont know the answer for that question it's hard question isn't it

5 0

3 years ago

Read 2 more answers

When the pressure that a gas exerts on a sealed container changes from 1100 bar to 75.5 bar, the temperature changes from k to 2

Serjik [45]

Gay-Lussac's law gives the relationship between pressure and temperature of gas. For a fixed amount of gas, pressure is directly proportional to temperature at constant volume.
P/T = k
where P - pressure , T - temperature and k - constant
$\frac{P1}{T1} = \frac{P2}{T2}$
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
substituting the values in the equation
$\frac{1100 bar}{T} = \frac{75.5 bar}{298 K}$
T = 4342 K
initial temperature was 4342 K

7 0

3 years ago