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denis23 [38]
3 years ago
7

3ehehehehe first gets brainlist

Chemistry
2 answers:
Diano4ka-milaya [45]3 years ago
5 0

Answer:

4ehehehehehehe

Explanation:

Thanks......

notsponge [240]3 years ago
5 0

Answer:

Hey wanna tälk with me :)

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You are given three cubes, A, B, and C; one is magnesium, one is aluminum, and the third is silver. All three cubes have the sam
katovenus [111]

Answer:

The cube A is magnesium, the cube B is aluminum and the cube C is silver.

Explanation:

Density is defined by the expression d=\frac{m}{V} where m is the mass and V is the volume, therefore:

- Density of the cube A:

d_{A}=\frac{m_{A}}{V_{A}}

- Density of the cube B:

d_{B}=\frac{m_{B}}{V_{B}}

- Density of the cube C:

d_{C}=\frac{m_{C}}{V_{C}}

Solving for mass:

m_{A}=d_{A}*V_{A}

m_{B}=d_{B}*V_{B}

m_{C}=d_{C}*V_{C}

And all the three cubes have the same mass, so:

m_{A}=m_{B}=m_{C}

Therefore:

d_{A}*V_{A}=d_{B}*V_{B} (Eq.1)

d_{A}*V_{A}=d_{C}*V_{C} (Eq.2)

Solving for d_{1} in Eq.1:

d_{A}=d_{B}\frac{V_{B}}{V_{A}}

Replacing values for the volume:

d_{A}=d_{B}\frac{16.7mL}{25.9mL}

d_{A}=d_{B}*0.64

As we know the density of the aluminum is 2.7\frac{g}{cm^{3}}, so replacing this value for d_{B}:

d_{A}=2.7\frac{g}{mL}*0.64

d_{A}=1.728\frac{g}{mL}

that is the density of the magnesium.

Solving for d_{C} in Eq.2:

d_{C}=d_{A}\frac{V_{A}}{V_{C}}

d_{C}=d_{A}\frac{25.9mL}{4.29mL}

d_{C}=d_{A}*6.04

d_{C}=1.728\frac{g}{mL}*6.04

d_{C}=10.4\frac{g}{mL}

That is the density of the silver.

Therefore the cube A is magnesium, the cube B is aluminum and the cube C is silver.

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