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3 years ago

What will an object moving at a constant velocity do?

Chemistry

1 answer:

Len [333]3 years ago

3 0

Answer:B

Explanation: an object in motion will stay in motion until acted on by another force and an object in rest will stay in test until acted on by another force

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How would one convert 36 yd to centimeters (2.54 cm = 1 in)

ValentinkaMS [17]

I would go yards to feet, feet to inches, inches to centimeters. So 36 yards is 108 feet. There are 12 inches in a foot so 1,296 inches in 108 feet. Multiply that by 2.54 centimeters to get 3,291.84 centimeters in 36 yards.

3 0

4 years ago

Read 2 more answers

Consider the following multistep reaction:

chubhunter [2.5K]

The question is incomplete, here is the complete question:

Consider the following multistep reaction:

C+D⇌CD (fast)

CD+D→CD₂ (slow)

CD₂+D→CD₃ (fast)

C+3D→CD₃

Based on this mechanism, determine the rate law for the overall reaction.

Answer: The rate law for the reaction is $\text{Rate}=k'[C][D]^2$

Explanation:

Rate law is the expression which is used to express the rate of the reaction in terms of the molar concentration of reactants where each term is raised to the power their stoichiometric coefficient respectively from a balanced chemical equation.

In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.

For the given chemical reaction:

$C+3D\rightarrow CD_3$

The intermediate reaction of the mechanism follows:

Step 1: $C+D\rightleftharpoons CD;\text{ (fast)}$

Step 2: $CD+D\rightarrow CD_2;\text{(slow)}$

Step 3: $CD_2+D\rightarrow CD_3;\text{(fast)}$

As, step 2 is the slow step. It is the rate determining step

Rate law for the reaction follows:

$\text{Rate}=k[CD][D]$ ......(1)

As, [CD] is not appearing as a reactant in the overall reaction. So, we apply steady state approximation in it.

Applying steady state approximation for CD from step 1, we get:

$K=\frac{[CD]}{[C][D]}$

$[CD]=K[C][D]$

Putting the value of [CD] in equation 1, we get:

$\text{Rate}=k.K[C][D]^2\\\\\text{Rate}=k'[C][D]^2$

Hence, the rate law for the reaction is $\text{Rate}=k'[C][D]^2$

5 0

3 years ago

Inside a star, the force of gravity is balanced by the:

xenn [34]

C. Outward pressure caused by heat

7 0

3 years ago

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A chemist needs 10 liters of a 25% acid solution. The solution is to be mixed from three solutions whose concentrations are 10%,

mojhsa [17]

Answer:

a) 1 litre of 10% solution and 7 litre of 20% solution

b) 1.67 litres of 50% solution and 8.33 litres of the 20% solution

c) 3.75 litres of 50% solution and 6.25 litres of 20% solution

Explanation:

Given:

chemist needs = 10 liters of a 25% acid solution

Concentration of three solutions that are to be mixed = 10%, 20% and 50%.

Solution:

A) Use 2 liters of the 50% solution

Let us mix this with 10% and 20% solution

They will have to equal 8 litres

Let x=20% solution

Then (8-x) =10%

So the equation becomes,

10%(8-x)+ 20%x+50%(2)=25(10)

(0.1)(8-x) +0.2x+0.50(2)= 0.25(10)

0.8-0.1x+0.2x+1.0=2.5

0.2x-0.1x=2.5-0.8-1.0

0.1x=0.7

$x=\frac{0.7}{0.1}$

x= 7

so, 8-x = 8 -7= 1 litre of 10% solution and 7 litre of 20% solution

B)Use as little as possible of the 50% solution

Let x be the amount of 50% solution.

Then(10-x) be the 20% solution

Now the equation becomes,

50%(x)+20%(10-x)=25%(10)

0.50x+0.2(10-x)=0.25(10)

0.5x+2.0-0.2x=2.5

0.3x=2.5-2.0

0.3x=0.5

$x=\frac{0.5}{0.3}$

x=1.67

now (10-x)=(10-1.67)=8.33

so there will be 1.67 litres of 50% solution and 8.33 litres of the 20% solution

c) ) Use as much as possible of the 50% solution

Let x be the amount of 50% solution.

Then(10-x) be the 20% solution

Now the equation becomes,

50%(x)+10%(10-x)=25%(10)

0.50x+0.1(10-x)=0.25(10)

0.5x+1.0-0.1x=2.5

0.4x=2.5-1.0

0.4x=1.5

$x=\frac{1.5}{0.3}$

x=3.75

Now, (10-x)=(10- 3.75)=6.26

So there will be 3.75 litres of 50% solution and 6.25 litres of 20% solution

5 0

3 years ago

A sample of ethanol (C2H6O) has a mass of 0.2301 g. Complete combustion of this sample causes the temperature of a bomb calorime

Keith_Richards [23]

Answer: $4.994\times 10^{-3}$ moles

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass , occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

Given mass of ethanol = 0.2301

Molar mass of ethanol = 46.07 g/mol

$\text{Number of moles of ethanol}=\frac{0.2301g}{46.07g/mol}=4.994\times 10^{-3}$

Thus there are $4.994\times 10^{-3}$ moles of ethanol are present in the sample.

7 0

4 years ago

Read 2 more answers