What is the name of the element with the atomic number 79

This is for chemistry:/

Andreyy89

Answer:

Its the bottom one :) ......

5 0

3 years ago

A solution was prepared by mixing 50.0 g

frez [133]

Answer:

4.78 %.

Explanation:

mass percent is the ratio of the mass of the solute to the mass of the solution multiplied by 100.

mass % = (mass of solute/mass of solution) x 100.

mass of MgSO₄ = 50.0 g,

mass of water = d.V = (0.997 g/mL)(1000.0 mL) = 997.0 g.

mass of the solution = mass of water + mass of MgSO₄ = 997.0 g + 50.0 g = 1047.0 g.

∴ mass % = (mass of solute/mass of solution) x 100 = (50.0 g/1047.0 g) x 100 = 4.776 % ≅ 4.78 %.

4 0

4 years ago

given two equations representing reactions: which type of reaction is represented by each of these equations?

Sergeeva-Olga [200]

Answer:

Equation 1 - nuclear fission

Equation 2 - nuclear fusion

Explanation:

Nuclear fission is a reaction in which a large nucleus is split into smaller nuclei when it is bombarded by neutrons. The process produces more neutrons to continue the chain reaction. This is clearly depicted in equation 1 as shown in the question.

Nuclear fusion is a reaction in which two light nuclei combine in order to form a larger nuclei. This is clearly depicted in equation 2 as shown in the question.

7 0

3 years ago

Why did early people most likely choose to settle in Mesopotamia?

Zigmanuir [339]

Answer:

People wanted to settle there because it had good living conditions. There was rich soil, and lots of water sources.

I hope this helps!

4 0

3 years ago

Here are some data from a similar experiment, to determine the empirical formula of an oxide of tin. Calculate the empirical for

eduard

Answer:

Empirical formula of the Tin oxide sample is SnO₂

Explanation:

Tin reacts with combines with oxygen to form an oxide of tin.

Mass of crucible with cover = 19.66 g

Mass of crucible, cover, and tin sample = 22.29 g

Mass of crucible and cover and sample, after prolonged heating gives constant weight = 21.76 g

Mass of Tin oxide sample = 22.29 - 19.66 = 2.63 g

Mass of ordinary tin, after heating to breakdown the tin and oxygen = 21.76 - 19.66 = 2.1 g

Meaning that, mass of oxygen in the tin oxide sample = 2.63 - 2.1 = 0.53 g

Mass of Tin in the Tin Oxide sample = 2.1 g

Mass of Oxygen in the Tin oxide sample = 0.53 g

Convert these to number of moles

Number of moles of Tin on the Tin oxide sample = 2.1/118.71 = 0.0177

Number of moles of Oxygen in the Tin oxide sample = 0.53/16 = 0.0335

divide the number of moles by the lowest number

0.0177:0.0335

It becomes,

1:2

SnO₂

Hence, the empirical formula for the Tin oxide sample = SnO₂

7 0

4 years ago