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1 year ago

Calculate the grams of so2 gas present at stp in a 5. 9 l container. (atomic weights: s=32, o=16).

Chemistry

1 answer:

Solnce55 [7]1 year ago

8 0

The 5.9 L container at STP contains 16.83 g of SO2.

To begin, we must determine how many moles of SO2 are contained in the 5.9 L container.

At standard temperature and pressure (STP), one mole of SO2 equates to 22.4 liters.

5.9 L = 5.9 / 22.45.9 L = 0.26 moles of SO2 as a result.

The result is that 0.263 moles of SO2 are present in the container.

Following that, we shall determine the mass of 0.263 moles of SO2.

The following are ways to do this:

The mass of SO2 is 32 + (16 2) = 32 + 32= 64 g/mol,

where SO2 Mole = 0.263 Molar.

Molar + Molar Mass Equals Mass

SO2 mass is 16.83 g, or 0.263 of a kilogram.

Therefore, 16.83 g of SO2 are present in the 5.9 L container at STP.

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The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

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3 years ago

Give an example of gravitational potential

larisa86 [58]

A raised weight.

Water that is behind a dam.

Have a nice day!

4 0

3 years ago

Read 2 more answers

0.50 mol A, 0.60 mol B, and 0.90 mol C are reacted according to the following reaction

algol [13]

Reactant C is the limiting reactant in this scenario.

Explanation:

The reactant in the balanced chemical reaction which gives the smaller amount or moles of product is the limiting reagent.

Balanced chemical reaction is:

A + 2B + 3C → 2D + E

number of moles

A = 0.50 mole

B = 0.60 moles

C = 0.90 moles

Taking A as the reactant

1 mole of A reacted to form 2 moles of D

0.50 moles of A will produce $\frac{2}{1}$ = $\frac{x}{0.50}$

thus 0.50 moles of A will produce 1 mole of D

Taking B as the reactant

2 moles of B reacted to form 2 moles of D

0.60 moles of B reacted to form x moles of D

$\frac{2}{2}$ = $\frac{x}{0.6}$

x = 2 moles of D is produced.

Taking C as the reactant:

3 moles of C reacted to form 2 moles of D

O.9 moles of C reacted to form x moles of D

$\frac{2}{3}$ = $\frac{x}{0.9}$

= 0.60 moles of D is formed.

Thus C is the limiting reagent in the given reaction as it produces smallest mass of product.

5 0

3 years ago

- How many grams are in 4.6 x 1016 molecules of phosphorus trichloride (PCI)?

MariettaO [177]

Answer:

0.0000101 grams.

Explanation:

First of all, use Avogadro's number to convert from molecules to moles.

Avogadro's number is 6.02 x 10^23.

Second of all, to convert from moles to grams, find the substance formula of phosphorus trichloride. After finding the substance formula, then find the molar mass of phosphorus trichloride.

Phosphorus trichloride is PCl3

Now find the molar mass of PCl3

P = 31.0 amu

Cl = 35.5 amu

They are three chlorine atoms and one phosphorus atom. So you do this.

31.0(1) + 35.5(3) = 137.5 g/mol

Now use dimensional analysis to show your work

4.6 x 10^16 grams of PCl3 * 1 mol/ 6.02 x 10^23 * 137.5 g/mol / 1 mol

The grams and the moles cancel out.

So now do [(4.6 x 10^16) / (6.02 x 10^23)] x 137.5

and you will get 0.000010507

4.6 has the fewest digits which is 2.

So use sig figs and the answer rounds to the nearest millionths.

0.000010507 rounds to 0.0000101

So the final answer is 0.0000101 grams(don't forget the units)

Hope it helped!

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2 years ago

Stronger acids are those that —

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F. hold on to their protons more strongly

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3 years ago