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3 years ago

Calculate the molarity of a solution that contains 3.00 grams Na2SO4 in 25 mL of solution.

Chemistry

1 answer:

Naddik [55]3 years ago

3 0

Answer:

M Na2SO4 sln = 0.8448 M

Explanation:

molarity (M) [=] mol/L

∴ mass Na2SO4 = 3.00 g

∴ volume soln = 25 mL = 0.025 L

∴ molar mass Na2SO4 = 142.04 g/mol

⇒ mol Na2SO4 = (3.00 g)*(mol/142.04 g) = 0.02112 mol

⇒ M Na2SO4 sln = (0.02112 mol/0.025 L ) = 0.8448 M

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Under standard-state conditions, which of the following half-reactions occurs at the cathode during the electrolysis of aqueous

Slav-nsk [51]

Answer:

The following reaction will occur at cathode:

$Ni^{+2}(aq)+2e--->Ni(s)$

Explanation:

The two half reaction during electrolysis of aqueous nickel sulfate will be

a) anode reaction :

Water will undergo oxidation and will evolve oxygen gas at anode as shown in the given reaction:

$2H_{2}O(l) ----> O_{2}(g) + 4H^{+}(aq) + 4e$

b) Cathode reaction: The reduction of Nickel ion will occur by gain of two electrons as shown in the given equation:

$Ni^{+2}(aq)+2e--->Ni(s)$

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2 years ago

What type of reaction is shown below? <br> Ca + 2H2O - Ca(OH)2 + H2

krek1111 [17]

A + BC ---> AB + C
So here one reactant (A) is accepting a group which is being given by another compound (BC) however the A is not giving any group / element or ion
So this single displacement
Similarly in the given reaction
the anion OH- is only being replaced
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So the correct answer is
Single replacement also known as single displacement.

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2 years ago

Consider the reaction 2 NO + O2 → 2 NO2

Sonbull [250]

Answer:

(a) Rate at which $NO_2$ is formed is 0.050 M/s

(b) Rate at which $O_2$ is consumed is 0.0250 M/s.

Explanation:

The given reaction is:-

$2NO+O_2\rightarrow 2NO_2$

The expression for rate can be written as:-

$-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}$

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$-\frac{1}{2}\frac{d[NO]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}$

$-\frac{d[NO]}{dt}=\frac{d[NO_2]}{dt}$

$-(-0.050\ M/s)=\frac{d[NO_2]}{dt}$

$\frac{d[NO_2]}{dt}=0.050\ M/s$

(a) Rate at which $NO_2$ is formed is 0.050 M/s

$-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}$

$-\frac{1}{2}\times -0.050\ M/s=-\frac{d[O_2]}{dt}$

$\frac{d[O_2]}{dt}=0.0250\ M/s$

(b) Rate at which $O_2$ is consumed is 0.0250 M/s.

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3 years ago

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