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velikii [3]
2 years ago
8

What volume will 50.2 grams of co2 (g) occupy at stp?

Chemistry
1 answer:
Genrish500 [490]2 years ago
8 0
Number of moles of CO2 =
Mass /Ar
= 50.2 / (12 + 32)
1.14 mols

For every 1 mol of gas, there will be
24000 cm^3 of gas

Vol. = 1.14 x 24 dm^3
= 27.36 dm^3
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Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown c
Alexus [3.1K]

Answer:

0.3229 M HBr(aq)

0.08436M H₂SO₄(aq)

Explanation:

<em>Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.</em>

<em />

Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).

NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)

When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.

18.45 \times 10^{-3} L NaOH.\frac{0.3500molNaOH}{1LNaOH} .\frac{1molHBr}{1molNaOH} .\frac{1}{20.00 \times 10^{-3} LHBr} =\frac{0.3229molHBr}{1LHBr} =0.3229M

<em>Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?</em>

<em />

Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).

2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)

When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.

42.18 \times 10^{-3} LNaOH.\frac{0.1000molNaOH}{1LNaOH} .\frac{1molH_{2}SO_{4}}{2molNaOH} .\frac{1}{25.00\times 10^{-3}LH_{2}SO_{4}} =\frac{0.08436molH_{2}SO_{4}}{1LH_{2}SO_{4}} =0.08436M

6 0
2 years ago
Which statement accurately compares ionic and covalent bonding?
Travka [436]

Answer:

Conduct electricity when they are molten, while covalent compounds usually do not conduct electricity when they are molten.

5 0
2 years ago
Read 2 more answers
What is the percent yield of this reaction if 22.o g of Mgl2 is produced by the reaction of 25.0 g of Mg with 25.0 g of l2?
expeople1 [14]

% yield = 80.719

<h3>Further explanation</h3>

Given

22.0 g of Mgl₂

25.0 g of Mg

25.0 g of l₂

Required

The percent yield

Solution

Reaction

Mg + I₂⇒ MgI₂

mol Mg = 25 g : 24.305 g/mol = 1.029

mol I₂ = 25 g : 253.809 g/mol = 0.098

Limiting reactant = I₂

Excess reactant = Mg

mol MgI₂ based on I₂, so mol MgI₂ = 0.098

Mass MgI₂ (theoretical):

= mol x MW

= 0.098 x 278.114

= 27.255 g

% yield = (actual/theoretical) x 100%

% yield = (22 / 27.255) x 100%

% yield = 80.719

8 0
1 year ago
Chlorine is used to disinfect swimming pools. the accepted concentration for this purpose is 1.00 ppm chlorine, or 1.00 g of chl
frez [133]

The accepted concentration of chlorine is 1.00 ppm that is 1 gram of chlorine per million of water.

The volume of water is 2.29\times 10^{4} gal.

Since, 1 gal= 3785.41 mL

Thus, 2.29\times 10^{4} gal=2.29\times 10^{4}\times 3785.41 mL=8.66\times 10^{7}mL

Density of water is 1 g/mL thus, mass of water will be 8.66\times 10^{7}g.

Since, 1 grams of chlorine →10^{6} grams of water.

1 g of water →10^{-6} g of chlorine and,

8.66\times 10^{7}g of water →86.6 g of chlorine

Since, the solution is 9% chlorine by mass, the volume of solution will be:

V=\frac{100}{9}\times 86.6 mL=9.62\times 10^{2} mL

Thus, volume of chlorine solution is 9.62\times 10^{2} mL.

6 0
2 years ago
How many moles of sodium bicarbonate is needed to neutralize 0.8 ml of sulphuric acid?
svet-max [94.6K]

Answer:

n NaHCO3 = 9.6 E-3 mol

Explanation:

balanced reaction:

  • 2 NaHCO3(s) + H2SO4(ac) ↔ Na2SO4(ac) + 2 CO2(g) + 2 H2O(l)
  • assuming a concentration of H2SO4 6M....normally worked in the lab

⇒ n H2SO4 = 8 E-4 L * 6 mol/L = 4.8 E-3 mol H2SO4

according to balanced reaction, we have that for every mol of H2SO4 there are two mol of NaHCO3 ( sodium bicarbonate)

⇒ mol NaHCO3 = 4.8 E-3 mol H2SO4 * ( 2 mol NaHCO3 / mol H2SO4 )

⇒ ,mol NaHCO3 = 9.6 E-3 mol

So 9.6 E-3 mol NaHCO3,  are the minimun moles necessary to neutralize the acid.

6 0
2 years ago
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