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AlexFokin [52]
3 years ago
15

Jason has a balanced chemical equation and a rate law for a chemical reaction. What should he do to determine the overall order

of the reaction?
A)add the exponents of the concentrations in the rate law
B)add the coefficients of the reactants from the balanced equation
C)add the coefficients of the reactants and products from the balanced equation
D)add the number of reactants to the rate constant
Chemistry
2 answers:
Afina-wow [57]3 years ago
6 0
The correct answer would be A. To determine the overall order of the reaction, Jason should add the exponents of the concentrations in the rate law. The overall order of a reaction is the sum of all the index or exponent of the concentration terms in the rate law.
Novosadov [1.4K]3 years ago
3 0

A on nudity

some extra words to get to 20 letters

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Atmospheric pressure at sea level is equal to a column of 760 mm Hg. Oxygen makes up 21 percent of the atmosphere by volume. The
Neporo4naja [7]

<u>Answer:</u> The partial pressure of oxygen is 160 mmHg

<u>Explanation:</u>

We are given:

Percent of oxygen in air = 21 %

Mole fraction of oxygen in air = \frac{21}{100}=0.21

To calculate the partial pressure of oxygen, we use the equation given by Raoult's law, which is:

p_{O_2}=p_T\times \chi_{O_2}

where,

p_{O_2} = partial pressure of oxygen = ?

p_T = total pressure of air = 760 mmHg

\chi_{O_2} = mole fraction of oxygen = 0.21

Putting values in above equation, we get:

p_{O_2}=760mmHg\times 0.21\\\\p_{O_2}=160mmHg

Hence, the partial pressure of oxygen is 160 mmHg

8 0
3 years ago
A food substance kept at 0°C becomes rotten (as determined by a good quantitative test) in 8.3 days. The same food rots in 10.6
ZanzabumX [31]

Answer:

1.   67.2 kJ/mol

Explanation:

Using the derived expression from Arrhenius Equation

In \ (\frac{k_2}{k_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})

Given that:

time t_1 = 8.3 days = (8.3 × 24 ) hours = 199.2 hours

time t_2 = 10.6 hours

Temperature T_1 = 0° C = (0+273 )K = 273 K

Temperature T_2 = 30° C = (30+ 273) = 303 K

Rate = 8.314 J / mol

Since (\frac{k_2}{k_1}=\frac{t_2}{t_1})

Then we can rewrite the above expression as:

In \ (\frac{t_2}{t_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})

In \ (\frac{199.2}{10.6}) = \frac{E_a}{8.314}(\frac{303-273}{273*303})

2.934 = \frac{E_a}{8.314}(\frac{30}{82719})

2.934 = \frac{30E_a}{687725.766}

30E_a = 2.934 *687725.766

E_a = \frac{2.934 *687725.766}{30}

E_a =67255.58 \ J/mol

E_a =67.2 \ kJ/mol

7 0
3 years ago
How many moles of solid naf would have to be added to 1.0 l of 2.39 m hf?
sp2606 [1]
Molarity is one way of expressing concentration and is equal to the number of moles of the solute per liter of the solution. Therefore,

Molarity = 2.39 mol / L solution

2.39 ( 1.0) = 2.39 mol HF

Hope this answers the question.
6 0
3 years ago
If 1.80 moles of NaCl was dissolved in enough water to make 3.60 L of solution, what is the Molarity?
sweet-ann [11.9K]

Answer:

0.5M

Explanation:

The equation for molarity is:

  • M = \frac{mol}{liters} ; where the "M" stands for molarity, the "mol" stands for moles of solute and the "liters" means the volume in liters of solution.

We are given that there are:

  • 1.80 moles of NaCl (the moles of solute)
  • 3.60 Liters of solution (the volume in liters of solution)

Now we just plug those numbers into the formula and get our answer:

  • M= \frac{1.80mol}{3.60L}= 0.5M

After doing the math and dividing the moles of solute by the liters of solution, we get that the molarity of the solution is 0.5M.

8 0
3 years ago
The mass of 0.10 mole of methane
saveliy_v [14]

Answer:

A methane molecule is made from one carbon atom and four hydrogen atoms. Carbon has a mass of 12.011 u and hydrogen has a mass of 1.008 u. This means that the mass of one methane molecule is 12.011 u + (4 × 1.008u), or 16.043 u. This means that one mole of methane has a mass of 16.043 grams.

メタン分子は、1つの炭素原子と4つの水素原子から作られています。炭素の質量は12.011uで、水素の質量は1.008uです。これは、1つのメタン分子の質量が12.011 u +(4×1.008u)、つまり16.043uであることを意味します。これは、1モルのメタンの質量が16.043グラムであることを意味します。^>^

5 0
3 years ago
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