All categories

2 years ago

How is the composition of a meteorite relevant to finding out the composition of earth’s core?

2 answers:

7 0

Since the belief is that the Earth was formed by an
enormous number of collisions of meteors, the
iron would have collected at the core due to its
mass. Many meteorites are about the same age as
the solar system so they may resemble the
material at the core.

boyakko [2]2 years ago

5 0

Explanation:

Since scientists think planets and meteorites were made at the same time and in the same place, it seems logical that whatever a meteorite is made of is also what planets are made of.

You might be interested in

Is there any evidence that the particles in a solid and liquid move? Provide examples.

BartSMP [9]

Answer:

maybe not solid but a liquid yes

6 0

3 years ago

Read 2 more answers

Would mean alot to help

Oksana_A [137]

Answer:

Explanation:

nobody knows

5 0

2 years ago

Read 2 more answers

Perform the following

strojnjashka [21]

Answer:

4 sig figs: 20.18

Explanation:

the smallest amount of digits was 4 in the expression

7 0

3 years ago

Bromine (Br2) is produced by reacting HBr with O2, with water as a byproduct. The O2 is part of an air (21 mol % O2, 79 mol % N2

Karolina [17]

Answer:

The mole fractions:

$x_{HBr}=\frac{100mol}{318.5}=0.314$

$x_{Br_2}=\frac{78mol}{318.5}=0.245$

$x_{H_2O}=\frac{78mol}{318.5}=0.245$

$x_{O_2}=\frac{62.5mol}{318.5}=0.196$

Explanation:

The reaction described is:

$2 HBr (g) + 1/2 O_2 (g) \longrightarrow Br_2 (g) + H_2O (g)$

The limiting reactant is the HBr (oxygen is in excess).

a) The mass (in moles) balance for this sistem:

$n_{Br_2}=\frac{ 1 mol Br_2}{1 mol HBr} *n_{HBr}*0.78$

(the 0.78 is because of the fractional conversion)

$n_{O_2}=\frac{ 0.5 mol O_2}{1 mol HBr} *n_{HBr}*1.25$

(the 1.25 is because of the oxygen excess)

$n_{H_2O}=\frac{ 1 mol H_2O}{1 mol Br_2} *n_{Br_2}$

There is only one degree of freedom in this sistem, you can either deffine the moles of HBr you have or the moles of Br2 you want to produce. The other variables are all linked by the equations above.

b) Base of calculation 100 mol of HBr:

$nn_{HBr}=100 mol HBr$