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2 years ago

At STP, which 2.0-gram sample of matter uniformly fills a 340-milliliter closed container?

Chemistry

1 answer:

Xelga [282]2 years ago

6 0

Answer;

D, Xe (g)

Solution and explanation;

If 2g has a volume of 340ml.

Density is 1000/340*2 = 5.88g/litre.

-This rules out the two solids, choices 2) &3)

If 1 litre has mass 5.88g,

then 22.4 liters (volume at STP) has mass 5.88*22.4 = 131.8g/mol

molar mass Br2 = 80*2 = 160g/mol NO

molar mass Xe = 131.3g/mol = YES.

Answer is Xe

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How does an object become electrically charged?will give brainliest

ICE Princess25 [194]

Answer:

It is Though the transfer of charges from one object to another, or (A).

3 0

2 years ago

How many moles of N2 are need to fill a 35 L tank at standard temperature and pressure?

Tems11 [23]

1.56 moles of N2 are needed to fill a 35 L tank at standard temperature and pressure. Details about moles can be found below.

<h3>How to calculate number of moles?</h3>

The number of moles of a substance can be calculated using the following formula:

PV = nRT

Where;

P = pressure
V = volume
n = number of moles
R = gas law constant
T = temperature

At STP;

T = 273K
P = 1 atm
R = 0.0821 Latm/molK

1 × 35 = n × 0.0821 × 273

35 = 22.41n

n = 35/22.41

n = 1.56mol

Therefore, 1.56 moles of N2 are needed to fill a 35 L tank at standard temperature and pressure.

Learn more about number of moles at: brainly.com/question/14919968

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4 0

2 years ago

You are given 100 g of a compound. The compound is composed of 37% hydrogen and 63% oxygen. How many grams of hydrogen are prese

Reil [10]

B is the best answer

4 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!

4 0

2 years ago

Calculate the solubilities of the following compounds in a 0.02 M solution of barium nitrate using molar concentrations, first i

Law Incorporation [45]

Answer:

a. 1.7 × 10⁻⁴ mol·L⁻¹; b. 5.5 × 10⁻⁹ mol·L⁻¹

c. 2.3 × 10⁻⁴ mol·L⁻¹; 5.5 × 10⁻⁸ mol·L⁻¹

Explanation:

a. Silver iodate

Let s = the molar solubility.

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq); Ksp = 3.0 × 10⁻⁸

E/mol·L⁻¹: s s

$K_{sp} =\text{[Ag$^{+}$][IO$_{3}$$^{-}$]} = s\times s = s^{2} = 3.0\times 10^{-8}\\s = \sqrt{3.0\times 10^{-8}} \text{ mol/L} = 1.7 \times 10^{-4} \text{ mol/L}$

b. Barium sulfate

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq); Ksp = 1.1 × 10⁻¹⁰

I/mol·L⁻¹: 0.02 0

C/mol·L⁻¹: +s +s

E/mol·L⁻¹: 0.02 + s s

$K_{sp} =\text{[Ba$^{2+}$][SO$_{4}$$^{2-}$]} = (0.02 + s) \times s \approx 0.02s = 1.1\times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{0.02} \text{ mol/L} = 5.5 \times 10^{-9} \text{ mol/L}$

c. Using ionic strength and activities

(i) Calculate the ionic strength of 0.02 mol·L⁻¹ Ba(NO₃)₂

The formula for ionic strength is

$\mu = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\\mu = \dfrac{1}{2} (\text{[Ba$^{2+}$]}\cdot (2+)^{2} + \text{[NO$_{3}$$^{-}$]}\times(-1)^{2}) = \dfrac{1}{2} (\text{0.02}\times 4 + \text{0.04}\times1)= \dfrac{1}{2} (0.08 + 0.04)\\\\= \dfrac{1}{2} \times0.12 = 0.06$

(ii) Silver iodate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(1)^{2}\sqrt{0.06} = -0.51\times 0.24 = -0.12\\\gamma = 10^{-0.12} = 0.75$

b. Calculate the solubility

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq)

$K_{sp} =\text{[Ag$^{+}$]$\gamma_{Ag^{+}}$[IO$_{3}$$^{-}$]$\gamma_{IO_{3}^{-}}$} = s\times0.75\times s \times 0.75 =0.56s^{2}= 3.0 \times 10^{-8}\\s^{2} = \dfrac{3.0 \times 10^{-8}}{0.56} = 5.3 \times 10^{-8}\\\\s =2.3 \times 10^{-4}\text{ mol/L}$

(iii) Barium sulfate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(2)^{2}\sqrt{0.06} = -0.51\times16\times 0.24 = -0.50\\\gamma = 10^{-0.50} = 0.32$

b. Calculate the solubility

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq

$K_{sp} =\text{[Ba$^{2+}$]$\gamma_{ Ba^{2+}}$[SO$_{4}$$^{2-}$]$\gamma_{ SO_{4}^{2-}}$} = (0.02 + s) \times 0.32\times s\times 0.32 \approx 0.02\times0.10s\\2.0\times 10^{-3}s = 1.1 \times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{2.0 \times 10^{-3}} \text{ mol/L} = 5.5 \times 10^{-8} \text{ mol/L}$

7 0

3 years ago