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Mariulka [41]
3 years ago
14

What is the volume of an object with a density of 2.6 g/cm³ and a mass of 30.5g?

Chemistry
1 answer:
inessss [21]3 years ago
5 0

Answer:

The answer is

<h2>11.73 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density}

From the question

mass of object = 30.5 g

Density = 2.6 g/cm³

The volume is

volume =  \frac{30.5}{2.6}  \\  = 11.7307692...

We have the final answer as

<h3>11.73 mL</h3>

Hope this helps you

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5. Which of these elements has the greatest atomic radius? *<br> Be<br> Mg<br> Ra<br> Ba
trapecia [35]
The answer is Ra
Atomic number for Be is 4

Atomic number for Mg 12

Atomic number for Ra 88

Atomic number for Ba 56
5 0
3 years ago
the chloride of a metal M contains 47.25% of metal 1.0 gram of metal would be displaced from a compound by 0.88 gram of another
GenaCL600 [577]

Answer:

The equivalent weight of M is approximately 31.8 g

The equivalent weight of N is approximately 27.98 g

Explanation:

The given parameters are;

The percentage of the the metal M in in the chloride = 47.25%

Where by the chemical formula for the metal chloride is MClₓ, we have;

47.25% of the mass of MClₓ = Mass of M = W

Therefore, we have;

\dfrac{0.4725}{W} = \dfrac{1}{W + 35.5 \cdot x}

0.4725 × (W +  35.5·x) = W

0.4725·W + 0.4725×35.5×x = W

W - 0.4725·W  = 16.77·x

0.5275·W = 16.77·x

W/x = 16.77/0.5275 = 31.799 = The equivalent weight of M

The equivalent weight of M = 31.799 ≈ 31.8 g

Given that 1 gram of M is displaced by 0.88 gram of N, then the equivalent weight of N that will displace 31.799 = 0.88 × 31.799 ≈ 27.98 g

The equivalent weight of N = 27.98 g.

7 0
2 years ago
What is more electronegative? <br> O or Br
cluponka [151]

Answer:

O

Explanation:

O = 3.44

Br = 2.96

8 0
2 years ago
If 0.255 moles of AgNO₃ react with 0.155 moles of H₂SO₄ according to this UNBALANCED equation below, how many grams of Ag₂SO₄ co
Ber [7]

Answer: 6.162g of Ag2SO4 could be formed

Explanation:

Given;

0.255 moles of AgNO3

0.155 moles of H2SO4

Balanced equation will be given as;

2AgNO3(aq) + H2SO4(aq) -> Ag2SO4(s) + 2HNO3(aq)

Seeing that 2 moles of AgNO3 is required to react with 1 moles of H2SO4 to produce 1 mole of Ag2SO4,

Therefore the number of moles of Ag2SO4 produced is given by,

n(Ag2SO4) = 0.255 mol of AgNO3 ×

[0.155mol H2SO4 ÷ 2 mol AgNO3] x

[ 1 mol Ag2SO4 ÷ 1 mol H2SO4]

= 0.0198 mol of Ag2SO4.

mass = no of moles x molar mass

From literature, molar mass of Ag2SO4 = 311.799g/mol.

Thus,

Mass = 0.0198 x 311.799

= 6.162g

Therefore, 6.162g of Ag2SO4 could be formed

4 0
3 years ago
Picture shown! <br><br> At 3 mmHg and 500C water exists as a<br> Liquid<br> Solid<br> Gas
mestny [16]

Answer:

At 3 mmHg and 50°C water exists as a <em><u>gas</u></em>.

8 0
3 years ago
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