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aliya0001 [1]
3 years ago
7

A sample of a substance is found to have a mass of 223.5 grams. A graduated cylinder contains 45.6 ml of water. After a metal cy

linder was added to the cylinder, the water level reads 68.3 ml. What is the density of the metal cylinder?
Chemistry
1 answer:
Paraphin [41]3 years ago
5 0

Answer:

We have the final answer as

<h2>9.85 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass of substance = 223.5 g

volume = final volume of water - initial volume of water

volume = 68.3 - 45.6 = 22.7

volume = 22.7 mL

It's density is

density =  \frac{223.5}{22.7}  \\  = 9.84581497797...

We have the final answer as

<h3>9.85 g/mL</h3>

Hope this helps you

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Which of the following is not the same as 0.00684 kg? 0.0684 dkg,6.84 g, 684 cg, 6,840 mg,
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The Answer would be C.) 684 cg


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4 years ago
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How much energy is required to raise the temperature of 3 kg of iron from 20° C to 25°C? Use the table below and this equation:
ElenaW [278]

The energy required to raise the temperature of 3 kg of iron from 20° C to 25°C is 6,750 J( Option B)

<u>Explanation:</u>

Given:

Specific Heat capacity of Iron= 0.450 J/ g °C

To Find:

Required Energy to raise the Temperature

Formula:

Amount of energy required is given by the formula,

Q = mC (ΔT)

Solution:

M = mass of the iron in g

So 3 kg = 3000 g

C = specific heat of iron = 0.450 J/ g °C [ from the given table]

ΔT = change in temperature = 25° C - 20°C = 5°C

Plugin the values, we will get,

Q = 3000 g ×  0.450 J/ g °C ×  5°C

 = 6,750 J

So the energy required is 6,750 J.

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4 years ago
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Where do the sun rays hit Earth's surface nearly at right angles ​
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The Sun Ray's hit earths surface at Earths Equator

8 0
3 years ago
). The molar mass of an organic acid, a compound composed of carbon, hydrogen, and oxygen, is 194.14 g/mol. Combustion of a 1.50
Nikitich [7]

Answer:

The empirical formula is C₆H₁₀O₇.

Step-by-step explanation:

1. Calculate the masses of C, H, and O from the masses given.

Mass of C =  2.0402 g CO₂ × (12.01 g C/44.01 g CO₂) = 0.5568  g C

Mass of H = 0.6955 g H₂O  × (2.016 H/18.02 g H₂O)  = 0.077 81 g H

Mass of O = Mass of compound - Mass of C - Mass of H = (1.500 – 0.5568 – 0.077 81) g = 0.8654 g O

=====

2. Convert these masses to moles.

Moles  C = 0.5568  × 1/12.01  = 0.046 36

Moles H = 0.077 81 × 1/1.008 = 0.077 19

Moles O = 0.8654   × 1/16.00 = 0.054 09

=====

3. Find the molar ratios.

Moles  C = 0.046 36/0.046 36 = 1

Moles H = 0.077 19/0.046 36   = 1.665

Moles O = 0.054 09/0.046 36 = 1.167

======

4. Multiply the ratios by a number to make them close to integers

C  = 1        × 6 = 6

H = 1.665 × 6 = 9.991

O = 1.167 × 6  = 7.001

=====

5. Round the ratios to integers

C:H:O =6:10:7

=====

6. Write the empirical formula

The empirical formula is C₆H₁₀O₇.

=======

7. Calculate the empirical formula mass

C₆H₁₀O₇ = 6×12.01 + 10×1.008 + 7×16.00

C₆H₁₀O₇ = 72.01 + 10.08+ 112.0

C₆H₁₀O₇ = 194.09

=====

8. Divide the molecular mass by the empirical formula mass.  

MM/EFM = 194.14/194.09 = 1.000 ≈ 1

=====

9. Determine the molecular formula

MF = (EF)ₙ = (C₆H₁₀O₇)₁ = C₆H₁₀O₇

7 0
3 years ago
How many molecules of CBr4 are in 250 grams of CBr4
Kazeer [188]

Answer:- 4.54*10^2^3 molecules.

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It is a two step unit conversion problem. In the first step, grams are converted to moles on dividing the grams by molar mass.

In second step, the moles are converted to molecules on multiplying by Avogadro number.

Molar mass of CBr_4  = 12+4(79.9)  = 331.6 g per mol

let's make the set up using dimensional analysis:

250g(\frac{1mol}{331.6g})(\frac{6.022*10^2^3molecules}{1mol})

= 4.54*10^2^3 molecules

So, there will be 4.54*10^2^3 molecules in 250 grams of CBr_4 .


8 0
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