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3 years ago

I need help on 4-6 pleasee

Chemistry

1 answer:

GenaCL600 [577]3 years ago

6 0

4. It is an endothermic process. The body of someone taking a shower would be absorbing heat from the environment, another example of this would be cooking an egg

5. Ice melting is endothermic. The ice is absorbing heat from the environment, water freezing would be an exothermic reaction as it is transferring heat into the environment

6 however i do not have an answer for. I may just be a little dense but I'm not entirely sure what it is asking when it asks what the system is.

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Calculate how many times more soluble Mg(OH)2 is in pure water Based on the given value of the Ksp, 5.61×10−11, calculate the ra

maks197457 [2]

Answer:

molar solubility in water = 2.412 * 10^-4 mol/L

molar solubility of NaOH in 0.130M = 3.32 * 10^-9 mol/L

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

Explanation:

The Ksp refers to the partial solubilization of a mostly insoluble salt. This is an equilibrium process.

The equation for the solubilization reaction of Mg(OH)2 can be given as:

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

Ksp can then be given as followed:

Ksp = [Mg^2+][OH^–]²

Step 2: Calculate the solubility in water

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X of OH-

The concentration at equilibrium will be XM Mg^2+ and 2X OH-

Ksp = [Mg^2+][OH^–]²

5.61*10^-11 = X * (2X)² = X *4X² = 4X³

X = 2.412 * 10^-4 mol/L = solubility in water

Step 3: Calculate solubility in 0.130 M NaOH

The initial concentration of Mg^2+ = 0 M

The initial concentration of OH- = 0.130 M

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X +0.130 for OH-

The concentration at equilibrium will be XM Mg^2+ and 0.130 + 2X OH-

The value of "[OH–] + 2X" is, because the very small value of X, equal to the value of [OH–] .

Let's consider:

[Mg+2] = X

[OH] = 0.130

Ksp = [Mg^2+][OH^–]²

5.61*10^-11 = X *(0.130)²

5.61*10^-11 = X * (0.130)^2

X = 3.32*10^-9 = solubility in 0.130 M NaOH

Step 4: Calculate how many times Mg(OH)2 is better soluble in pure water.

(2.412*10^-4)/ (3.32*10^-9) = 0.73 * 10^5

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

4 0

3 years ago

Suppose the reaction between nitrogen and hydrogen was run according to the amounts presented in Part A, and the temperature and

andrew11 [14]

Explanation:

Assuming that moles of nitrogen present are 0.227 and moles of hydrogen are 0.681. And, initially there are 0.908 moles of gas particles.

This means that, for $N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}$

moles of $N_{2}$ + moles of $H_{2}$ = 0.908 mol

Since, 2 moles of $N_{2}$ = $2 \times 0.227$ = 0.454 mol

As it is known that the ideal gas equation is PV = nRT

And, as the temperature and volume were kept constant, so we can write

$\frac{P(_in)}{n_(in)}$ = $\frac{P_(final)}{n_(final)}$

$\frac{10.4}{0.908}$ = $\frac{P_(final)}{0.454
}$

$P_(final)$ = $10.4 \times \frac{0.454}{0.908}$

= 5.2 atm

Therefore, we can conclude that the expected pressure after the reaction was completed is 5.2 atm.

7 0

3 years ago

I COULD REALLY USE SOME HELP! I WILL RATE BRAINLIEST!

Travka [436]

0.000169 mol/g citric acid

3 0

3 years ago

For molecules with more than one ionizable groups, such as glycine, which of the following statements is true? For molecules wit

tester [92]

Answer: Option (a) is the correct answer.

Explanation:

The isoelectric point (pI) is actually the pH where a particular amino acid exists as a neutral molecule. This means that if pH equals pI then there will be no electric charge on the molecule.

Therefore, the statement at pH near the pI, nearly all the molecules carry no net charge, is true.

A buffer is composed of a solution of weak acid or base and salt of weak acid/base.So, when pH equals pI then there occurs no net charge. Hence, there will be no existence of conjugate acid-base pair.

Therefore, the statement when the pH is near the pI, the solution is near its maximum buffering ability, is false.

It is known that when pH is less than pI then there will be a net positive charge on the amino acid.

Hence, the statement at a pH near the pI, nearly all the molecules carry a net positive charge, is false.

When pH is greater than pI then amino acid will have a net negative charge. Hence, the statement at a pH near the pI, nearly all the molecules carry a net negative charge, is false.

Thus, we can conclude that out of the given options the statement at pH near the pI, nearly all the molecules carry no net charge, is true.

5 0

3 years ago

During a titration the following data were collected. A 10. mL portion of an unknown monoprotic acid solution was titrated with

Liula [17]

Answer:

8.0 moles

Explanation:

Since the acid is monoprotic, 1 mole of the acid will be required to stochiometrically react with 1 mole of NaOH.

Using the formula: $\frac{concentration of acid X volume of acid}{concentration of base X volume of base} = \frac{mole of acid}{mole of base}$

Concentration of acid = ?

Volume of acid = 10 mL

Concentration of base = 1.0 M

Volume of base = 40 mL

mole of acid = 1

mole of base = 1

Substitute into the equation:

$\frac{concentration of acid X 10}{1.0 X 40} = \frac{1}{1}$

Concentration of acid = 40/10 = 4.0 M

To determine the number of moles of acid present in 2.0 liters of the unknown solution:

Number of moles = Molarity x volume

molarity = 4.0 M

Volume = 2.0 Liters

Hence,

Number of moles = 4.0 x 2.0 = 8 moles

8 0

3 years ago