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Rasek [7]
3 years ago
5

If a sample has a mass of 1.25 x 102 g and a volume of 51 mL, what is its density?

Chemistry
1 answer:
Dafna11 [192]3 years ago
3 0

Answer:

<h2>2.45 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

d =  \frac{m}{v}  \\

We have

d =  \frac{1.25 \times  {10}^{2} }{51}  \\  = 2.45098..

We have the final answer as

<h3>2.45 g/mL</h3>

Hope this helps you

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At a high temperature, equal concentrations of 0.160 mol/L of H2(g) and I2(g) are initially present in a flask. The H2 and I2 re
Nikitich [7]

Explanation:

The given reaction is as follows.

                                H_{2} + I_{2} \rightarrow 2HI

Initial :                  0.160    0.160          0  

Change :                  -x           -x              2x

Equilibrium:        0.160 - x    0.160 - x       x

It is given that [H_{2}] = [0.160 - x] = 0.036 M

and,                [I_{2}] = [0.160 - x] = 0.036 M      

so,                             x = (0.160 - 0.036) M

                                    = 0.124 M

As, [HI] = 2x.

So,           [HI] = 2 \times 0.124

                       = 0.248 M

As it is known that expression for equilibrium constant is as follows.

               K_{eq} = \frac{[HI]^{2}}{[H_{2}][I_{2}]}

                                  = \frac{(0.248)^{2}}{(0.036)(0.036)}

                                  = 47.46

Thus, we can conclude that the equilibrium constant, Kc, for the given reaction is 47.46.

                               

7 0
3 years ago
Place the following compounds in order of increasing vapor pressure. methane, CH4; pentane, CH3CH2CH2CH2CH3; 1-propanol, CH3CH2C
ella [17]

Explanation:

A compound that contains more stronger bonds will need more amount of heat in order to break the bonds so that it changes into vapor state.

In 1-propanol, there is hydrogen bonding and it is stronger in nature. As a result, more amount of heat is required to break the bonds between molecules of 1-propanol.

Whereas in propanone, there will be dipole-dipole interactions which are less stronger than hydrogen bonding. Hence, propanone molecule will need less amount of energy than 1-propanol.

On the other hand, pentane will need more amount of heat as it has longer chain of carbon atoms as compared to methane.

Thus, we can conclude that given compounds are arranged in order of increasing vapor pressure as follows.

         methane < pentane < propanone < 1-propanol

5 0
3 years ago
Read 2 more answers
Suppose you have an atom composed of 19 protons, 20 neutrons and 19 electrons. what is the mass of this atom?
bazaltina [42]
Mass = PROTONS + neutrons. So, 19 +20 = 39.
(Don't confuse yourself between electrons and protons when they are equal because it could be an ion)
7 0
4 years ago
The repeating subunits that are responsible for the chance of a crystal are know as
Naddik [55]
Unit cells.
They are the smallest group of atoms that take form of a crystal and can repeat in 3 dimensions.
7 0
3 years ago
What mass of CO is needed to react completely with55.0 g of Fe2O3(s)+CO(g) yield Fe(s)+CO2(g)?
Katarina [22]

Answer:

28.9 g

Explanation:

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

<em>Gather all the information in one place</em> with molar masses above the formulas and masses below them.  

M_{r}:     159.69    28.01

              Fe₂O₃ + 3CO ⟶ 2Fe + 3CO₂

Mass/g:  55.0

1. Use the molar mass of Fe₂O₃ to calculate the moles of Fe₂O₃.

\text{Moles of Fe$_{2}$O$_{3}$} =\text{55.0 g Fe$_{2}$O$_{3}$} \times \frac{\text{1 mol Fe$_{2}$O$_{3}$}}{\text{159.69 g Fe$_{2}$O$_{3}$}}= \text{0.3444 mol Fe$_{2}$O$_{3}$}

2. Use the molar ratio of CO:Fe₂O₃ to calculate the moles of CO.

\text{Moles of CO} = \text{0.3444 mol Fe$_{2}$O$_{3}$} \times \frac{\text{3 mol CO}}{\text{1 mol Fe$_{2}$O$_{3}$}}= \text{1.033 mol CO}

3.Use the molar mass of CO to calculate the mass of CO.

\text{Mass of CO} = \text{1.033 mol CO}  \times \frac{\text{28.01 g CO} }{\text{1 mol CO}}= \textbf{28.9 g CO}  

3 0
3 years ago
Read 2 more answers
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