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3 years ago

Balance the equations

Chemistry

1 answer:

bearhunter [10]3 years ago

6 0

1. 2 H2 + O2 = 2 H2O

2. 6 K + B2O3 = 3 K2O + 2 B

3. 10 Na + 2 NaNO3 = 6 Na2O + N2

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Explain the meaning of the terms "saturated," "unsaturated," and "supersaturated"

dmitriy555 [2]

Saturated Solution: A solution with solute that dissolves until it is unable to dissolve anymore, leaving the undissolved substances at the bottom. Unsaturated Solution: A solution ( with less solute than the saturated solution )that completely dissolves, leaving no remaining substances. Supersaturated Solution.

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2 years ago

Match each statement with a layer of the atmosphere. There will be two statements for each layer.

LuckyWell [14K]

The troposphere:

H. This layer can have thunderstorms or clear, sunny skies.

A. The biosphere interacts most with this layer.

The stratosphere:

B. It is the second layer from Earth's surface.

G. Winds are strong and steady in this layer.

The mesosphere:

E. It is heated by the ozone layer beneath it.

D. This layer is where most meteor showers occur.

The thermosphere :

F. It contains the ionosphere and exosphere.

C. It contains layers of single, unmixed gas.

Explanation:

Depending on the Earth's temperature the atmosphere can be separated into layers. The troposphere, the stratosphere, the mesosphere and the thermosphere are those layers. The lowest layer is named as Troposphere (0-10 km from the Earth outer surface), it comprises about 75% of the atmosphere's total air and nearly most the water vapor.

Stratosphere (10-30) includes much of the surface ozone. The change in height temperature arises as this ozone absorbs ultraviolet (UV) radiation from the sun. The temperature in Mesosphere (30-50 Km) declines again with height, hitting a minimum of about -90 ° C at the "mesopause." Above this thermosphere (50-400 Km) is settled which is a area where temperatures rise with height once again. The penetration of intense UV and X-ray radiation from the sun induces this temperature rise.

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3 years ago

Which will most likely cause an earthquake?

ArbitrLikvidat [17]

C: plates slipping along a fault

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2 years ago

Read 2 more answers

What royal scientist used the 29th

olchik [2.2K]

J. Dorman probably was the scientist.

I have a question too. Can you please help me with that? I have a test tomorrow. Thank you!

1. True or False? When you charge your laptop, you are separating charges in the laptop battery.

2. True or False? When you use your laptop, the battery becomes more neutral as it runs low on power.

8 0

3 years ago

Flag this question question 8 10 pts use the δh°f and δh°rxn information provided to calculate δh°f for if: δh°f (kj/mol) if7(g)

GarryVolchara [31]

$\Delta H\textdegree{}_f(\text{IF} \; (g)} = -95 \; \text{kJ} \cdot \text{mol}^{-1}$

Explanation

$\text{IF}_7 \; (g) + \text{I}_2 \; (s) \to \text{IF}_5 \; (g) + 2\; \text{IF} \; (g)$

$\Delta H\textdegree{}_\text{rxn} = -89\; \text{kJ} \cdot \text{mol}^{-1}$
$\Delta H\textdegree{}_f (\text{IF}_7 \; (g) ) = -941 \; \text{kJ} \cdot \text{mol}^{-1}$ (from the question)
$\Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) = -840 \; \text{kJ} \cdot \text{mol}^{-1}$ (from the question)

As an the most stable allotrope under standard conditions, $\Delta H\textdegree{}_f (\text{I}_2) = 0\; \text{kJ} \cdot \text{mol}^{-1}$

By definition,

$\Delta H\textdegree{}_\text{rxn} = \Delta H\textdegree{}_f (\text{all products}) - \Delta H\textdegree{}_f (\text{all reactants})$

$\Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) + 2 \; \Delta H\textdegree{}_f (\text{IF} \; (g) ) - \Delta H\textdegree{}_f (\text{IF}_7 \; (g) ) - \Delta H\textdegree{}_f (\text{I}_2 \; (s) ) \\ = \Delta H\textdegree{}_{\text{rxn}}$

$\begin{array}{ccc} \Delta H\textdegree{}_f (\text{IF} \; (g) )& = & 1/2\; ( \Delta H\textdegree{}_{\text{rxn}} - \Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) + \Delta H\textdegree{}_f (\text{IF}_7 \; (g) ) + \Delta H\textdegree{}_f (\text{I}_2 \; (s) ) )\\ & = & 1/2 \; (-89 - (-840) + (-941))}\\ & = & - 95 \; \text{kJ} \cdot \text{mol}^{-1} \end{array}$

Note, that iodine on the reactant side is stated as a gas in the equation given in the question whereas under standard conditions it is expected to be under the solid state; the $\Delta H\textdegree{} _f$ given in the question seemingly corresponds to the one in which the reactant iodine exists as a solid rather than as a gas. Evaluating the last expression using data from an external source

$\Delta H\textdegree{}_f (\text{I}_2 \; (g) ) = \Delta H\textdegree{}_f(\text{I}_2 \; (s)) + \Delta H\textdegree{}_{\text{sublimation}}(\text{I}_2) = 62.42 \; \text{kJ} \cdot \text{mol}^{-1}$ (Cox, Wagman, et al., 1984)

... yields $\Delta H\textdegree{}_f (\text{IF} \; (g) ) \approx -64 \; \text{kJ}\cdot \text{mol}^{-1}$ , which deviates significantly from the experimental value of $-94.76 \; \text{kJ}\cdot \text{mol}^{-1}$ (Chase, 1998.) It is thus assumed that the $\Delta H\textdegree{}_\text{rxn}$ value provided requires a reaction with $\text{I}_2 \; (s)$ rather than $\text{I}_2 \; (g)$ as a reactant.

3 0

3 years ago

Balance the equations​

Balance the equations