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Salsk061 [2.6K]
3 years ago
10

PLEASE HELP BEFORE 7 A.M. PACIFIC TIME SEE ATTACHED.

Chemistry
2 answers:
pickupchik [31]3 years ago
7 0

Answer:

(i) 32.9 mL; (ii)  37.5 mL; (iii) 21.73 mL

Explanation:

You should always try to read a measuring instrument to a tenth of the smallest scale division.

Here, you are measuring liquids, so you take the scale reading from the bottom of the meniscus.

(i) Graduated cylinder

There are 10 divisions between 30 mL and 40 mL, so each division represents 1 mL.

The level of the liquid appears to be between 32 mL and 33 mL. It is much closer to 33 mL (perhaps right on 33 mL).

You should report the volume to the nearest 0.1 mL. I would read the volume as 32.9 mL, but 32.8 and 33.0 are also acceptable.

Note: If you think the level is right on the 33 mark, you report the volume as 33.0 mL (NOT 33 mL).

(ii) Thermometer

The reading is about half-way between 87 ° and 88 °.

I would report the temperature as 87.5 °, but 87.4 ° and 87.6 ° would also be acceptable.

(iii) Buret

There are 10 divisions between 21 mL and 22 mL, so each division represents 0.1 mL.

You should estimate to the nearest 0.01 mL.

The liquid level is about a third of the way from 21.7 mL to 21.8 mL.

I would report the volume as 21.73 mL, but 21.72 mL and 21.74 mL are also acceptable.

 

anastassius [24]3 years ago
5 0
For i: 33mL
For ii: 87-88mL
For iii:22.3mL
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3 years ago
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The answer is:
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b. 0.210 M
c. 0.336 M

Molarity is a measure of the concentration of solute in a solution.
It can be expressed as moles of solute ÷ volume of solution:
c = n ÷V
where:
c - concentration of solute,
n - moles of solute
V - volume of solution

n can be expressed as:
<span>n = m ÷ Mr
</span>where:
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</span>m - mass of solute
Mr - relative molecular mass

a. We know volume:
V = 289.2 mL = 0.2892 L
We need n and c.

n = m ÷ Mr
m = 15.4 g
Mr (<span>KCl) = 74.55 g/mol
n = </span>15.4 g ÷ <span>74.55 g/mol
n = 0.206 mol</span>

Thus, 
c = 0.206 mol ÷ <span>0.2892 L
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</span>b. We know volume:
V = 0.614 L
We need n and c.

n = m ÷ Mr
m = 14.4 g
Mr (CaCl₂<span>) = 110.98 g/mol
n = </span>14.4 g ÷ <span>110.98 g/mol
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Thus, 
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c = 0.210 mol/L = 0.210 M
</span>
c. We can use formula:
m₁V₁ = m₂V₂
<span>m₁ = 3 M
</span><span>V₁ = 28 mL= 0.028 L
</span><span>m₂ = ?
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</span>Thus:
3 M × 0.028 L = m₂× <span>0.250 L
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4 0
3 years ago
how do i caculate the trmperature change that 725 grams of aluminum will undergo when 2.35x10^4 Joules of thermal energy are add
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Answer:

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To solve this problem, we simply use the expression below:

  The quantity of energy is given as:

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Δt is the change in temperature

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Answer:

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