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Kamila
3 days ago
14

To save time you can approximate the initial volume of water to ±1 mL and the initial mass of the solid to ±1 g. For example, if

you are asked to add 23 mL of water, add between 22 mL and 24 mL. Which metals in each of the following sets will have equal density?
1. 20.2 g gold placed in 21.6 mL of water and 12.0 g copper placed in 21.6 mL of water.
2. 20.2 g silver placed in 21.6 mL of water and 12.0 g silver placed in 21.6 mL of water.
3. 15.2 g copper placed in 21.6 mL of water and 50.0 g copper placed in 23.4 mL of water.
4. 15.4 g gold placed in 20.0 mL of water and 15.7 g silver placed in 20.0 mL of water.
5. 20.2 g silver placed in 21.6 mL of water and 20.2 g copper placed in 21.6 mL of water.
6. 11.2 g gold placed in 21.6 mL of water and 14.9 g gold placed in 23.4 mL of water.
Chemistry
1 answer:
castortr0y [927]3 days ago
7 0

Answer:

The correct options include choice 2, 3, and 6.

Explanation:

Density is identified as the mass of a substance per unit volume occupied by that substance.

Density=\frac{Mass}{Volume}

The density remains constant for a given substance, regardless of variations in mass and volume hence it is considered an intensive property.

2. 20.2 g of silver in 21.6 mL of water and 12.0 g of silver also in 21.6 mL of water.

3. 15.2 g of copper in 21.6 mL of water and 50.0 g of copper in 23.4 mL of water.

6. 11.2 g of gold in 21.6 mL of water and 14.9 g of gold in 23.4 mL of water.

The same metals in both instances will yield consistent densities due to the fixed density of the metal.

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Consider butter (density= 0.860 g/mL) and sand (density= 2.28 g/mL). If 1.00 mL of butter were mixed with 1.00 mL of sand and mi
Anarel [852]

The mixture’s density is 1.57 g/cm³.


Step 1: Determine the mass of the butter.


\text{Mass} = \text{1.00 cm}^{3 } \times \frac{\text{0.680 g} }{\text{1 cm}^{3 }} = \text{0.860 g}\\

Step 2: Determine the mass of the sand.


\text{Mass} = \text{1.00 cm}^{3 } \times \frac{\text{2.28 g} }{\text{1 cm}^{3 }} = \text{2.28 g}\\

Step 3: Determine the density of the mixture.

Total mass = 0.860 g + 2.28 g = 3.14 g.

Total volume = 1 cm³ + 1 cm³ = 2 cm³

\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{\text{3.14 g} }{\text{2 cm}^{3 }} = \textbf{1.57 g/cm}{^{3}\\

6 0
1 day ago
How much volume (in cm3) is gained by a person who gains 11.1 lbs of pure fat?
Anarel [852]

The density of human fat is 0.918 g/cm^{3}. The mass of the pure fat is 11.1 lbs.

First, convert the mass from pounds to grams as follows:

1 lb=453.592 g

Thus,

11.1 lb\rightarrow 11.1\times 453.592 g=5034.875 g

Density is defined as mass per unit volume, meaning volume can be calculated as:

V=\frac{m}{d}

By substituting the values,

V=\frac{5034.875 g}{0.918 g/cm^{3}}=5484.61 cm^{3}

Consequently, the volume gained by the individual will be 5484.61 cm^{3}.

6 0
8 days ago
Now that Snape and Dumbledore has taught you the finer points of hydration calculations they have a slightly more challenging pr
eduard [944]

Answer:

The integer value of x in the hydrate is 10.

Explanation:

Molarity=\frac{Moles}{Volume(L)}

Molar concentration of the solution = 0.0366 M

Volume of the solution = 5.00 L

Moles of hydrated sodium carbonate = n

0.0366 M=\frac{n}{5.00 L}

n=0.0366 M\times 5 mol=0.183 mol

Weight of hydrated sodium carbonate = n = 52.2 g

Molar mass of hydrated sodium carbonate = 106 g/mol + x * 18 g/mol

n=\frac{\text{mass of Compound}}{\text{molar mass of compound}}

0.183 mol=\frac{52.2 g}{106 g/mol+x\times 18 g/mol}

106 g/mol+x\times 18 g/mol=\frac{52.2 g}{0.183 mol}

By solving for x, we arrive at:

x = 9.95, approximating to 10

The integer x in the hydrate equals 10.

6 0
15 hours ago
A water tank can hold 1 m3 of water. When it’s empty, how much liters is needed to refill it?
Alekssandra [971]

Response: 1000

Rationale: because 5 cubic meters equals 5000 liters

3 0
5 days ago
A student is given a sample of a blue copper sulfate hydrate. He weighs the sample in a dry covered porcelain crucible and got a
KiRa [971]

Answer:

There are 5.5668 moles of water for every mole of CuSO₄.

Explanation:

The mass of anhydrous CuSO₄ is:

23.403g - 22.652g = 0.751g.

mass of crucible + lid + CuSO₄ - mass of crucible + lid

Given that the molar mass of CuSO₄ is 159.609g/mol, we calculate the moles:

0.751g ×\frac{1mol}{159,609g} = 4.7052x10⁻³ moles CuSO₄

The mass of water in the initial sample is:

23.875g - 0.751g - 22.652g = 0.472g.

mass of crucible + lid + CuSO₄ hydrate - CuSO₄ - mass of crucible + lid

As the molar mass of H₂O is 18.02g/mol, we find the moles:

0.472g ×\frac{1mol}{18,02g} = 2.6193x10⁻² moles H₂O

The mole ratio of H₂O to CuSO₄ is:

2.6193x10⁻² moles H₂O / 4.7052x10⁻³ moles CuSO₄ = 5.5668

This indicates there are 5.5668 moles of water per mole of CuSO₄.

I hope this is helpful!

5 0
8 days ago
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