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2 years ago

As temperature increase and the volume stays constant, how will the pressure change

Chemistry

1 answer:

crimeas [40]2 years ago

5 0

Answer:

There will be more collisions and so a greater pressure. The number of particles is proportional to pressure, if the volume of the container and the temperature remain constant. ... Volume is inversely proportional to pressure, if the number of particles and the temperature are constant.

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Be sure to answer all parts. Consider the reaction A + B → Products From the following data obtained at a certain temperature, d

worty [1.4K]

Answer : The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.

Explanation :

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:

$A+B\rightarrow Products$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first observation:

$3.20\times 10^{-1}=k(1.50)^a(1.50)^b$ ....(1)

Expression for rate law for second observation:

$3.20\times 10^{-1}=k(1.50)^a(2.50)^b$ ....(2)

Expression for rate law for third observation:

$6.40\times 10^{-1}=k(3.00)^a(1.50)^b$ ....(3)

Dividing 1 from 2, we get:

$\frac{3.20\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(1.50)^a(2.50)^b}{k(1.50)^a(1.50)^b}\\\\1=1.66^b\\b=0$

Dividing 1 from 3, we get:

$\frac{6.40\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(3.00)^a(1.50)^b}{k(1.50)^a(1.50)^b}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[A]^1[B]^0$

$\text{Rate}=k[A]$

Thus,

The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.

7 0

3 years ago

3. Which of the following most likely describes an isotope? *

Nimfa-mama [501]

I think the answer might be B but i’m not positive

5 0

3 years ago

Here is the electron configuration for Magnesium. How many total electrons are in the 2nd energy level?

balandron [24]

Answer:

8 electrons

Explanation:

Magnesium is present on group 2.

It has 2 valence electrons.

Electronic configuration of magnesium:

Mg₁₂ = 1s² 2s² 2p⁶ 3s²

1st energy level contain 2 electrons.(1s²)

2nd energy level contain 8 electrons. (2s² 2p⁶)

3rd energy level contain 2 electrons. (3s²)

3rs energy level of magnesium is called valence shell. It contain two valance electrons. Magnesium can easily donate its two valance electrons and get stable electronic configuration.

It react with halogens and form salt. For example,

Mg + Cl₂ → MgCl₂

7 0

3 years ago

In the partition coefficient experiment 4A this week, after thorough mixing of the reagents, phase separation will occur. The to

Andreas93 [3]

Answer:

The top layer is the Aqueous layer, and the benzoic acid is contained in the non-aqueous layer/oil phase.

Explanation:

A separating funnel is a very important piece of laboratory glassware that is used to separate the components of liquid-liquid mixtures which are immiscible. This technique is used in the extraction of the components of mixtures.

The liquids separate into two phases. The separation is based on the differences in the liquids' densities, where the denser liquid settles below and the lower density liquid stays afloat. Liquids used for this kind of separation are usually different liquids, one is the aqueous layer and the other, a non-aqueous layer.

Partition coefficient or distribution coefficient is defined as the ratio of the concentrations of a compound in two immiscible solvents at equilibrium.

Organic solvents (except halogenated organic compounds) with densities greater than that of water i.e 1g/mL ( usually called the oil phase) settle at the bottom of the aqueous phase.

Benzoic acid. will settle at the bottom layer ( i.e the lower phase).

7 0

3 years ago

Determine the formula weights of each of the following compounds.Part A) Nitrous oxide, N2O, known as laughing gas and used as a

densk [106]

Answer:

See explanation

Explanation:

a) Nitrous oxide (N2O) has a molar mass of 44.014 amu. It has 2 nitrogen atoms each with a mass of 14.007 amu and 1 oxygen atom with a mass of 16.0 amu.

Percentage nitrogen = (2*14.007 amu/ 44.014 amu ) *100% = 63.6%

Percentage oxygen = (16 amu/44.014 amu) * 100% = 36.4 %

63.6% + 36.4% = 100%

b) Benzoic acid (C7H6O2) has a molar mass of 122.13 amu. It has 6 hydrogen atoms each with a mass of 1.01 amu; it has 7 carbon atoms each with a mass of 12.01amu and 2 oxygen atoms with a mass of 16.0 amu.

Percentage hydrogen = (6*1.01 amu / 122.13 amu)*100% = 4.96%

Percentage carbon = (7*12.01 amu/ 122.13 amu)*100% = 68.8%

Percentage oxygen = (2*16 amu/ 122.13 amu) *100% = 26.2%

c) Magnesium hydroxide (Mg(OH)2) has a molecular mass of 58.32 amu. It has 2 hydrogen atoms each with a mass of 1.01 amu; it has 1 magnesium atom with a mass of 24.3 amu and two oxygen atoms with a mass of 16.0 amu.

Percentage hydrogen = (2*1.01 amu/ 58.32 amu) *100% = 3.46 %

Percentage magnesium = (24.3 amu/58.32 amu)*100% = 41.7%

Percentage oxygen = (2*16 amu/58.32 amu)*100% = 54.9%

d) Urea CO(NH2)2 has a molecular mass of 60.064 amu. It has 2 Nitrogen atoms each with a mass of 14.007 amu, 4 hydrogen atoms each with a mass of 1.01 amu,1 carbon atom with a mass of 12.01 amu and 1 oxygen atom with a mass of 16.0 amu.

Percentage nitrogen = (2*14.007 amu/ 60.064amu)*100% = 46.6%

Percentage hydrogen = (4*1.01 amu/60.064amu)*100% = 6.72%

Percentage carbon = (12.01 amu/60.064amu)*100% = 20.0%

Percentage oxygen = (16 amu/60.064amu)*100% = 26.6%

e) Osopentyl acetate (C7H14O2) has a molecular mass of 130.2 amu. It has 14 hydrogen atoms each with a mass of 1.01 amu,7 carbon atoms each with a mass of 12.01 amu and 2 oxygen atom with a mass of 16.0 amu.

Percentage hydrogen = (14*1.01 amu/130.2 amu)*100% = 10.8%

Percentage carbon = (7*12.01 amu/130.2 amu)*100% = 64.6%

Percentage oxygen =(2*16 amu/130.2 amu)*100% = 24.6%

7 0

3 years ago