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Lena [83]
2 years ago
13

I. Choose the letter of the correct answer. Write your answer on a separate sheet of paper.

Chemistry
1 answer:
omeli [17]2 years ago
6 0

Answer:

,,,

Explanation:

..

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The body also loses heat by the evaporation of sweat. How is a tea kettle boiling similar to the evaporation of sweat from the b
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In both cases, water is being left out of a “container” due to heat
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2 years ago
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Is there a difference between a homogeneous mixture of hydrogen and oxygen in a 2:1 ratio and a sample of water vapor?
Feliz [49]

Answer:

Yes

Explanation:

There is a difference between the homogeneous mixture of the hydrogen and the oxygen in a 2:1 ratio and the sample of the water vapor.

In the homogeneous mixture of the hydrogen and the oxygen which are present in the ratio, 2:1 , the elements are not chemically combined. They are explosive also as both shows their specific properties. They can be separated by physical means (Condensation, diffusion).

On the other hand, in water vapor, the two elements are chemically bonded in a specific mixture which cannot be separated via physical means. Water has its unique properties and they can be separated by chemical means only.

5 0
3 years ago
Someone help me with question pls
gizmo_the_mogwai [7]

Explanation:

the conductors are the three u have checked

8 0
2 years ago
When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t
OleMash [197]

Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

Mn(s)+2HCl(aq)\rightarrow  MnCl_2(aq)+H_2(g)

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

m=1.00 g/mL\times 100.0 mL = 100 g

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

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

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = 4.18 J/^oC

T_{final} = final temperature = 23.1^oC

T_{initial} = initial temperature = 28.9^oC

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

q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC

q=2,242.4 J=2.242 kJ

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

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

where,

\Delta H = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = \frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol

\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

8 0
3 years ago
In which picture(s) will you find a state of matter that has particles who have more energy than liquid water?
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