2H2 + O2 → 2H2O

Chemistry

1 answer:

Rudiy273 years ago

6 0

Answer: 88g

Explanation:

11 moles H2 = 11 mole H2O = 5.5 moles O2 = 16*5.5 g

You might be interested in

Can someone help me

harina [27]

11 LIMITING react this is the answer

7 0

3 years ago

Less energy was entering Earth's

ololo11 [35]

Answer:

The temperatures on Earth increase

Explanation:

more energy results in more heat.

6 0

3 years ago

When a writer uses “common knowledge” in an essay, why does it NOT require a citation nor is it considered plagiarism?

krek1111 [17]

Plagiarism is when you just copy exactly what someone else says, without their permission, or without citing where you got it from.
Common knowledge should be your own words. unless you take it from somewhere else, it doesnt need anything.

6 0

3 years ago

Read 2 more answers

Calculate the [H+] and pH of a 0.0040 M hydrazoic acid solution. Keep in mind that the Ka of hydrazoic acid is 2.20×10−5. Use th

Vera_Pavlovna [14]

Answer:

$[H^+]=0.000285$

$pH=3.55$

Explanation:

In this, we can with the ionization equation for the hydrazoic acid ( $HN_3$ ). So:

$HN_3~~H^+~+~N_3^-$

Now, due to the Ka constant value, we have to use the whole equilibrium because this is not a strong acid. So, we have to write the Ka expression:

$Ka=\frac{[H^+][N_3^-]}{[HN_3]}$

For each mol of $H^+$ produced we will have 1 mol of $N_3^-$ . So, we can use "X" for the unknown values and replace in the Ka equation:

$Ka=\frac{X*X}{[HN_3]}$

Additionally, we have to keep in mind that $HN_3$ is a reagent, this means that we will be consumed. We dont know how much acid would be consumed but we can express a subtraction from the initial value, so:

$Ka=\frac{X*X}{0.004-X}$