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3 years ago

Mole conversions. Find the number of moles of argon in 607g of argon?

Chemistry

1 answer:

bogdanovich [222]3 years ago

5 0

Answer:

24249.65 mol

Explanation:

n=MM × m

n= 39.95 ×607

n=24249.65

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Any body tell me Iron salt and water produced what

kolezko [41]

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7 0

3 years ago

Consider a sample of calcium carbonate in the form of a cube measuring 2.805 in. on each edge.

Naya [18.7K]

Answer:

The answer to your question is: 8.82 x 10 ²⁴ atoms of oxygen

Explanation:

Data

Cube measuring : 2.805in on each edge

density = 2.7 g/cm3

# of oxygen atoms in the cube = ?

Process

Volume of the cube = (2.805)³ = 22.07 in³

Convert in³ to cm³ 1 in³ ------------------- 16.39 cm³

22.07 in³ -------------------- x

x = (22.07 x 16.39) / 1 = 361.72 cm³

Density = mass / volume

Mass = density x volume

Mass = (2.7)(361.72) = 976.66 g

Atomic mass CaCO3 = 40 + 12 + (16 x 3) = 100 g

100 g of CaCO3 -------------------- 48 g of O2

976.66 g of CaCO3 --------------- x

x = (976.66 x 48) / 100 = 468.8 g of O2

32 g of O2 ---------------------- 1 mol

468.8 g of O2 -------------------- x

x = (468.8 x 1) / 32 = 14.65 mol

1 mol ----------------------- 6.023 x 10 ²³ atoms

14.65 mol ------------------- x

x = (14.65 x 6.023 x 10 ²³) / 1 = 8.82 x 10 ²⁴ atoms of oxygen

3 0

3 years ago

What physical property makes metal pot good for cooking

inn [45]

It's a conductor meaning metal is good for radiating heat.

5 0

3 years ago

How many grams of AgNO3 are needed to make 250 ml of a solution that is 0.135 M

myrzilka [38]

Answer:

( About ) 5.7 grams

Explanation:

Take a look at the attachment below for a proper explanation;

5 0

4 years ago

How much volume will 8.8 moles of gas fill at 0.12 atm at 56*C

Neporo4naja [7]

Answer:

2000 L

General Formulas and Concepts:

Atomic Structure

Moles
Temperature Conversion: K = °C + 273.15

Gas Laws

Ideal Gas Law: PV = nRT

P is pressure
V is volume (in L)
n is number of moles
R is gas constant
T is temperature (in K)

Explanation:

Step 1: Define

[Given] 8.8 moles gas

[Given] 0.12 atm

[Given] 56 °C = 329.15 K

Step 2: Solve for V

Substitute in variables [Ideal Gas Law Formula]: $\displaystyle (0.12 \ atm)V = (8.8 \ mol)(0.0821 \ \frac{L \cdot atm}{mol \cdot K})(329.15 \ K)$
Isolate V: $\displaystyle V = \frac{(8.8 \ mol)(0.0821 \ \frac{L \cdot atm}{mol \cdot K})(329.15 \ K)}{(0.12 \ atm)}$
Multiply/Divide [Cancel out units]: $\displaystyle V = 1981.7 \ L$

Step 3: Check

Follow sig fig rules and round. We are given 2 sig figs.

1981.7 L ≈ 2000 L

3 0

3 years ago