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4 years ago

Problems such as crazing cracks and skill are the result of improper

Chemistry

1 answer:

evablogger [386]4 years ago

6 0

Answer:

The question is incomplete.(not enough data provided).

Explanation:

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What caused wind to blow

Arlecino [84]

Answer:

Different atmospheric pressure. When there is a different atmospheric pressure, air moves from the higher pressure to the lower pressure area which results in what you call WIND but can result in various speeds and pressure.

Hope this helped and if it did, please give my answer a brainliest.

7 0

4 years ago

Read 2 more answers

For the reaction A + B + C → D + E, the initial reaction rate was measured for various initial concentrations of reactants. The

Alik [6]

Answer : The rate for trial 5 will be $4.25\times 10^{-2}Ms^{-1}$

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:

$A+B+C\rightarrow D+E$

Rate law expression for the reaction:

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

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

$6.0\times 10^{-4}=k(0.20)^a(0.20)^b(0.20)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-3}=k(0.20)^a(0.20)^b(0.60)^c$ ....(2)

Expression for rate law for third observation:

$2.4\times 10^{-3}=k(0.40)^a(0.20)^b(0.20)^c$ ....(3)

Expression for rate law for fourth observation:

$2.4\times 10^{-3}=k(0.40)^a(0.40)^b(0.20)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-3}}{2.4\times 10^{-3}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^0[C]^1$

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

Putting values in equation 1, we get:

$6.0\times 10^{-4}=k(0.20)^2(0.20)^0(0.20)^1$

$k=7.5\times 10^{-2}M^{-2}s^{-1}$

Thus, the value of the rate constant 'k' for this reaction is $7.5\times 10^{-2}M^{-2}s^{-1}$

Now we have to calculate the rate for trial 5 that starts with 0.90 M of reagent A, 0.60 M of reagents B and 0.70 M of reagent C.

$\text{Rate}=k[A]^2[B]^0[C]^1$

$\text{Rate}=(7.5\times 10^{-2})\times (0.90)^2(0.60)^0(0.70)^1$

$\text{Rate}=4.25\times 10^{-2}Ms^{-1}$

Therefore, the rate for trial 5 will be $4.25\times 10^{-2}Ms^{-1}$

3 0

3 years ago

Which of the following statements is TRUE?

Karo-lina-s [1.5K]

Answer:

The correct option is E.

Explanation:

Average speed of the gas molecules = $v_a$

$v_a=\sqrt{\frac{2RT}{M}}$

T = Temperature of the gas molecule

M = molar mass of gas molecule

R = gas constant

$v_a\propto \frac{1}{M}$

Higher the molar mass of the molecule lesser will be the speed and vice versa.

Hence, statement A is not true.

Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

Larger the molar mass that larger the molecule lessor will be the effusion and vice versa.

Hence, statement B is not true.

According to Boyle's law, pressure of the gas is inversely proportional o the volume occupied by the gas at constant temperature. And same can also be observed from ideal gas law:

$PV=nRT$

$P=\frac{nRT}{V}$

$P\propto \frac{1}{V}$

Higher the pressure of the gas lower will be the volume occupied by the gas and vice versa.

Hence, statement C is not true.

With rise in temperature, effusion of gas molecule increase as kinetic energy of gas molecules increases in with increase in temperature.

Average kinetic energy is defined as the average of the kinetic energies of all the particles present in a system. It is determined by the equation:

$K=\frac{3RT}{2N_A}$

where,

K = Average kinetic energy

R = Gas constant

T = Temperature of the system

$N_A$ = Avogadro's number

$K\propto T$

Higher the temperature more will be kinetic energy of gas molecule more easily it will effuse.

Hence, statement D is not true.

From this we can conclude that out of five statements first 4 statements are not true which means that none of these statements are true.

3 0

4 years ago

If a sample of carbon monoxide is at 57 degreees celsius and under 67.88 kPa of pressure and takes up 85.3 L of space how many m

Goryan [66]

57 degrees celcius is equal to 330 degrees kelvin
67.88 kPa is equal to 67880 Pa
85.3 liters is equal to 0.0853 m^3

Now, the equation we will use to solve this question is:
PV = nRT where:
P is the pressure of gas = 67880 Pa
V is the volume of gas = 0.0853 m^3
n is the number of moles we are looking for
R is the gas constant = 8.31441 J K-1 mol-1
T is the temperature of gas = 330 degrees kelvin

Substitute with the givens in the above equation to get n as follows:
n = (PV) / (RT)
n = (67880*0.0853) / (8.31441*330)
n = 2.11 moles

4 0

4 years ago

How do I find the correct answer for this chemistry question

horsena [70]

Answer:

letter C. Solid iron and chlorine gas react to produce solid iron (lll) chloride

Explanation:

i hope it helps

6 0

3 years ago