The RNA and DNA backbone differ because the DNA sugar is missing what element?

A sample of neon occupies a volume of 461 mL at 1 atm and 273 K. What will be the volume of the neon when the pressure is reduce

Softa [21]

Answer:

0.501 L

Explanation:

To solve this problem we will use Boyle,s Law. According to this law "The volume of given amount of gas is inversely proportional to applied pressure at constant temperature".

V∝ 1/P

V= K/P

VP=K

Here the K is proportionality constant.

so,

P1V1 = P2V2

P= pressure

V= volume

Given data:

P1= 1 atm

V1= 461 mL

P2= 0.92 atm

V2= ? (L)

To solve this problem we have to convert the mL into L first.

1 L = 1000 mL

461/1000= 0.461 L

Now we will put the values in the equation,

P1V1 = P2V2

V2= P1V1/ P2

V2= 1 atm × 0.461 L / 0.92 atm

V2= 0.501 L

4 0

3 years ago

Which of the following molecules contains a nonpolar covalent bond?

mezya [45]

Answer:

H2

Explanation:

6 0

3 years ago

Read 2 more answers

How do I find percent mass composition

castortr0y [4]

Answer:

1. Find the molar mass of all the elements in the compound in grams per mole.

2. Find the molecular mass of the entire compound.

3. Divide the component's molar mass by the entire molecular mass.

4. You will now have a number between 0 and 1. Multiply it by 100% to get percent composition.

Explanation:

5 0

3 years ago

Identify which location in the periodic table you would have the<br> largest atomic radii.

bearhunter [10]

Answer:

left to right across a period when it decreases and when it increases top to bottom in a group,

hope i helped

5 0

3 years ago

My professor gave me two questions to solve using the Van Der Waals Equation. She told us to solve for P and the second one we h

Fed [463]

Answer:

P=atm

$b=\frac{L}{mol}$

Explanation:

The problem give you the Van Der Waals equation:

$(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT$

First we are going to solve for P:

$(P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}$

$P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}$

Then you should know all the units of each term of the equation, that is:

$P=atm$

$n=mol$

$R=\frac{L.atm}{mol.K}$

$a=atm\frac{L^{2}}{mol^{2}}$

$b=\frac{L}{mol}$

$T=K$

$V=L$

where atm=atmosphere, L=litters, K=kelvin

Now, you should replace the units in the equation for each value:

$P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}$

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

$P=\frac{L.atm}{L-L}-atm$

Then operate the fraction subtraction:

P= $P=\frac{L.atm-L.atm}{L}$

$P=\frac{L.atm}{L}$

And finally you can find the answer:

P=atm

Now solving for b:

$(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT$

$(V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}$

$nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}$

$b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}$

Replacing units:

$b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}$

Multiplying and dividing units,(please see the second photo below), we have:

$b=\frac{L-\frac{L.atm}{atm}}{mol}$

$b=\frac{L-L}{mol}$

$b=\frac{L}{mol}$

7 0

3 years ago