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2 years ago

Help pls you get 22 points if you answer me

Chemistry

1 answer:

snow_tiger [21]2 years ago

5 0

Answer:

the 2nd one

Explanation:

All other symbols are for Mathematics.

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Chocolate candy melting in your hand is what type of heat transfer

gregori [183]

The answer to that is: Conduction

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3 years ago

Cobalt-60 is a radioactive isotope used to treat cancers. A gamma ray emitted by this isotope has an energy of 1.33 MeV (million

Natalka [10]

Answer:

E = 1.33 MeV = 2.13 x $10^{-13}$ J

v = wavelength = E / h = 2.13 x $10^{-13}$ / 6.626 x $10^{-34}$ = 3.2 x $10^{20}$ m

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3 years ago

In a chemical reaction, the mass of the products ____.

Pani-rosa [81]

In a chemical reaction, the mass of the products is less than the mass of the reactants.
the first step in most stoichiometry problems is to convert given quantities to moles.

Hope this helps! If it did, pls marky me brainly thank, and 5star me

5 0

3 years ago

Read 2 more answers

HALP PLZ 50 points brainlest and thanks if u halp

Novay_Z [31]

I Would Say The Answer Is C.

8 0

4 years ago

Read 2 more answers

For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox

ololo11 [35]

Answer:

1) Maximun ammount of nitrogen gas: $m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent: $m_{N2}=4.274 g$

Explanation:

The reaction

$2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)$

Moles of nitrogen monoxide

Molecular weight: $M_{NO}=30 g/mol$

$n_{NO}=\frac{m_{NO}}{M_{NO}}$

$n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol$

Moles of hydrogen

Molecular weight: $M_{H2}=2 g/mol$

$n_{H2}=\frac{m_{H2}}{M_{H2}}$

$n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol$

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) Maximun ammount of nitrogen gas => when all NO reacted

$m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}$

$m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent:

$m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}$

$m_{N2}=4.274 g$

8 0

4 years ago