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3 years ago

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From your solubility curve determine the solubility of ammonium chloride in g/100 mL of H2O at 60 Degrees Celsius.

1 answer:

Natasha_Volkova [10]3 years ago

4 0

To determine the solubility of ammonium chloride, we first find the curve for ammonium chloride in the chart for solubility measured in grams of solute per 100 mL of water solvent. Reading the chart at 60 degrees Celsius, approximately 55 grams of ammonium chloride will dissolve in 100 mL of water. Therefore, the solubility of ammonium chloride is 55 grams NH4Cl/100mL H2O at 60°C.

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Convert 15.78 g of H2O2 to mols. <br> Convert 15.3 mols Au to g

iren2701 [21]

Answer:

.46 moles H2O2

3,014 grams Au

Explanation:

H2O2:

15.78g (1 mol/34g) = .46 moles H2O2

Au:

15.3mol (197g/mol) = 3,014 grams Au

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3 years ago

Why can't polyatomic ions ever stand alone?

natta225 [31]

I believe its because oxygen and carbon cant be sperated

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3 years ago

Be sure to answer all parts. what are the concentrations of hso4−, so42−, and h in a 0. 31 m khso4 solution? (hint: h2so4 is a s

disa [49]

At equilibrium the concentrations of:

[HSO₄⁻] = 0.10 M;

[SO₄²⁻] = 0.037 M;

[H⁺] = 0.037 M;

There is initially very little H+ and no SO₄²⁻ in the solution. A salt is KHSO₄⁻. All KHSO₄⁻ will split apart into K⁺ and HSO₄⁻ ions. [HSO₄⁻] will initially be present at a concentration of 0.14 M.

HSO₄⁻ will not gain H⁺ to produce H₂SO₄ since H₂SO₄ is a strong acid. HSO₄⁻ may act as an acid and lose H⁺ to form SO₄²⁻. Let the final H⁺ concentration be x M. Construct a RICE table for the dissociation of HSO₄²⁻.

R $HSO_4^-$ ⇄ $H^+ + SO_4^2^-$

I $0.14$

C $- x$ $+x$ $+x$

E $0.14-x$ $x$ $x$

$K_a = 1.3$ × $10^-^2$ for $HSO^-_4$ . As a result,

$\frac{[H^+]. [SO_4^2^-]}{HSO_4^-} = K_a$

$K_a$ is large. It is no longer valid to approximate that $[HSO^-_4]$ at equilibrium is the same as its initial value.

$\frac{x^2}{0.14-y} = 1.3 * 10^-^2$

$x^2+1.3*10^-^2x - 0.14$ × $1.3$ × $10^-^2= 0$

Solving the quadratic equation for $x , x \geq 0$ since $x$ represents a concentration;

$x=0.0366538$

Then, round the results to 2 significant figure;

$[SO_4^2^-] = x = 0.037 mol. L ^-^1$
$[H^+] = x = 0.037 mol. L ^-^1$
$[HSO_4^-] = 0.14 - x = 0.10 mol. L ^-^1$

Learn more about concentration here:

brainly.com/question/14469428

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3 0

1 year ago

Which of the following is most likely to be a conductor

aleksandr82 [10.1K]

Answer:

A compound containing metallic bonds.

Explanation:

8 0

3 years ago

For the following reaction, 9.30 grams of glucose (C6H12O6) are allowed to react with 13.8 grams of oxygen gas. glucose (C6H12O6

amid [387]

Answer:

13.7 g of CO₂

Limiting reactant: C₆H₁₂O₆

3.81 g of O₂

Explanation:

We convert the mass of the reactants to moles, in order to find out the limiting reactant and the excess reagent

9.30 g / 180 g/mol = 0.052 moles of glucose

13.8 g / 32 g/mol = 0.431 moles of oxygen

The equation is: C₆H₁₂O₆(s) + 6O₂ (g) → 6CO₂ (g) + 6H₂O (l)

Ratio is 1:6. Let's consider this rule of three:

1 mol of glucose reacts with 6 moles of oxygen

Then, 0.052 moles of glucose must react with (0.052 . 6) /1 = 0.312 moles

We have 0.431 moles of oxygen and we only need 0.312 moles. This means that an amount of oxygen still remains after the reaction is complete:

0.431 - 0.312 = 0.119 moles. We convert the moles to mass:

0.119 mol . 32 g / 1mol = 3.81 g

In conclussion, the limiting reactant is the glucose.

6 moles of oxygen react with 1 mol of glucose

0.431 moles of O₂ will react with (0.431 . 1) /6 = 0.072 moles of glucose

We only have 0.052 moles, so it is ok to say, that glucose is the limiting cause we do not have enough glucose.

Let's verify, the maximum amount of carbon dioxide that can be formed:

1 mol of glucose can produce 6 moles of CO₂

Therefore 0.052 moles of glucose will produce (0.052 . 6) /1 = 0.312 moles

We convert the moles to mass → 0.312 mol . 44 g /1 mol = 13.7 g

6 0

3 years ago