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3 years ago

How many argon (ar) atoms are there in 1.5 × 102 g of argon?

1 answer:

8 0

Answer is: there is 2,25·10²⁴ argon atoms.
m(Ar) = 1,5·10² g = 150 g.
n(Ar) = m(Ar) ÷ M(Ar).
n(Ar) = 150 g ÷ 40 g/mol.
n(Ar) = 3,75 mol.
N(Ar) = n(Ar) · Na.
N(Ar) = 3,75 mol · 6·10²³ 1/mol.
N(Ar) = 2,25·10²⁴.
n - amount of substance.
Na - Avogadro number.
M - molar mass of atom.
m - mass of atom.

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A car with a mass of 1.1 × 103 kilograms hits a stationary truck with a mass of 2.3 × 103 kilograms from the rear end. The initi

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The answer is C : 15.7 m/s
Use the idea of : momentum before collision = momentum after collision

Before collision;
For car:mass=1.1×10^3, velocity=22
For truck:mass=2.3×10^3, velocity=0
After collision;
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For truck:mass=2.3×10^3, velocity=V
(1.1×10^3 × 22) + (2.3×10^3 × 0) = (1.1×10^3 × -11) + (2.3×10^3 × V)
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3 years ago

Read 2 more answers

Certain compound contains 7.3% carbon, 4.5% hydrogen, 36.4% oxygen, and 31.8% nitrogen. It’s reality molecular mass is 176.0. Fi

erastovalidia [21]

Answer:

The answer to your question is:

Explanation:

Data

carbon 7.3% = 7.3g

hydrogen 4.5% = 4.5g

oxygen 36.4% = 36.4 g

nitrogen 31.8% = 31.8 g

Now

For carbon

12 g --------------------1 mol

7.3 g ------------- x

x = 7.3/12 = 0.608 mol

For hydrogen

1 g -------------------- 1 mol

4.5 g ------------------ x

x = 4.5 mol

For oxygen

16 g ------------------- 1 mol

36.4 g ---------------- x

x = 2.28 mol

For nitrogen

14 g ---------------- 1 mol

31.8 g --------------- x

x = 2.27 mol

Now divide by the lowest result, the is 0.608 from carbon

carbon 0.608/0.608 = 1

hydrogen 4.5/ 0.608 = 7.4

oxygen 2.28/0.608 = 3.75

nitrogen 2.27/0.608 = 3.73

Empirical formula = CH₇O₄N₄

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2 years ago

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Given the following data:

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$176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

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