Which of the following best describes what happens when hydrogen and oxygen to form water?

It is common to mix polar solvents (e.g., acetone) with non-polar solvents (e.g., hexane) to obtain an eluting solvent of interm

hammer [34]

Answer:

No, it is not appropriate to mix water and DMSO

Explanation:

We have to realize that DMSO is a highly polar solvent and water is a highly polar solvent. The question explicitly says that our target is to produce a solvent of intermediate polarity.

We can only do this by mixing a polar and a nonpolar solvent. We have been given the example of the mixture of acetone/hexane which is quite a perfect mixture.

Thus, it is inappropriate to mix DMSO and water.

6 0

2 years ago

In an experiment, a student was assigned to find the formula of an unknown hydrate that weighed 15.67g. The anhydrate, XCO3 (mol

meriva

Answer:

5.0 moles of water per one mole of anhydrate

Explanation:

To solve this question we must find the moles of the anhydrate. The difference in mass between the dry and the anhydrate gives the mass of water. Thus, we can find the moles of water and the moles of water per mole of anhydrate:

Moles Anhydrate:

7.58g * (1mol / 84.32g) = 0.0899 moles XCO3

Moles water:

15.67g - 7.58g = 8.09g * (1mol / 18.01g) = 0.449 moles H2O

Moles of water per mole of anhydrate:

0.449 moles H2O / 0.0899 moles XCO3 =

5.0 moles of water per one mole of anhydrate

6 0

2 years ago

A chemist must dilute 61.9 mL of 548. nM aqueous sodium carbonate (Na2CO3) solution until the concentration falls to 484. nM . H

maw [93]

Answer:

The final volume will be "70.08 mL".

Explanation:

The given values are:

Molar mass,

M1 = 548 nM

or,

= $5.48\times 10^{-7} \ M$

M2 = 484 nM

or,

= $4.84\times 10^{-7} \ M$

Volume,

V1 = 61.9 mL

V1 = ?

By using the expression, we get

⇒ $M1\times V1=M2\times V2$

or,

⇒ $V2=\frac{M1\times V1}{M2}$

By substituting the values, we get

$=\frac{5.48\times 10^{-7}\times 61.9}{4.84\times 10^{-7}}$

$=\frac{339.212}{4.84}$

$=70.08 \ mL$

7 0

2 years ago

Temperature is a measure of how much heat you have in, or around you. is it true or false

allochka39001 [22]

Your answer is

true

hope it helps you

7 0

2 years ago

Concerning Boyle's Law, if you had a gas at a pressure of 101 kPa and decreased the volume of the container holding the gas to o

Viefleur [7K]

Answer:

$P_2=404 kPa$

Explanation:

Hello,

In this case, the Boyle's is mathematically defined via:

$P_1V_1=P_2V_2$

Which stands for an inversely proportional relationship between volume and pressure, it means the higher the volume the lower the pressure and vice versa. In such a way, since the volume is decreased to one quarter, we can write:

$V_2=\frac{1}{4} V_1$

We can compute the new pressure:

$P_2=\frac{P_1V_1}{V_2} =\frac{P_1V_1}{\frac{1}{4} V_1} =\frac{101kPa*V_1}{\frac{1}{4} V_1} \\\\P_2=4*101kPa\\\\\\P_2=404 kPa$

Which means the pressure is increased by a factor of four.

Regards.

7 0

2 years ago