According to the cup wall diagrams, why does the double wall vacuum

Look at the graph. Which part of the line shows a time<br> when the object was not moving?

marysya [2.9K]

Answer:

$\bold \blue{b)B}$

hope it helps :)

7 0

2 years ago

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After the death of living material, the ratio of carbon-12 to carbon-14 isotopes in the material;

Ad libitum [116K]

Decrease on edg 2020

3 0

3 years ago

Iron(III) oxide and hydrogen react to form iron and water, like this: Fe_2O_3(s) + 3H_2(g) rightarrow 2Fe(s) + 3H_2O(g) At a cer

Sergeeva-Olga [200]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The equilibrium constant is $K_c= 2.8*10^{-4}$

Explanation:

From the question we are told that

The chemical reaction equation is

$Fe_{2} O_{3}_{(s)} + 3H_{2}_{(g)} -----> 2Fe_{(s)} + 3H_{2} O_{(g)}$

The voume of the misture is $V_m = 5.4L$

The molar mass of $Fe_{2} O_{3}_{(s)}$ is a constant with value of $M_{Fe_{2} O_{3}_{(s)} } = 160g/mol$

The molar mass of $H_{2}_{(g)}$ is a constant with value of $H_2 = 2g/mol$

The molar mass of $H_{2}O$ is a constant with value of $H_2O = 18g/mol$

Generally the number of moles is mathematically given as

$No \ of \ moles \ = \frac{mass}{molar\ mass}$

For $Fe_{2} O_{3}_{(s)}$

$No \ of\ moles = \frac{3.54}{160}$

$= 0.022125 \ mols$

For $H_{2}$

$No \ of\ moles = \frac{3.63}{2}$

$= 1.815 \ mols$

For $H_{2}O$

$No \ of\ moles = \frac{2.13}{18}$

$= 0.12 \ mols$

Generally the concentration of a compound is mathematicallyrepresented as

$Concentration = \frac{No \ of \ moles }{Volume }$

For $Fe_{2} O_{3}_{(s)}$

$Concentration[Fe_2 O_3] = \frac{0.222125}{5.4}$

$= 4.10*10^{-3}M$

For $H_{2}$

$Concentration[H_2] = \frac{1.815}{5.4}$

$= 0.336M$

For $H_{2}O$

$Concentration [H_2O] = \frac{0.12}{5.4}$

$= 0.022M$

The equilibrium constant is mathematically represented as

$K_c = \frac{[concentration \ of \ product]}{[concentration \ of \ reactant ]}$

Considering $H_2O \ for \ product$

And $H_2 \ for \ reactant$

At equilibrium the

$K_c = \frac{0.022}{0.336}$

$K_c= 2.8*10^{-4}$

3 0

3 years ago

Which ion has the lower ratio of charge to volume? Explain.<br> (b) Sc³⁺ or Ca²⁺

WINSTONCH [101]

Ca²⁺ion has a lower ratio of charge to volume.

The charge density of an ion is defined as the ratio of the charge of an ion to its volume. Scandium ion has high charge density than calcium ions. The charge density of an ion is defined as the ratio of the charge of an ion to its volume.

Charge density also depends on the size of the ion and valence electrons. The volume of an ion increases with its size. It is inversely proportional to ion volume and directly proportional to charge magnitude. Scandium and calcium are IV periodic elements with atomic numbers of 21 and 20 respectively. Scandium loses three electrons and has a +3 charge and calcium is a divalent cation. Hence, the Scandium ion has high charge density than the calcium ion.

Learn more about charge density here: brainly.com/question/12968377

#SPJ4

4 0

1 year ago

You wish to make a 0.289 M hydroiodic acid solution from a stock solution of 3.00 M hydroiodic acid. How much concentrated acid

nataly862011 [7]

Answer:

$V_1=9.63mL$

Explanation:

Hello,

In this case, we work on a dilution process in which we can state that the moles remain the same after the dilution process. In such a way, we can write:

$n_1=n_2\\$

That in terms of molarities and volumes is:

$M_1V_1=M_2V_2$

Whereas $M_1$ is the initial molarity (3.00 M) of the stock solution, $M_2$ the molarity of the diluted solution (0.289 M), $V_1$ the aliquot of the stock (concentrated) solution and $V_2$ the volume of the diluted solution (100 mL), thus, we compute $V_1$ as required:

$V_1=\frac{M_2V_2}{M_1} =\frac{0.289M*100mL}{3.00M} \\\\V_1=9.63mL$

Best regards.

7 0

3 years ago