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Rainbow [258]
3 years ago
11

20 points! So I got this lab report to do and I need some more info it's about lab safety got any info I can use! :) If so, Than

k you!
Chemistry
1 answer:
FromTheMoon [43]3 years ago
8 0
Don’t eat or drink in labs

Dress for the lab; don’t wear open toed shoes

Dispose of lab waste properly

No horse play

Don’t taste or sniff things in the lab

Tire your hair back
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Predict how many grams of KCI is produced from 40 grams of K?
Nutka1998 [239]

Explanation:

firstly find for the molar mass of kcl and molar mass of k

and then

molar mass of k = x

molar mass of kcl= 40

cross mutiply and then simplify you will get your answer

5 0
4 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
Determine the formula of chromic acid.
Katyanochek1 [597]

<u>Answer:</u> The formula of chromic acid is H_2CrO_4

<u>Explanation:</u>

We are given an ionic compound named as chromic acid. This acid is formed by the combination of hydrogen ion and chromate ion.

An acid is defined as the substance which releases hydrogen ion when dissolved in water.

Hydrogen is the 1st element of periodic table having electronic configuration of 1s^1.

This element will loose 1 electron to form H^{+} ion.

Chromate ion is a polyatomic ion having chemical formula of CrO_4^{2-}

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for the given compound is H_2CrO_4

Thus, the formula of chromic acid is H_2CrO_4

8 0
3 years ago
Nitrogen gas is being withdrawn at the rate of 4.5 g/s from a 0.15-m3 cylinder, initially containing the gas at a pressure of 10
faust18 [17]

Answer:

Final temperature = 152.57K,

Pressure = 0.6907 bar.

dT/dt = - 1,151 K/s.

Explanation:

The first thing to do here is to write out the equation for mass balance as given below:

dN/dt = N -------------------------------------------------------------------------------------------(1).

N = P/T, then, substitute the values given in the question into:

d[p/T]/ dt = [- 4.5/28 × 8.314]/0.15 = - 8.9 × 10⁻⁵ bar/K.s.

Thus, there is the need to integrate, Integrate [p/T]f = 10/320 - 8.9 × 10⁻⁵ bar/K.s. ------------------------------------(2).

NB; fT = final temperature, fP = final pressure and iT = initial temperature.

Also, [ fT]³⁰/₈.₃₁₄/ [fP] = [iT]³⁰/₈.₃₁₄/ Pi] = [ 320]³⁰/₈.₃₁₄/ 10.

Therefore, [fT]³⁰/₈.₃₁₄ = 109.52 × 10⁶.

Final temperature=  [fP]³⁰/₈.₃₁₄ × 169.05.

Note that fP/ [fP]³⁰/₈.₃₁₄ × 169.05 = 10/320 - 8.9 × 10⁻⁵.

Therefore, [fP]¹ ⁻ ³⁰/₈.₃₁₄ = 0.7651.

Hence, Final temperature = 152.57K,

Pressure = 0.6907 bar

dT/ dt = N[RT]² / Cv . PV.

R = 30 - 8.314 = 21.86 J/mol K.

Then, the rate of change of the gas temperature at this time = dT/dt = - 1,151 K/s.

4 0
3 years ago
I need help with this worksheet please
IrinaVladis [17]

starting with q.1 its the 2nd and 7th day

q.2 the 12th

q.3 is 8 days

q.4 its the second 6 days

q.5 9 days

q.6 4 days

q.7 5 of the days

4 0
3 years ago
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