All categories

2 years ago

Explain why, when performing stoichiometric calculations, it is

Chemistry

1 answer:

pshichka [43]2 years ago

3 0

Answer:

Stoichiometric Coefficients

The balanced equation makes it possible to convert information about one reactant or product to quantitative data about another element. Understanding this is essential to solving stoichiometric problems

Explanation:

You might be interested in

What is the molar mass of Ca(OH)2?

liubo4ka [24]

Answer: the molar mass is 74.092 g/mol

Explanation:

8 0

3 years ago

I don't know how to solve this

Misha Larkins [42]

For the first part, use the question M=mol/vol (liters)

To do this, you have the given 1.6 M solution

divide the 360g by the molar mass of ethanol (44.07) to get moles

360/44.07=8.16 mol

1.6M = 8.16 mol/x vol

volume: 5.1 Liters

8 0

3 years ago

The boiling point of chloroform is 61.7 C. The enthalapy of vaporization is 31.4 kj/mol. Caculate the entropy of vaporization. D

dedylja [7]

Answer:

Δ S = 93.8 J/mol-K

Explanation:

Given,

Boiling point of chloroform = 61.7 °C

= 273 + 61.7 = 334.7 K.

Enthalapy of vapourization = 31.4 kJ/mol.

Using Gibbs free energy equation

Δ G = Δ H - T (ΔS)

at equilibrium (when the liquid is boiling), Δ G = 0

so, 0 = ΔH - T (Δ S)

T (Δ S) = Δ H

and ΔS = ΔH / T

Δ S = (31400 J/mol.) / 334.7 K

Δ S = 93.8 J/mol-K

6 0

3 years ago

What is the half-life of an isotope that decays to 6.25% of its original activity in 18.9 hours?

inessss [21]

Radioactive material obeys 1st order decay kinetics,
For 1st order reaction, we have
k = $\frac{2.303}{t}Xlog \frac{\text{initial conc.}}{\text{final conc.}}$
where, k = rate constant of reaction

Given: Initial conc. 100, Final conc. = 6.25, t = 18.9 hours

∴ k = $\frac{2.303}{18.9} X log \frac{100}{6.25}$ = 0.1467 hours^(-1)

Now, for 1st order reactions: half life = $\frac{0.693}{k} = \frac{0.693}{0.1467}$ = 4.723 hours.

8 0

3 years ago

A particular form of electromagnetic radiation has a frequency of 8.11 x 10^14 hz. What is the energy in joules of one quantum o

quester [9]

<span>E = hxf, où h = constante de Planck = 6,626 x 10 ^ -34Js

E = 6,626 x 10 ^ -34 x 8.11 x 10 ^ 14 = 5.373 10^ -19 J </span><span>

Hope this answers your question, Kimmyers14!</span>

5 0

3 years ago

Read 2 more answers