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Jlenok [28]
4 years ago
11

5 Fe2+(aq) + MnO4−(aq) + 8 H+(aq) ⟶ 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l) determine if each of the following statements is True or Fa

lse. 1) MnO4−(aq) is the oxidizing agent. 2) Fe2+(aq) gains electrons. 3) Mn2+(aq) is the reducing agent. 4) H+(aq) undergoes oxidization.
Chemistry
1 answer:
aalyn [17]4 years ago
3 0

Explanation:

Fe2+ lose electrons to become Fe 3+ hence oxidized

maO4- is the oxidizing agent

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Why does the moon spin around the earth and the earth spins around the sun
liberstina [14]

This easy bro there is only one answer Gravity

but here my advance word

Sometimes the gravity of big objects would capture smaller ones in orbit. This could be one way the planet acquired the moon

7 0
3 years ago
Name this compound please ​
jasenka [17]

Answer:

Ethanol

Explanation:

2 C atoms and a single OH group.

Formula: C2H5OH

4 0
4 years ago
At a given temperature the vapor pressure of pure liquid benzene and toluene are 745 torr and 290 torr, respectively. A solution
IRISSAK [1]

Answer:

Vapour pressure of benzene over the solution is 253 torr

Explanation:

According to Raoult's law for a mixture of two liquid component A and B-

vapour pressure of a component (A) in solution = x_{A}\times P_{A}^{0}

vapour pressure of a component (B) in solution = x_{B}\times P_{B}^{0}

Where x_{A},x_{B} are mole fraction of component A and B in solution respectively

P_{A}^{0},P_{B}^{0} are vapour pressure of pure A and pure B respectively

Here mole fraction of benzene in solution is 0.340 and vapour pressure of pure benzene is 745 torr

So, vapour pressure of benzene in solution = 0.340\times 745 torr

                                                                         = 253 torr

7 0
3 years ago
An analytical chemist is titrating of a solution of nitrous acid with a solution of . The of nitrous acid is . Calculate the pH
Burka [1]

Answer:

pH = 2.69

Explanation:

The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>

<em />

The reaction of HNO₂ with KOH is:

HNO₂ + KOH → NO₂⁻ + H₂O + K⁺

Moles of HNO₂ and KOH that react are:

HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>

KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>

That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:

NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻

HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂

It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:

pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]

pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]

<h3>pH = 2.69</h3>
8 0
3 years ago
Exactly how much time must elapse before 16 grams of potassium-42decays, leaving 2 grams of the original isotope?(1) 8 × 12.4 ho
AleksandrR [38]
The answer is <span>(3) 3 × 12.4 hours
</span>
To calculate this, we will use two equations:
(1/2) ^{n} =x
t_{1/2} = \frac{t}{n}
where:
<span>n - number of half-lives
</span>x - remained amount of the sample, in decimals
<span>t_{1/2} - half-life length
</span>t - total time elapsed.

First, we have to calculate x and n. x is <span>remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams </span>× 100 percent ÷ 16 grams
x = 12.5% = 0.125

Thus:
<span>(1/2) ^{n} =x
</span>(0.5) ^{n} =0.125
n*log(0.5)=log(0.125)
n= \frac{log(0.5)}{log(0.125)}
n=3

It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
<span>t_{1/2} = 12.4
</span><span>t_{1/2} *n = t
</span>t= 12.4*3

Therefore, it must elapse 3 × 12.4 hours <span>before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope</span>
7 0
3 years ago
Read 2 more answers
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