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Sedbober [7]
3 years ago
11

3. What amount of

Chemistry
1 answer:
Anton [14]3 years ago
7 0

Hey there!:

Density = 30 mg/mL

Mass = 150 mg

Volume = ??

Therefore:

D = m / V

30 = 150 / V

V = 150 / 30

V = 5.0 mL

converts volume into liters :

5.0 mL / 1000 => 0.005 L or 5.0 x 10⁻³ L

Hope this helps!

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How much energy is needed to completely boil a 5.05g sample of water?
Keith_Richards [23]

Given what we know, we can confirm that the amount of heat energy that would be required in order to boil 5.05g of water is that of 11.4kJ of heat.

<h3>Why does it take this much energy to boil the water?</h3>

We arrive at this number by taking into account the energy needed to boil 1g of water to its vaporization point. This results in the use of 2260 J of heat energy. We then take this number and multiply it by the total grams of water being heated, in this case, 5.05g, which gives us our answer of 11.4 kJ of energy required.

Therefore, we can confirm that the amount of heat energy that would be required in order to boil 5.05g of water is that of 11.4kJ of heat.

To learn more about the behavior of water visit:

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8 0
2 years ago
Copy and balance these chemical equstions​
lyudmila [28]

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4 0
3 years ago
Read 2 more answers
How many of grams of CuSO4 are in 475ml of a 2.0M aqueous solution
Anettt [7]

Answer:

151.63 g

Explanation:

We first get the number of moles;

Moles = Molarity × volume

          = 2.0 × 0.475

          = 0.95 moles

1 mole of CuSO4 = 159.609 g/mol

Therefore;

Mass = moles × molar mas

         = 0.95 moles × 159.609 g/mol

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8 0
4 years ago
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A solution of diamminesilver(I) chloride is treated with dilute nitric acid. What products you get? [Ag(NH3)2]Cl(aq) + HNO3(aq)
Anit [1.1K]
The reaction is a double displacement one which means the cation of one reactant is substituted  to the cation of the other reactant to identify the products. Hence in this reaction, the products are silver ammonium nitrate (Ag(NH3)2NO3) and hydrochloric acid (HCl).
8 0
3 years ago
What is the concentration of magnesium bromide, in ppm, if 133.4 g MgBr2 dissolved in 1.84 L water. Then solve for the bromine c
Serhud [2]

Answer: 0.0725ppm

Explanation:

133.4g of MgBr2 dissolves in 1.84L of water.

Therefore Xg of MgBr2 will dissolve in 1L of water. i.e

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The concentration of MgBr2 is 72.5g/L = 0.0725mg/L

Recall,

1mg/L = 1ppm

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7 0
3 years ago
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