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3 years ago

What are the steps to balance a chemical equation?

Chemistry

1 answer:

aleksandrvk [35]3 years ago

5 0

Answer:

Step 1: The Unbalanced Chemical Equation. The unbalanced chemical equation is given to you. ...

Step 2: Make a List. ...

Step 3: Identifying the Atoms in Each Element. ...

Step 4: Multiplying the Number of Atoms. ...

Step 5: Placing Coefficients in Front of Molecules. ...

Step 6: Check Equation. ...

Step 7: Balanced Chemical Equation.

Explanation:

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What conclusion can you make about a hypothesis that contradicts an existing theory?

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What are you making a hypothesis for

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3 years ago

Using henry's law, calculate the molar concentration of o2 in the surface water of a mountain lake saturated with air at 20 ∘c a

s2008m [1.1K]

Answer: The molar concentration of oxygen gas in water is $1.43\times 10^{-7} mol/L$ .

Explanation:

Partial pressure of the $O_2$ gas = 685 torr = 0.8905 bar

1 torr = 0.0013 bar

According Henry's law:

$p_{o_2}=K_H\times\chi_{O_2}$

Value of Henry's constant of oxygen gas at 20 °C in water = 34860 bar

$0.0013=34860 bar\times \chi_{O_2}$

$\chi_{O_2}=\frac{0.8905 bar}{34680 bar}=2.56\times 10^{-5}$

Let the number of moles of $O_2$ gas in 1 liter water be n.

1 Liter water = 1000 g of water

Moles of water in 1 L $n_w=\frac{1000 g}{18 g/mol}=55.55 mol$

$\chi_{O_2}=\frac{n}{n+n_w}$

$2.56\times 10^{-5}=\frac{n}{n+55.55}$

$n=1.43\times 10^{-7} moles$

$Molarity=\frac{\text{Moles of}O_2}{Volume}$

Molar concentration of oxygen gas in 1 L of water:

$=\frac{1.43\times 10^{-7} moles}{1 L}=1.43\times 10^{-7} mol/L$

The molar concentration of oxygen gas in water is $1.43\times 10^{-7} mol/L$ .

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3 years ago

What is the mass in grams of 1.204x10^24 molecules of hydrogen gas (H2) ?

Katen [24]

Answer:

68 g/mol. 34 g/mol.

Explanation:

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2 years ago

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How might two lithospheric plates interact with each other?

Natalija [7]

Answer:

Ur answer is B. I, II, and III

Explanation:

Tectonic plates can interact with each other.

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2 years ago

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The amount of radioactive carbon-14 in a sample is measured using a Geiger counter, which records each disintegration of an atom

sergij07 [2.7K]

Answer:

4121 years

Explanation:

From;

0.693/t1/2 = 2.303/t log No/N

t1/2= half life of the carbon-14

No= count rate of the living tissue

N= count rate of the sample

t = age of the sample

0.693/5730 =2.303/t log (13.5/8.2)

1.21 * 10^-4 = 2.303/t * 0.2165

1.21 * 10^-4 = 0.4986/t

t = 0.4986/1.21 * 10^-4

t = 4121 years

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3 years ago