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3 years ago

Which is an intensive property of a substance? density volume length mass

Chemistry

2 answers:

lidiya [134]3 years ago

5 0

Answer:

A.) Density

Explanation:

kupik [55]3 years ago

3 0

Density is an intensive property of a substance. An intensive property means that it is a physical property that does not depend on the size or the amount of material in the matter. For example, if a diamond is cut in half, the density does NOT change, therefore it is an intensive property.

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With the given chemical compounds, what is the balanced chemical equation when lit with fire?

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3 years ago

Agnesium, calcium, and strontium are examples of ____________.

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D: alkaline earth metals

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3 years ago

At constant pressure, which of these systems do work on the surroundings? Check all that apply.

lana66690 [7]

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3 years ago

What is the rate law for the reaction 2A + 2B + 2C --> products

-Dominant- [34]

Answer:

R = 47.19 [A]*([B]^2)*[C]

Explanation:

The rate law for the reaction 2A + 2B + 2C --> products

Is being sought.

The reaction rate R could be expressed as

R = k ([A]^m)*([B]^n)*([C]^p) (1)

where m, n, and p are the reaction orders with respect to (w.r.t.) components A, B and C respectively. This could be reduced to

R = ka ([A]^m) (2)

Where ka=(k[B]^n)*([C]^p);

R = kb ([B]^n) (3)

Where kb=(k[A]^m)*([C]^p); and

R = kc ([C]^p) (4)

Where kc=(k[A]^m)*([B]^n).

Equations (2), (3) and (4) are obtained for cases when the concentrations of two components are kept constant, while only one component’s concentration is varied. We can determine the reaction wrt each component by employing these equations.

The readability is very much enhanced when the given data is presented in the following manner:

Initial [A] 0.273 0.819 0.273 0.273

Initial [B] 0.763 0.763 1.526 0.763

Initial [C] 0.400 0.400 0.400 0.800

Rate 3.0 9.0 12.0 6.0

Run# 1 2 3 4

Additional row is added to indicate the run # for each experiment for easy reference.

First, we use the initial rate method to evaluate the reaction order w.r.t. each component [A], [B] and [C] based on the equations (2), (3) and (4) above.

Let us start with the order wrt [A]. From the given data, for experimental runs 1 and 2, the concentrations of reactants B and C were kept constant.

Increasing [A] from 0.273 to 0.819 lead to the change of R from 3.0 to 9.0, hence we can apply the relation based on equation (2) between the final rate R2, the initial rate R1 and the final concentration [A2] and the initial concentration [A1] as follows:

R2/R1=ka[A2]^m/ka[A1]^m=([A2]/[A1])^m

9.0/3.0 = (0.819/0.273)^m

3 = (3)^m = 3^1 -> m = 1

Similarly, applying experimental runs 1 and 3 could be applied for the determination of n, by employing equation (3):

R3/R1=kb[B3]^n/kb[B1]^n=([B3]/[B1])^n

12/3= (1.526/0.763)^n

4= 2^n, -> n = 2

And finally for the determination of p we have using runs 4 and 1:

R4/R1=kc[C4]^p/kc[C1]^p=([C4]/[C1])^p

6/3= (0.8/0.4)^p

2= 2^p , -> p = 1

Therefore, plugging in the values of m, n and p into equation (1), the rate law for the reaction will be:

R = k [A]*([B]^2)*[C]

The value of the rate constant k could be estimated by making it the subject of the formula, and inserting the given values, say in run 1:

k = R /( [A]*([B]^2)*[C]) = 3/0.273*(0.763^2)*0.4 =

47.19

Finally, the rate law is

R = 47.19 [A]*([B]^2)*[C]

7 0

4 years ago

Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide. Ca) + H2O --> Ca(OH)2 In a particular

mars1129 [50]

Answer:

Percent yield = 92.5%

Explanation:

The question asks for the percent yield which can be defined as:

$\frac{actual yield}{theoretical yield} .100$

Where the actual yield is how much product was obtained, in this case 6.11 g of Ca(OH)₂, and the theoretical yield is how much product could be obtained with the given reactants theoretically, that is if the reaction would work perfectly. So we need to calculate first the theoretical yield.

1. First lets write the chemical equation reaction correctly and check that it is balanced:

CaO + H₂O → Ca(OH)₂

2. Calculate the amount of product Ca(OH)₂ that can be obtained with the given reactants (theoretical yield), which are 5.00g of CaO and excess of water. So the amount of CaO will determined how much Ca(OH)₂ we can obtained.

For this we'll use the molar ratio between CaO and Ca(OH)₂ which we see it is 1:1. For every mol of CaO we'll obtain a mol of Ca(OH)₂. So lets convert the 5.00 g of CaO to moles:

Molar Mass of CaO: 40.078 + 15.999 = 56.077 g/mol

moles of CaO = 5.00 g / 56.077 g/mol = 0.08916 moles

As we said before from the molar ratio moles of Ca(OH)₂ = moles of CaO

So the moles of Ca(OH)₂ that can be obtained are 56.077 g/mol

We need to convert this value to grams:

Molar Mass of Ca(OH)₂ = 40.078 + (15.999 + 1.008)*2 = 74.092 g/mol

Theoretical yield of Ca(OH)₂ = 0.08916 moles x 74.092 g/mol = 6.606 g

3. Calculate the percent yield:

$\frac{actual yield}{theoretical yield} .100$

Percent yield = (6.11 g / 6.606g) x 100 = 92.5 %

5 0

3 years ago

Read 2 more answers