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Levart [38]
2 years ago
15

A homogenous mixture of two or more substances is known as a(n) Select one: a. element b. compound c. solution d. atom

Chemistry
1 answer:
babymother [125]2 years ago
7 0
The correct answer is c. solution
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The reaction of nitrogen gas and oxygen gas to form nitrogen monoxide gas is shown below. If the measured concentrations of all
Anettt [7]

Answer:

The forward reaction is occurring at a faster rate than the reverse reaction.

Explanation:

Reaction quotient (Q) of the given reaction is -

                                     Q=\frac{[NO]^{2}}{[N_{2}][O_{2}]}

where [NO], [N_{2}] and [O_{2}] represents concentrations of respective species at a certain time.

Here [N_{2}]= 0.80 M, [O_{2}]= 0.050 M and [NO] = 0.10 M

So, Q=\frac{(0.10)^{2}}{(0.80)\times (0.050)}=2

Hence Q> K

It means that forward reaction is faster than reverse reaction at that point. Because then only concentration of NO is higher than concentrations of [N_{2}] and [O_{2}] which makes Q higher than K.

4 0
2 years ago
Can frozen ice transfer thermal energy to another substance? if so, how?
Studentka2010 [4]

Answer:

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7 0
2 years ago
A bomb calorimetric experiment was run to determine the enthalpy of combustion of methanol. The reaction is CH3OH(l)+3/2O2(g)→CO
dimaraw [331]

Answer:

The change in internal energy during the combustion reaction is- 545.71 kJ/mol.

Explanation:

First we have to calculate the heat gained by the calorimeter.

q=c\times (T_{final}-T_{initial})

where,

q = heat gained = ?

c = specific heat = 250.0 J/^oC

T_{i} = Initial temperature = 21.50^oC=294.65 K

T_{f} = Final temperature = 23.41^oC=296.56 K

Now put all the given values in the above formula, we get:

q=250.0 J/K\times (296.56 -294.65 )K

q=477.5 J

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat gained = -477.5 J

n = number of moles methanol = \frac{\text{Mass of methanol}}{\text{Molar mass of methanol}}=\frac{0.028 g}{32 g/mol}=0.000875 mol

\Delta H=-\frac{477.5 }{0.000875 mol}=-545,714.28 J/mol=-545.71 kJ/mol

Therefore, the change in internal energy during the combustion reaction is- 545.71 kJ/mol.

7 0
3 years ago
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