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sdas [7]
3 years ago
14

What will happen to the salt when mixed to liquid?​

Chemistry
2 answers:
Pepsi [2]3 years ago
8 0

Answer:

the salt dissolves in water because of covalent bonds.

Explanation:

mark me brainliest.

Anestetic [448]3 years ago
8 0
Answer,
When salt is mixed with water, the salt dissolves because the covalent bonds of water are stronger than the ionic bonds in the salt molecules. ... Once this happens, the salt is dissolved, resulting in a homogeneous solution.
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A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce
spin [16.1K]

Explanation:

The given data is as follows.

       Volume of lake = 15 \times 10^{6} m^{3} = 15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}

        Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = 5.6 \times 15 \times 10^{9} mg

                                                                    = 84 \times 10^{9} mg

                                                                    = 84 \times 10^{3} kg

Flow rate of river is 50 m^{3} sec^{-1}

Volume of water in 1 day = 50 \times 10^{3} \times 86400 liter

                                          = 432 \times 10^{7} liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are 2.9792 \times 10^{10} mg or 2.9792 \times 10^{4} kg

Flow rate of sewage = 0.7 m^{3} sec^{-1}

Volume of sewage water in 1 day = 6048 \times 10^{4} liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = 1.8144 \times 10^{10} mg or 1.8144 \times 10^{4}kg

Therefore, total concentration of lake after 1 day = \frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l

                                        = 6.8078 mg/l

                 k_{D} = 0.2 per day

       L_{o} = 6.8078

Hence, L_{liquid} = L_{o}(1 - e^{-k_{D}t}

             L_{liquid} = 6.8078 (1 - e^{-0.2 \times 1})  

                             = 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

                                                             = 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

5 0
4 years ago
___________________is the rate that an object moves in a certain direction. Question 3 options:
Nesterboy [21]

Answer: Velocity

Explanation:

i just took the quiz for k12

3 0
3 years ago
What does density have to do with the cycling of matter in earths interior
Anastaziya [24]

Answer:

The rocks of higher density inter the mantle. Due to density they sink and melt. When they melt the density decreases and rises to the top. They then are recycled through volcanoes.  

Explanation:

8 0
3 years ago
1. Based on the observed performance of the air bag models and the amounts of sodium bicarbonate and acetic acid (vinegar) neede
Ira Lisetskai [31]

Sodium bicarbonate and acetic acid are not good substitute for sodium azide in airbags since the require more mass and produce less gas.

<h3>Which is the better chemical for an airbag?</h3>

The chemical equation for the production of nitrogen gas from sodium azide is given below:

  • NaN₃ → 2 Na + 3 N₂

1 mole or 66 go of sodium azide produces 3 moles or 67.2 L of nitrogen gas.

The equation for the production of carbon dioxide from sodium bicarbonate and acetic acid is given below:

  • Na₂CO₃ + CH₃COOH → CH₃COONa + CO₂ + H₂O

1 mole, 106 g of Na₂CO₃ and 1 mole, 82 g of CH₃COOH are required to produce 1 mole or 22.4 L of CO₂.

The mass of sodium azide required is less than that of sodium bicarbonate and acetic acid required. Also, sodium azide produces a greater volume of gas. Therefore, sodium bicarbonate and acetic acid are not good substitute for sodium azide in airbags.

In conclusion, sodium azide is a better choice in airbags.

Learn more about airbags at: brainly.com/question/14954949

#SPJ1

7 0
2 years ago
1. Which of the following nuclear equations is correctly balanced? (choices attached)
lawyer [7]

1. <em>Balancing nuclear equations </em>

Answer:

_{18}^{37}\text{Ar} + _{-1}^{0}\text{e} \rightarrow _{17}^{37}\text{Cl}

Explanation:

The main point to remember in balancing nuclear equations is that the <em>sums of the superscripts</em> (the mass numbers) and the <em>subscripts</em> (the nuclear charges) <em>must balance</em>.

Mass numbers: 37 + 0 = 37; balanced.

Charges: 18 + 1 = 17; balanced

B is <em>wrong</em>. Mass numbers not balanced. 6 +2(1) ≠ 4 + 3.

C is <em>wron</em>g. Mass numbers not balanced. 254 + 4 ≠ 258 + 2(1).

D is <em>wron</em>g. Mass numbers not balanced. 14 + 4 ≠ 17 + 2.

===============

2. <em>Amount remaining </em>

Answer:

D. 5.25 g

Explanation:

The half-life of Th-234 (24 da) is the time it takes for half the Th to decay.  

After one half-life, half (50 %) of the original amount will remain.  

After a second half-life, half of that amount (25 %) will remain, and so on.

We can construct a table as follows:  

  No. of                  Fraction       Amount  

<u>half-lives</u>   <u>t/(da</u>)  <u>remaining</u>  <u>remaining/g</u>  

      1             24            ½                21.0  

      2            48            ¼                10.5  

      3             72           ⅛                 5.25  

      4             96          ¹/₁₆                 2.62  

We see that 72 da is three half-lives, and the amount of Th-234 remaining is 5.25 g.  

===============

3.<em> Calculating the half-life </em>

Answer:

a. 2.6 min

Explanation:

The fraction of the original mass remaining is 1.0 g/4.0 g ≈ ¼.

We saw from the previous table that it takes two half-lives to decay to ¼ of the original amount.

2 half-lives = 5.2 min       Divide both sides by 2

  1 half-life = 5.2 min/2 = 2.6 min

7 0
3 years ago
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