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Andrew [12]
3 years ago
9

The majority of the mass of the atom is located in the _______. A) nucleus B) core electrons C) pions and quarks D) valence elec

trons
Chemistry
2 answers:
NeX [460]3 years ago
5 0

The majority of the mass of the atom is located in the nucleus. Remember that the nucleus contains both protons and neutrons and therefore, most of the mass of the atom.

Helen [10]3 years ago
3 0

Answer:

A

Explanation:

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A liquid was analyzed to be 54.5% c, 9.10% h, and 36.4% o. an empty flask, whose mass was 45.32 g, when filled with the vapor of
nataly862011 [7]
<span>number Moles of C = 54.5 g / 12.011 = 4.54 number Moles of H = 9.10 / 1.008 = 9.02 number Moles of O = 36.4 / 16 = 2.28 if we want to divide by the smallest number 4.54 / 2.28 = 2 => C 9.02 / 2.28 = 4 => H 2.28 / 2.28 = 1 => O Empirical formula will be = C2H4O</span>
7 0
4 years ago
In the equation below
yulyashka [42]

Answer:

6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.

Explanation:

We are given the chemical equation:

\displaystyle 2\text{NH$_3$}_\text{(g)} + 3\text{CuO}_\text{(s)} \longrightarrow \text{N$_2$}_\text{(g)}  + 3\text{Cu}_\text{(s)}+3\text{H$_2$O}_\text{(g)}

And we want to determine the amount of products produced when 12.5 moles of NH₃ is reacted with excess CuO.

Compute using stoichiometry. From the equation, we can see the following stoichiometric ratios:

  • The ratio between NH₃ and N₂ is 2:1. (i.e. One mole of N₂ is produced from every two moles of NH₃.)
  • The ratio between NH₃ and Cu is 2:3.
  • The ratio between NH₃ and H₂O is 2:3. (i.e. Three moles of H₂O or Cu is produced frome every two moles of NH₃.)

Dimensional Analysis:

  • The amount of N₂ produced:

\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{1\text{ mol N$_2$}}{2\text{ mol NH$_3$}} = 6.25\text{ mol N$_2$}

  • The amount of Cu produced:

\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol Cu}}{2\text{ mol NH$_3$}} = 18.8\text{ mol Cu}

  • And the amount of H₂O produced:

\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol H$_2$O}}{2\text{ mol NH$_3$}} = 18.8\text{ mol H$_2$O}

In conclusion, 6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.

8 0
2 years ago
Hydrogen and oxygen react chemically to form water how much water would form if 14.8grams of hydrogen reacted with 34.8 grams of
pishuonlain [190]

Answer:

There will be formed 39.1935 grams H2O formed

Explanation:

<u>Step 1:</u> The balanced equation

2H2 + 02 → 2H20

<u>Step 2</u>: Given data

mass of hydrogen = 14.8 grams

Molar mass of hydrogen = 2.02 g/mole

mass of oxygen = 34.8 grams

Molar mass of oxygen = 32 g/mole

<u>Step 3: </u>Calculate moles

moles = mass / Molar mass

moles of hydrogen = 14.8g/ 2.02 g/mole = 7.33 moles

moles of oxygen = 34.8g / 32g/mole = 1.0875 moles

For 2 moles hydrogen consumed, we need 1 mole of oxygen.

This means oxygen is the limiting reagens and will be consumed completely. Hydrogen is the reactant in excess, there will remain 5.155 moles of hydrogen

<u>Step 4:</u> Calculate moles of H2O

We see that for 2 moles of H2 consumed, there is needed 1 mole of O2, to produce 2 moles of H2O.

For 1.0875 moles of oxygen consumed, there will be produced 2.175 moles of H2O

<u>Step 5:</u> Calculate mass of water

Mass of H2O = moles of H2O * Molar mass of H2O

Mass of H2O = 2.175 moles * 18.02 g/moles 39.1935 grams

There will be formed 39.1935 grams H2O formed

4 0
4 years ago
Wich of these equations are balanced? a) H2SO4+2AL---&gt;AL2(SO4) b)2KCL+Pb(NO3)2---&gt;2KNO3+PbCL2
Zolol [24]

The answer is b since on a it says H2 but on the right side there is no H idk if you forgot to put H there so im guessing b. Have a good day

4 0
3 years ago
As a scuba diver descends under water, the pressure increases. At a total air pressure of 2.71 atm and a temperature of 25.0 C,
Semenov [28]

Answer:

1.32\times 10^{-3} mol/Lis the solubility of nitrogen gas in a diver's blood.

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{N_2}=K_H\times p_{liquid}

where,

K_H = Henry's constant = 6.26 \times 10^{-4}mol/L.atm

p_{N_2} = partial pressure of nitrogen

p_{N_2}= P\times \chi_{N_2} (Raoult's law)

=2.71 atm\times 0.78=2.1138 atm

C_{N_2}=6.26 \times 10^{-4}mol/L.atm\times 2.1138 atm

C_{N_2}=1.32\times 10^{-3} mol/L

1.32\times 10^{-3} mol/Lis the solubility of nitrogen gas in a diver's blood.

3 0
4 years ago
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