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3 years ago

What volume will 2.5 mol of a gas at STP occupy ?

Chemistry

1 answer:

kvv77 [185]3 years ago

6 0

Answer:

Explanation:

1 mole of gas at STP occupies 22.4 L of the gas

2.5 mole of the gas at STP occupies 22.4×2.5 L of the gas

so 2.5 mole of the gas at STP occupies 56 L of the gas .

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A student calculates the volume of a graduated cylinder to be 43.26 ml. The actual volume is 42.32 ml. What is the percent error

Tpy6a [65]

Answer:

2.2%

Explanation:

Percentage error,

You apply the formula,

[(Estimated value - Actual value)/Actual value] × 100%

; [(43.26 - 42.32)/42.32] × 100

; (0.94/42.32) × 100

; 0.022 × 100

Percent error = 2.2%

7 0

3 years ago

The solubility of o2 at 20c is 1.38 x10^-3. the partial presure of o2 in the air at sea level is 0.27 atm. using henery;s law, c

netineya [11]

<u>Answer:</u> The solubility of oxygen at 682 torr is $4.58\times 10^{-3}M$

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{A}=K_H\times p_{A}$

Or,

$\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}$

where,

$C_1\text{ and }p_1$ are the initial concentration and partial pressure of oxygen gas

$C_2\text{ and }p_2$ are the final concentration and partial pressure of oxygen gas

We are given:

Conversion factor used: 1 atm = 760 torr

$C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm$

Putting values in above equation, we get:

$\frac{1.38\times 10^{-3}}{C_2}=\frac{0.27atm}{0.897atm}\\\\C_2=\frac{1.38\times 10^{-3}\times 0.897atm}{0.27atm}=4.58\times 10^{-3}M$

Hence, the solubility of oxygen gas at 628 torr is $4.58\times 10^{-3}M$

4 0

3 years ago

The word root for a group of microorganisms living together is a. steriliz. b. prodrom. c. coloniza. d. nosocomi.

UkoKoshka [18]

The answer is C) Coloniza.

7 0

3 years ago

How many particles are present in0.24moles of carbon?

Ray Of Light [21]

Answer:

1.45 x 10²³ particles

Explanation:

Given parameters:

Number of moles of carbon = 0.24moles

Unknown:

Number of particles = ?

Solution:

A mole of a substance contains the Avogadro's number of particles.

The Avogadro's number of particles is 6.02 x 10²³

So;

0.24 moles of carbon will contain 0.24 x 6.02 x 10²³ = 1.45 x 10²³ particles

7 0

3 years ago

In a single displacement reaction between Sodium Phosphate and Barium, how much of each product (in grams) will be formed from 1

mixer [17]

Answer:

14.6 g of barium phosphate

3.35 g of sodium metal

Explanation:

2Na3PO4(aq) + 3Ba(s) -------> Ba3(PO4)2(aq) + 6Na(s)

The first step in any such reaction is to but down the balanced reaction equation according to the stoichiometry of the reaction.

The two products formed are barium phosphate and sodium metal.

Number of moles of barium corresponding to 10.0g of barium = mass of barium/ molar mass of barium

Molar mass of barium = 137.327 g

Number of moles of barium = 10/137.327

Number of moles of barium = 0.0728 moles

For barium phosphate;

3 moles of barium yields 1 mole of barium phosphate

0.0728 moles yields 0.0728 moles × 1/3 = 0.0243 moles of barium phosphate

Molar mass of barium phosphate = 601.93 g/mol

Therefore mass of barium phosphate = 0.0243 moles × 601.93 g/mol = 14.6 g of barium phosphate

For sodium metal

3 moles of barium yields 6 moles of sodium metal

0.0728 moles of barium yields 0.0728 × 6 / 3 = 0.1456 moles of sodium

Molar mass of sodium metal= 23 gmol-1

Mass of sodium metal= 0.1456g × 23 gmol-1 = 3.35 g of sodium metal

4 0

3 years ago