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Scrat [10]
2 years ago
6

What does internal combustion engines release?

Chemistry
2 answers:
In-s [12.5K]2 years ago
6 0
Internal combustion engines release exhaust
Pavlova-9 [17]2 years ago
4 0

Answer:

you should give options.

Explanation:

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List the first 20 elements chemistry
timama [110]

Answer:

h- Hydrogen

He - Helium

Li - Lithium

Be - Beryllium

B - Boron

C - Carbon

N - Nitrogen

O - Oxygen

Explanation:

Happy to help:)

8 0
3 years ago
In a heating curve, when is the temperature constant? need answers!
Varvara68 [4.7K]
During a phase change
3 0
3 years ago
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A solution of hydrochloric acid of unknown concentration was titrated with 0.10 M NaOH. If a 100.-mL sample of the HCl solution
madam [21]

<u>Answer:</u> The initial pH of the HCl solution is 3

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=?M\\V_1=100mL\\n_2=1\\M_2=0.10M\\V_2=1mL

Putting values in above equation, we get:

1\times M_1\times 100=1\times 0.10\times 1\\\\M_1=\frac{1\times 0.10\times 1}{1\times 100}=10M

1 mole of HCl produces 1 mole of H^+ ions and 1 mole of Cl^- ions

To calculate the pH of the solution, we use the equation:

pH=-\log[H^+]

We are given:

[H^+]=0.001M

Putting values in above equation, we get:

pH=-\log (0.001)\\\\pH=3

Hence, the initial pH of the HCl solution is 3

6 0
3 years ago
What is the molar mass of Ca(NO3)2?
Valentin [98]
The answer to the molar mass of Ca(NO3)2 is gonna be C. 164.1 g/mol



3 0
3 years ago
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Consider the reaction between iron (III) oxide, Fe2O3 and carbon monoxide, CO.
Lana71 [14]
<span>Fe2O3 + 3CO --> 2Fe + 3CO2
</span><span>
m(Fe2O3)=213 g
m(CO)=140 g
</span>_______________

<span>n(Fe2O3)=?
m(Fe)=?
n(Fe2O3)=?
n(CO)=?
n(CO2)=?
</span>
<span>n(Fe2O3)=m(Fe2O3) / M(Fe2O3)
n(Fe2O3)= 213 g / 159,7 gmol-1 = 1,33 mol
</span>
<span>n(CO)= m(CO) / M(CO) n(CO)= 140 g / 28,01 gmol-1 = 4,99 mol</span>
3 0
3 years ago
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