How many grams of calcium cyanide (Ca(CN)2) are contained in 0.79 mol of calcium cyanide?

 <span>(NH4)2CO3 -> 96.09 g/mol

(6.995g ammonium carbonate)(1mol ammonium carbonate/ 96.09 g ammonium carbonate) = 0.072796 mol ammonium carbonate

In this calculation, the unit 'grams' cancels out as it's present in both the numerator and the denominator, leading to 'mol' being the remaining unit.

Examining the formula (NH4)2CO3, it can be interpreted as:
2 mol (NH4) + 1 mol (CO3) = 1 mol (NH4)2CO3

This means every mole of ammonium carbonate yields one mole of carbonate ions and two moles of ammonium ions.

(0.072796 mol ammonium carbonate) = (0.072796 mol carbonate ion) + (0.363981 mol ammonium ion) </span>

Answer:

Is it true that you shouldn't wash your hair, drink cold water, eat ice cream, or exercise during your period? Please explain your answer.

No, this is not accurate; doing any of these activities is perfectly fine. None of them affects us because they are not connected to our bodily systems. Also, I apologize for any language errors as I utilized Google Translate.

I hope this is helpful :)

Answer: The mole fraction of nitrogen is calculated to be 0.4615.

Explanation: In the mixture of nitrogen () and hydrogen (), the mole ratio established is 1: 1.5.

At this juncture, we identify that () serves as the limiting reagent.

When 0.4 moles of () are produced, it will require 0.4 moles of () along with 3.4 moles of ().

The total amount of remaining () ends up being 0.6, while the amount of () left in the mixture is 0.3 moles.

The mole fraction of () is computed as 0.6 divided by the sum of 0.6, 0.4, and 0.3, resulting in 0.4615.

Response:

ΔH = -793.6 kJ

Reasoning:

The ΔH for this particular reaction can be calculated through Hess's law, which allows one to combine the ΔH values of the half-reactions to find the overall ΔH:

The half-reactions are as follows:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

The calculation involves summing (4) + 4×(3) - (2) - 2×(1) - 2×(5):

(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369.2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356.9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483.6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)

<pThus, ΔH is:

ΔH = -639.1 kJ -369.2 kJ -356.9 kJ +483.6 kJ +88.0 kJ

ΔH = -793.6 kJ

I trust this clarifies things!

Answer:

0.133

Explanation:

The reaction that occurs between KIO3 and KI in an acidic medium is described as

IO3⁻ +5I⁻ +6h⁺ → 3I₂ + 3H₂O

I₂ subsequently reacts with sodium thiosulfate

NaS₂O₃  → 2Na⁺ + S₂O₃²⁻

The overall reaction can be summarized as

IO⁻₃ + 6H⁺ + 6S₂O₃³⁻ → I⁻ + 3S₄O₆²⁻ + 3H₂O

The mole of KIO₃

is computed using molarity multiplied by volume

which equals 0.00002mol

One mole of KIO₃ reacts with 6 moles of S₂O₃²⁻

which gives 2x6x10⁻⁵

= 0.00012 mol

The volume is 0.90 ml

1 ml equals 0.001L

0.90ML  is 0.0009L

To find concentration,

molarity/volume

= 0.00012/0.0009

= 0.133m