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Fantom [35]
3 years ago
14

How many liters of 4.0 M NaOH solution will react with 1.2 mol H2SO4? (Remember to balance the equation.)

Chemistry
2 answers:
valina [46]3 years ago
8 0
The answer to this is D
nydimaria [60]3 years ago
6 0

Answer:

The answer is D. 0.60 L

Explanation:

The balanced reaction equation including states of matter is;

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

More simple:

H2SO4 + 2NaOH → Na2SO4 + 2H2O

Now, we can see from this reaction equation that the mole ratio of NaOH to H2SO4 is 2:1

Number of moles of H2SO4 reacted = 1.2 moles

Hence;

2 moles of NaOH reacts with 1 mole of H2SO4

x moles of NaOH reacts with 1.2 moles of H2SO4

x = 2 * 1.2/1 = 2.4 moles of NaOH

Recall that;

Number of moles = Concentration * Volume

Volume = number of moles/concentration

Volume of NaOH  is obtained from;

Volume = 2.4 moles/ 4.0 M

Volume =  0.60 L

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3 years ago
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Calcium has an atomic number of 20. How many electrons of a calcium atom are involved in chemical bonding?
NARA [144]

The answer is 2 electrons.


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Calcium is a metal. When metals react with non-metals, electrons are transferred from the metal atoms to the non-metal atoms forming ions. The resulting compound is known  as an ionic compound.

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3 0
3 years ago
The reaction C 4 H 8 ( g ) ⟶ 2 C 2 H 4 ( g ) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ / mol. 262 kJ/mol. At 600.0 K,
ludmilkaskok [199]

Answer: 4.3\times 10^{-13}s^{-1}

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 600.0K = 6.1\times 10^{-8}s^{-1}

K_2 = rate constant at 775.0 = ?

Ea = activation energy for the reaction = 262 kJ/mol = 262000J/mol

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 600.0K

T_2 = final temperature = 775.0K

Now put all the given values in this formula, we get

\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]

\log (\frac{6.1\times 10^{-8}s^}{K_2})=5.150

(\frac{6.1\times 10^{-8}}{K_2})=141253.8

Therefore, the value of the rate constant at 775.0 K is 4.3\times 10^{-13}s^{-1}

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