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elena-s [515]
3 years ago
10

Determine if the following compounds will be soluble or insoluble in water?

Chemistry
1 answer:
Molodets [167]3 years ago
7 0

Answer:

a) Insoluble

b) Soluble

c) Insoluble

d) Insoluble

e) Soluble

f) Soluble

Explanation:

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What has more particles: a mole of hydrogen or a mole of uranium?
algol [13]

Answer

hydrogen

Explanation:

7 0
2 years ago
Which of the following is an example of a covalent bond?
loris [4]

Answer:

C. H2O

yep this one is right

4 0
3 years ago
What can be said about a reaction with H = 62.4 kJ/mol and S = 0.145 kJ/(mol·K)?
Zigmanuir [339]
Answer:

At 430.34 K the reaction will be at equilibrium, at  T > 430.34 the reaction will be spontaneous, and at T < 430.4K the reaction will not occur spontaneously.

Explanation:

1) Variables:

G = Gibbs energy
H = enthalpy
S = entropy

2) Formula (definition)

G = H + TS

=> ΔG = ΔH - TΔS

3) conditions

ΔG < 0 => spontaneous reaction
ΔG = 0 => equilibrium
ΔG > 0 non espontaneous reaction

4) Assuming the data given correspond to ΔH and ΔS

ΔG = ΔH - T ΔS = 62.4 kJ/mol + T 0.145 kJ / mol * K

=>  T = [ΔH - ΔG] / ΔS

ΔG = 0 =>  T = [ 62.4 kJ/mol - 0 ] / 0.145 kJ/mol*K = 430.34K

This is, at 430.34 K the reaction will be at equilibrium, at  T > 430.34 the reaction will be spontaneous, and at T < 430.4K the reaction will not occur spontaneously.

3 0
3 years ago
Read 2 more answers
The conversation of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7x10-⁴s-¹. a) i
Rus_ich [418]

Answer: a)  The concentration after 8.8min is 0.17 M

b) Time taken for the concentration of cyclopropane to decrease from 0.25M to 0.15M is 687 seconds.

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) concentration after 8.8 min:

8.8\times 60s=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{a-x}

\log\frac{0.25}{a-x}=0.15

\frac{0.25}{a-x}=1.41

(a-x)=0.17M

b) for concentration to decrease from 0.25M to 0.15M

t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{0.15}\\\\t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\times 0.20

t=687s

7 0
3 years ago
Match these items.
mars1129 [50]
Here are the possible answers for the following questions above:
1. H-CC-H (name) - C<span>. ethyne
</span>2. cyclic compound with both saturated and unsaturated characteristics - G<span>. benzene
</span>3. CnH2n - E<span>. general formula for alkenes
</span>4. reaction typical of unsaturated hydrocarbons - A<span>. addition
</span>5. CnH2n-2 - F<span>. general formula for alkynes
</span>6. series name of hydrocarbons with triple bond - D<span>. alkyne
</span>7. CnH2n+2 - B<span>. general formula of alkanes</span>
6 0
3 years ago
Read 2 more answers
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