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Sloan [31]
2 years ago
14

How much mass of sodium chloride ( ) should be dissolved in water to make 1.5 L of 0.75 M aqueous solution? The molar mass of is

58.5 g.
Chemistry
1 answer:
iogann1982 [59]2 years ago
4 0

Answer:

Mass = 65.8 g

Explanation:

Given data:

Mass of sodium chloride = ?

Volume of solution = 1.5 L

Molarity of solution = 0.75 M

Solution:

Number of moles of sodium chloride:

Molarity = number of moles / volume in L

By putting values,

0.75 M = number of moles = 1.5 L

Number of moles = 0.75 M × 1.5 L

Number of moles = 1.125 mol

Mass of sodium chloride:

Mass = number of moles × molar mass

Mass = 1.125 mol × 58.5 g/mol

Mass = 65.8 g

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bogdanovich [222]

Answer:

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Explanation:

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6 0
3 years ago
One of the compounds used to increase the octane rating of gasoline is toluene (pictured). Suppose 43.3 mL of toluene (d = 0.867
Thepotemich [5.8K]

<u>Answer:</u>

(A)

Density = Mass / Volume

So  

Mass = Density × Volume

= 0.867 g/mL \times 43.3mL = 37.5411 g Toluene

1C_6 H_5 CH_3  + 9 O_2  > 7 CO_2  + 4 H_2 O

Mole ratio of toluene : Oxygen is 1 : 9

$37.5411 g \text { Toluene } \times \frac{1 \text {mol} \text {toluene}}{92 g \text { toluene}} \times \frac{9 {mol} O_{2}}{1 \text {mol} \text { toluene }} \times \frac{32 g O_{2}}{1 {mol} O_{2}}=117 g O_{2}(\text {Answer})$

(B)

1 mole of Toluene produces 7 moles of CO_2 gas and 4 moles of H_2 O Vapour

So the mole ratio is 1 : 11

37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene }} \times \frac{11 \mathrm{mol} \text { gas }}{1 \text { mol toluene }} $$\\\\=4.49 \text { mol gaseous products (Answer) } $

(C)

1mole contains 6.022\times10^{23} molecules

37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene}} \times \frac{4 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \text { toluene }} \times \frac{6.022 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}} $\\\\$=9.82 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O} \text { (Answer) } $

6 0
3 years ago
Write a balanced nuclear equation for the formation of 28 Si 14 from beta-minus emission.
Fantom [35]

Answer:

Phosphorus-28 undergoes beta-minus decay to produce

  • an electron,
  • a Silicon-28 nuclei, and
  • an electron antineutrino.

\rm ^{28}_{13} Al \to ^{28}_{14} Si + ^{\phantom{-}0}_{-1}e + \bar{\mathnormal{v}}_{\rm e}

Explanation:

In simple words, when a nucleus undergoes beta-minus decay, a neutron is converted to a proton. An electron and an electron antineutrino will be released.

\rm ^{1}_{0}n^{0} \to ^{1}_{1}p^{+} + ^{\phantom{-}0}_{-1}e^{-} + \bar{\mathnormal{v}}_{\rm e}.

One way to tell whether a neutron is converted to a proton, but not vice versa, is to check the sum charges on the two sides of this equation.

  • Left-hand side: 0. Neutron is neutral.
  • Right-hand side: 1 + (-1) = 0. Each proton carries a charge of +1. Each electron (beta-minus particle) carries a charge of -1. Antineutrinos are neutral.

The charges on the two sides of this equation is the same. Hence this nuclear equation is possible (but not necessarily correct; however, if the proton and the neutron are in the wrong place the charge won't even be the same.)

Since the mass number of a proton and a neutron are both 1, the overall mass number of the atom will stay the same.

The atomic number is the number of protons in each atom. That number determines the symbol and the chemical properties of the atom. When one neutron in an atom is converted to a proton, the atomic number of the atom will increase by 1.

The atomic number of the daughter nucleus, silicon, is 14. It takes a parent nucleus with atomic number 14 - 1 = 13 to produce a silicon atom. Refer to a modern periodic table. Atomic number 13 corresponds to the element aluminum.

Also, the mass number of the daughter nucleus is 28. Since the mass number would stay the same in a beta decay, the mass number of the parent nucleus would also be 28. In other words, it takes an aluminum-28 atom to undergo beta-decay to produce a silicon-28 atom.

Complete the other details (electron and electron antineutrino) to obtain the equation

\rm ^{28}_{13} Al \to ^{28}_{14} Si + ^{\phantom{-}0}_{-1}e + \bar{\mathnormal{v}}_{\rm e}.

6 0
3 years ago
Which place in the world has the highest average annual precipitation?
alekssr [168]

Answer: Colombia

Explanation:

6 0
3 years ago
Read 2 more answers
What is the minimum volume of 5.50 M HCl necessary to neutralize completely the hydroxide in 718.0 mL of 0.183 M NaOH
Deffense [45]

The volume of HCl required is 23.89 mL

Calculation of volume:

The reaction:

HCl + NaOH \rightarrow NaCl + H_2O

As HCl and NaOH react in 1 : 1 ratio.

Volume of NaOH= 718 mL

Concentration= 0.183M

Volume of HCl= ?

Concentration= 5.50M

Using the dilution formula:

M_1\times V_1(NaOH)= M_2\times V_2(HCl)

0.183\times 718= 5.50 \times V_2\\V_2=\frac{131.394}{5.50} \\V_2 = 23.89 \,mL

Therefore,

Volume of HCl required will be 23.89 mL.

Learn more about neutralization reaction here,

brainly.com/question/1822651

#SPJ4

6 0
1 year ago
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