When 5800 joules of energy are applied to a 15.2-kg piece of lead metal, how much does the temperature change by

Answer:

The nichrome wire has contaminants.

The sample solution might be tainted.

Explanation:

If the nichrome wire is contaminated, sodium impurities could be causing the yellow flame. The wire is initially placed in the flame without the sample to check for such impurities.

The testing solution could also be contaminated, causing it to display a color different from the anticipated shade of the test ion.

Answer:

The mass of 22-Na included in the sample amounts to 0.0599 g

Explanation:

The total mass of the isotope mixture is 1.8385g.

It has an apparent mass of 22.9573 u.

For 23-Na, the relative atomic mass is 22.9898 u, while for 22-Na it is 21.9944 u.

Let the relative abundance of 23-Na be denoted as X.

This means that the relative abundance of 22-Na can be expressed as (1-X).

The equation formed is 21.9944 (1-X) + 22.9898 X = 22.9573.

Rearranging gives: 21.9944 - 21.9944X + 22.9898X = 22.9573.

Which simplifies to 22.9898X - 21.9944X = 22.9573 - 21.9944.

Hence, 0.9954X = 0.9639, leading to X = 0.9674.

The relative abundance of 23-Na is now identified as 0.9674.

Consequently, the relative abundance of 22-Na is 1 - 0.9674 = 0.0326.

Now, the mass of 22-Na contained within the 1.8385g sample is determined by

Relative abundance of 22-Na multiplied by the mass of the total sample = 0.0326 × 1.8385g = 0.0599 g.

What is being removed during the wash is the solvent.

At 30°C, glucose has a solubility of 1.25 g per gram of water. Given that the density of water at this temperature is 1 g/mL, the mass corresponding to 400 mL of water is also 400 g. Therefore, the concentration of the solution is calculated as 550 g divided by 400 g of water, which gives 1.375 g of glucose per gram of water. Since this concentration exceeds the solubility limit for glucose at this temperature, the solution can be classified as SATURATED.

Answer:

C a B r 2 ( a q ) + N a 2 S O 4 ( a q ) ⟶ 2 N a B r ( a q ) + C a S O 4 ( s )

Explanation:

A precipitation reaction involves a displacement process where a solid precipitate forms. This precipitate, being solid, is distinct from the other products and can be separated.

C a 2 + ( a q ) + S O 4 2 − ( a q ) ⟶ C a S O 4 ( s )

This equation is incorrect as it results in only C a S O 4.

C a B r 2 ( a q ) + N a 2 S O 4 ( a q ) ⟶ 2 N a B r ( a q ) + C a S O 4 ( s )

This is the proper reaction where C a S O 4 precipitate is produced.

C a 2 + ( a q ) + 2 B r − ( a q ) + 2 N a + ( a q ) + S O 4 2 − ( a q ) ⟶ 2 N a + ( a q ) + 2 B r − ( a q ) + C a S O 4 ( s )

This equation illustrates the ionic details of the precipitation reaction.