Explain three main differences between bacteria and virus.

1. If the water is not completely removed from the salt during the heating process will the reported water-to-salt mole ratio (X

Vitek1552 [10]

Answer:

The water in the hydrate (referred to as "water of hydration") can be removed by heating the hydrate. When all hydrating water is removed, the material is said to be anhydrous and is referred to as an anhydrate.

Explanation:

4 0

3 years ago

2. What major alkene product is produced by the dehydration of the following alcohols?

weeeeeb [17]

2-Dimethylcyclohexanol major alkene product is produced by the dehydration of the alcohols.

4 0

3 years ago

Calculate the concentration in mol/L, M, of an aqueous sugar solution with a concentration of 23.5% (w/w) and density of 1.005 g

Temka [501]

Answer:

The concentration in mol/L is 0.683M

Explanation:

23.5% (w/w)

This data means that in 100 g of solution, we have 23.5 grams of solute.

From this point, we can calculate the moles of sugar.

Moles = Mass / Molar mass

Moles = 23.5 g /342.30 g/m

Moles = 0.068 moles

Density data make us know, the volume of our solution.

solution δ = solution mass / solution volume

δ = 1.005 g/mL = 100 g / solution volume

solution volume = 100g / 1.005 g/ml

solution volume = 99.5 mL

In conclusion, 0.068 moles are in 99.5 mL

Molarity (M) is mol/L

Let's convert 99.5 mL in L

99.5 mL / 1000 = 0.0995 L

0.068 m / 0.0995L = 0.683

If we convert moles in mmoles, we can also get Molarity (M)

mmoles / mL = M

0.068 moles . 1000 = 68 mmoles

68 mmmoles / 99.5mL = 0.683

8 0

3 years ago

Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm8 at 27 ∘C. (R =

maxonik [38]

Answer:

Total pressure = 56.77 bar

Explanation:

Given data:

Mass of dioxygen = 8 g

Mass of dihydrogen = 4 g

Volume of vessel = 1 dm³

Temperature = 27°C (27+273 = 300 K)

R = 0.083 bar.dm³ / mol.K

Total pressure = ?

Solution:

Number of moles of dioxygen:

Number of moles = mass/molar mass

Number of moles = 8 g/ 32 g/mol

Number of moles = 0.25 mol

Pressure of dioxygen:

PV = nRT

P = nRT/V

P = 0.25 mol × 0.083 bar.dm³ / mol.K × 300 K / 1 dm³

P = 6.97 bar.dm³ /1 dm³

P = 6.97 bar

Number of moles of dihydrogen:

Number of moles = mass/molar mass

Number of moles = 4 g/ 2 g/mol

Number of moles = 2 mol

Pressure of dihydrogen:

PV = nRT

P = nRT/V

P = 2 mol × 0.083 bar.dm³ / mol.K × 300 K / 1 dm³

P = 49.8 bar.dm³ /1 dm³

P = 49.8 bar

Total pressure of mixture in a vessel:

Total pressure = P (O₂) + P(H₂)

Total pressure = 6.97 bar + 49.8 bar

Total pressure = 56.77 bar

7 0

3 years ago

A solution containing a mixture of 0.0351 M potassium chromate ( K 2 CrO 4 ) and 0.0537 M sodium oxalate ( Na 2 C 2 O 4 ) was ti

ki77a [65]

Answer:

Explanation:

As it was stated in the comments by other user, the question is incomplete but luckily it was posted the rest of the question, so, I'm gonna answer it with the data of that. If you have another questions there, please submit it again or put it in the comments.

a) Which barium salt will precipite first?

In order to know this, we need to take a look at the Ksp of both salts. The given Ksp are:

Ksp 1: 2.1x10^-10 (BaCrO4)

Ksp 2: 1.3x10^-6 (BaC2O4)

Now, we can see that Ksp1 < Ksp 2, but what's this Ksp value means? a Ksp value means that an aqueous solution will form a precipite (So the solid formed, it's not soluble in water), As Ksp 1 is a smaller value than Ksp2, means that the concentrations of Ba and CrO4 are too small, and therefore, it takes more time to precipite. Therefore, the BaC2O4 will precipite first.

b) Concentration of Ba2+ present to BaCrO4 precipite

In this, we know that the reaction taking place is the following:

BaCrO4(s) <--------> Ba^2+ + CrO4^2- Ksp = 2.1x10^-10

The expression for Ksp is:

Ksp = [Ba][CrO4]

We know the concentration of CrO4 cause this comes from the K2CrO4 so, replacing here, we solve for Ba:

[Ba^2+] = Ksp / [CrO4^2-]

[Ba^2+] = 2.1x10^-10 / 0.0351

[Ba^2+] = 5.98x10^-9 M

c) Concentration of Ba^2+ to reduce 10% oxalate concentration

The 10% concentration of oxalate is:

[C2O4] = 0.0537 * 0.1 = 0.00537 M

So, we do the same thing we did in part b) but solving for C2O4:

BaC2O4(s) <--------> Ba^2+ + C2O4^2- Ksp = 1.3x10^-6

[Ba^2+] = 1.3x10^-6 / 0.00537

[Ba^2+] = 2.42x10^-4 M

d) ratio of oxalate and chromate when Ba concentration is 0.005 M

In this case, we calculate concentration of CrO4 and C2O4 with the value of Ba and it's respective Ksp, and then, calculate the ratio:

[CrO4^2-] = Ksp1 / [Ba^2+] = 2.1x10^-10 / 0.0050 = 4.2x10^-8 M

[C2O4^2-] = Ksp2 / [Ba^2+] = 1.3x10^-6 / 0.0050 = 2.6x10^-4 M

So the ratio is:

[C2O4^2-] / [CrO4^2-] = 2.6x10^-4 / 4.2x10^-8

[C2O4^2-] / [CrO4^2-] = 6.19x10^3

8 0

3 years ago