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mariarad [96]
3 years ago
11

2. What is the weight of the hydrochloric acid that fills a 144.5 mL container? The density of

Chemistry
1 answer:
hoa [83]3 years ago
8 0
I don’t understand the question sorry
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Calcium chloride has a melting point of 772°C, and water has a boiling point of 100°C. Which statement is true for calcium chlor
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The difference is due to the fact that calcium chloride has stronger intermolecular forces than water.
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4 years ago
Calculate the temperature in k of 3.05 moles of gas occupying 3.70 L at 4.12 atm
Gnoma [55]

Answer:

{ \bf{PV = nRT}} \\ { \tt{(4.12 \times 3700) = 3.05 \times 0.083 \times T }} \\ { \tt{15244 = 0.25315 \: T}} \\ { \tt{T = 6.02 \times  {10}^{4} \: kelvin }}

6 0
3 years ago
What happens to the chemical energy in methane's bonds
Alexus [3.1K]

Answer: chemical energy is converted to heat and light energy

Explanation: exothermic reaction

4CH3+7O2–>6 H2O+4CO2+energy

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3 years ago
Ca(OH)2(aq)+HNO3(aq)→H2O(l)+Ca(NO3)2(aq)
shtirl [24]

what are we doing in that question

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6 0
3 years ago
Determine the rate law and the value of k for the following reaction using the data provided.
blsea [12.9K]

<u>Answer:</u> The rate law expression is \text{Rate}=k[NO]^2[O_2]^1 and value of 'k' is 1.727\times 10^3M^{-2}s^{-1}

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

2NO(g)+O_2(g)\rightarrow 2NO_2(g)

Rate law expression for the reaction:

\text{Rate}=k[NO]^a[O_2]^b

where,

a = order with respect to nitrogen monoxide

b = order with respect to oxygen

  • <u>Expression for rate law for first observation:</u>

8.55\times 10^{-3}=k(0.030)^a(0.0055)^b       ....(1)

  • <u>Expression for rate law for second observation:</u>

1.71\times 10^{-2}=k(0.030)^a(0.0110)^b       ....(2)

  • <u>Expression for rate law for third observation:</u>

3.42\times 10^{-2}=k(0.060)^a(0.0055)^b      ....(3)

Dividing 1 from 2, we get:

\frac{1.71\times 10^{-2}}{8.55\times 10^{-3}}=\frac{(0.030)^a(0.0110)^b}{(0.030)^a(0.0055)^b}\\\\2=2^b\\b=1

Dividing 1 from 3, we get:

\frac{3.42\times 10^{-2}}{8.55\times 10^{-3}}=\frac{(0.060)^a(0.0055)^b}{(0.030)^a(0.0055)^b}\\\\2^2=2^a\\a=2

Thus, the rate law becomes:

\text{Rate}=k[NO]^2[O_2]^1

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

8.55\times 10^{-3}=k[0.030]^2[0.0055]^1\\\\k=1.727\times 10^3M^{-2}s^{-1}

Hence, the rate law expression is \text{Rate}=k[NO]^2[O_2]^1 and value of 'k' is 1.727\times 10^3M^{-2}s^{-1}

4 0
3 years ago
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