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frosja888 [35]
2 years ago
6

1. A student attempts to make a saline solution by adding salt to water. How much water did

Chemistry
1 answer:
dexar [7]2 years ago
8 0

111.1 mL of water

Explanation:

Weight per volume concentration (w/v %) is defined as

weight per volume concentration = (mass of solute (g) / volume of solution (mL)) × 100

volume of solution = (mass of solute × 100) / weight per volume concentration

volume of solution = (1 × 100) / 0.9 = 111.1 mL

volume of water = volume of solution = 111.1 mL

Learn more about:

weight per volume concentration

brainly.com/question/12721794

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vladimir2022 [97]
To fill the other element's shell.
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3 years ago
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Current passes through a solution of sodium chloride. In 1.00 second, 2.68×1016Na+ ions arrive at the negative electrode and 3.9
EleoNora [17]

Answer : The current passing between the electrodes is, 1.056\times 10^{-2}A

Explanation :

First we have to calculate the charge of sodium ion.

q=ne

where,

q = charge of sodium ion

n = number of sodium ion = 2.68\times 10^{16}

e = charge on electron = 1.6\times 10^{-19}C

Now put all the given values in the above formula, we get:

q=(2.68\times 10^{16})\times (1.6\times 10^{-19}C)=4.288\times 10^{-3}C

Now we have to calculate the charge of chlorine ion.

q'=ne

where,

q' = charge of chlorine ion

n = number of chlorine ion = 3.92\times 10^{16}

e = charge on electron = 1.6\times 10^{-19}C

Now put all the given values in the above formula, we get:

q'=(3.92\times 10^{16})\times (1.6\times 10^{-19}C)=6.272\times 10^{-3}C

Now we have to calculate the current passing between the electrodes.

I=\frac{q}{t}+\frac{q'}{t}

I=\frac{4.288\times 10^{-3}}{1.00}+\frac{6.272\times 10^{-3}}{1.00}

I=1.056\times 10^{-2}A

Thus, the current passing between the electrodes is, 1.056\times 10^{-2}A

4 0
3 years ago
A standard backpack is approximately 30cm x 30cm x 40cm. Suppose you find a hoard of pure gold (density = 19.3 g/cm3) while trea
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Answer:

30

Explanation:

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6 0
3 years ago
A solution has a pH of 2.5. Answer the following questions.
olga_2 [115]
PH + pOH = 14  ⇒  pOH = 14 - pH
pOH = 14 - 2.5
pOH = 11.5

[H⁺] = 10^(-pH) = 10^(-2.5)
[H⁺] = 0.003 M
[OH⁻] = 10^(-pOH) = 10^(-11.5) = 3 × 10⁻¹² M
[OH⁻] = 3 × 10⁻¹² M


pH = 2.5 implies one significant digit
6 0
2 years ago
How many calories is required to change the temperature of 2.18g of water from 15.3°C to 69.5°C. The specific heat of liquid wat
lozanna [386]

The number  of calories that are  required  to change the temperature  of 2.18 g of water from 15.3 c to 69.5 c is  <u>118.16 cal</u>


    <u><em> calculation</em></u>

  •    Heat in calories  = MCΔ T where,
  • M(mass)= 2.18 g
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 heat is therefore= 2.18 g x 1.00 cal/g/c  x 54.2 c=118.16  cal

7 0
3 years ago
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