All waves carry A) energy. B) light. C) matter. D) particles.

A sample of oxygen has a volume of 128 milliliter at a pressure of 500.0 mmhg calculate the volume this sample would occupy at s

leva [86]

Answer:

84.2mL

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 128mL

Initial pressure (P1) = 500mmHg

Final pressure (P2) = stp = 760mmHg

Final volume (V2) =..?

Thus, the new volume of the gas can be obtained by using the Boyle's law equation as follow:

P1V1 = P2V2

500 x 128 = 760 x V2

Divide both side by 760

V2 = (500 x 128)/760

V2 = 84.2mL

Therefore, the new volume of the gas is 84.2mL

4 0

4 years ago

Summarize the characteristics of each type of element below

aniked [119]

What are the elements?

7 0

3 years ago

a compound has 15.39 g of gold for every 2.77 g of chlorine. simplified there is _____ g of gold for every 1 g of chlorine

iren [92.7K]

Answer:

There is 5.56 g of gold for every 1 g of chlorine

Explanation:

The ratio is the relationship between two numbers, defined as the ratio of one number to the other. So, the ratio between two numbers a and b is the fraction $\frac{a}{b}$

You know that a compound has 15.39 g of gold for every 2.77 g of chlorine. This can be expressed by the ratio:

$\frac{15.39 g of gold}{2.77 g of chlorine}$

The proportion is the equal relationship that exists between two reasons and is represented by: $\frac{a}{b}=\frac{c}{d}$

This reads a is a b as c is a d.

To calculate the amount of gold per 1 g of chlorine, the following proportion is expressed:

$\frac{15.39 g of gold}{2.77 g of chlorine}=\frac{mass of gold}{1 g of chlorine}$

Solving for the mass of gold gives:

$mass of gold=1 g of chlorine*\frac{15.39 g of gold}{2.77 g of chlorine}$

mass of gold= 5.56 grams

So, there is 5.56 g of gold for every 1 g of chlorine

5 0

3 years ago

What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom

Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains $6.022\times 10^{23}$ atoms

So, 5.5 g of lead contains $\frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22}$ atoms

and,

As, 118.71 g of lead contains $6.022\times 10^{23}$ atoms

So, 94.5 g of lead contains $\frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23}$ atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

$\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%$

and,

$\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%$

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

6 0

3 years ago

Is this right??? If not gimme da right answer plz

NemiM [27]

Answer:

yeah

Explanation:

well, probably. they kicked me out of math class because I put a live chicken in the classroom and it pooped everywhere, so I had to clean it up and bring it back where I found it (which is the side of the road.)

3 0

3 years ago

Read 2 more answers