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Luda [366]
3 years ago
5

Write the rate laws for the following elementary reactions. (a) ch3nc(g) → ch3cn(g)

Chemistry
1 answer:
klasskru [66]3 years ago
4 0

Answer:

a. rate = k[CH3NC]

Explanation:

In elementary reactions you use the coefficients of the reactants for the exponents. Other than that its the same. In this elementary step, the coefficients of the reactant is 1.  

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AIR PRESSURE INCREASES AS ALTITUDE INCREASES
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What is the ph of a 0.01 m solution of the strong acid hno3 in water?
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Answer is: pH <span>of a 0,01 M solution is 2.
c(HNO</span>₃) = 0,01 M = 0,01 mol/L.
pH = -log(c(HNO₃).
pH = -log(0,01 mol/L).
pH = 2.
pH<span> is a numeric scale used to specify the </span>acidity<span> or </span>basicity<span> of an </span>aqueous solution<span>. If pH is less than seven, than solution is acidic and if pH is greater seven, solution is basic, if pH is equal seven, solution is neutral.</span>
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Determine what is missing from this neutralization reaction: AgNO3+KCl→AgCl+−−−−
lapo4ka [179]

Answer:

The answer to your question is: KNO₃

Explanation:

                                AgNO3  +  KCl   →   AgCl  +   −−−−

A. KNO3    this option is correct because it is a double replacement reaction then potassium must attached to NO₃.

B. KOH  this product is not possible because there is no water to form OH⁻ ions.

C. Ag2K  this product is not possible because both Ag and K are metals, then it is difficult that they attach.

D. KN2O This product is imposible to form, this option is wrong.

7 0
2 years ago
Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005 g sample of menthol is comb
Bogdan [553]

<u>Answer:</u> The molecular formula for the menthol is C_{10}H_{20}O

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=0.2829g

Mass of H_2O=0.1159g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.2829 g of carbon dioxide, \frac{12}{44}\times 0.2829=0.077g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.1159 g of water, \frac{2}{18}\times 0.1159=0.0129g of hydrogen will be contained.

  • Mass of oxygen in the compound = (0.1005) - (0.077 + 0.0129) = 0.0106 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.077g}{12g/mole}=0.0064moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.0129g}{1g/mole}=0.0129moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0106g}{16g/mole}=0.00066moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00066 moles.

For Carbon = \frac{0.0064}{0.00066}=9.69\approx 10

For Hydrogen  = \frac{0.0129}{0.00064}=19.54\approx 20

For Oxygen  = \frac{0.00066}{0.00066}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 10 : 20 : 1

Hence, the empirical formula for the given compound is C_{10}H_{20}O_1=C_{10}H_{20}O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{molecular mass}}{\text{empirical mass}}

We are given:

Mass of molecular formula = 156.27 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

n=\frac{156.27g/mol}{156g/mol}=1

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O

Thus, the molecular formula for the menthol is C_{10}H_{20}O

4 0
3 years ago
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Calculate a reasonable amount (mass in g) of your unknown acid to use for a titration. You will want about 30 mL of titrant to g
Vinil7 [7]

Answer:

"0.60 g" is the appropriate solution.

Explanation:

The given values are:

Volume of base,

= 30 ml

Molarity of base,

= 0.05 m

Molar mass of acid,

= 400 g/mol

As we know,

⇒  Molarity=\frac{Number \ of \ moles \ of \ base}{Number \ of \ solution}

On substituting the values, we get

⇒           0.05=\frac{Number \ of \ moles \ of \ base}{30\times 10^{-3}}

⇒  Number \ of \ moles \ of \ base=0.05\times 30\times 10^{-3}

⇒                                             =1.5\times 10^{-3}  

hence,

⇒  Moles \ of \ acid=\frac{Mass \ of \ acid}{Molar \ mass \ of \ acid}

On substituting the values, we get

⇒  1.5\times 10^{-3}=\frac{Mass \ of \ acid}{400}

⇒  Mass \ of \ acid=1.5\times 10^{-3}\times 400

⇒                         =0.60 \ g

8 0
3 years ago
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