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BartSMP [9]
3 years ago
5

Is it true that Organisms need water, vitamins and minerals. But they do not need salt to function.

Chemistry
1 answer:
Veseljchak [2.6K]3 years ago
5 0
All organisms need water, vitamins, minerals, and salt together to balance things out.
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What is the mole fraction of NaOH in an aqueous solution that
astraxan [27]

Answer:

The mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882

Explanation:

We are given that

Aqueous solution that contains 22.9% NaOH by mass means

22.9 g NaOH in 100 g solution.

Mass of NaOH(WB)=22.9 g

Mass of water =100-22.9=77.1

Na=23

O=16

H=1.01

Molar mass of NaOH(MB)=23+16+1.01=40.01

Number of moles =\frac{Given\;mass}{Molar\;mass}

Using the formula

Number of moles of  NaOH(n_B)=\frac{W_B}{M_B}=\frac{22.9}{40.01}

n_B=0.572moles

Molar mass of water=16+2(1.01)=18.02g

Number of moles of water(n_A)=\frac{77.1}{18.02}

n_A=4.279 moles

Now, mole fraction of NaOH

=\frac{n_B}{n_B+n_A}

=\frac{4.279}{0.572+4.279}

=0.882

Hence, the mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882

4 0
3 years ago
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:

<u>The reaction </u>

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) <u>Maximun ammount of nitrogen gas</u> => 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) <u>Limiting reagent</u>: NO

3) <u>Ammount of excess reagent</u>:

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
3 years ago
Explain why the larger the hydrocarbon molecule is the more
RideAnS [48]

Answer:

Longer hydrocarbon molecules have a stronger intermolecular force. More energy is needed to move them apart so they have higher boiling points . This makes them less volatile and therefore less flammable

6 0
2 years ago
Read 2 more answers
For the reaction 2 s (s) + 3 o2 (g) → 2 so3 (g), how much so3 can be produced from 4 g o2 and excess s? 1. 0.25 mol so3 2. 0.13
igor_vitrenko [27]
The balanced equation for the above reaction is as follows;
2S + 3O₂ --> 2SO₃
Stoichiometry of O₂ to SO₃ is 3:2
O₂ is the limiting reactant and S is provided in excess. since O₂ is the limiting reactant, the whole amount is consumed in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of O₂ moles reacted- 4 g / 32 g/mol = 0.125 mol
3 mol of O₂ forms 2 mol of SO₃
therefore when 0.125 mol of O₂ reacts number of SO₃ moles - 2/3 x 0.125 mol
Number of SO₃ moles formed - 0.0833 mol
Answer is 4) 0.08 mol
8 0
3 years ago
1. Use the following equation:
fenix001 [56]

Answer:

7.28 mol

Explanation:

2 NaOH +  H₂SO₄ =  2 H₂O +  Na₂SO₄ -------------------(1)

mole fraction for the reaction is;

2 : 1 = 2 : 1

Number of moles of H₂SO₄ = 7.28 mol

1 mol of H₂SO₄ shall form 1 mole of Na₂SO₄

therefore,

7.28 mol of H₂SO₄ shall form 7.28 mole of Na₂SO₄

8 0
3 years ago
Read 2 more answers
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