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3 years ago

PLEASE ANSWER

Chemistry

1 answer:

lana [24]3 years ago

3 0

Answer: option D. Their speed in a vacuum

Explanation:

Electromagnetic waves all have same speed in vacuum. This speed has been calculated to be 3x10^8m/s.

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Given the following data:

bagirrra123 [75]

$176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its $\Delta H$ can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with $\Delta H$ given be denoted as (1), (2), (3), and the last equation (4). Let $a$ , $b$ , and $c$ be letters such that $a \times (1) + b \times (2) + c \times (3) = (4)$ . This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, $3 + (-1) = 2$ shall resemble the number of $\text{H}_2$ left on the product side when the second equation is directly added to the third. Similarly

$\text{NH}_4 \text{Cl} \; (s)$ : $-2 \; a = 1$
$\text{NH}_3\; (g)$ : $-2 \; b = -1$
$\text{HCl} \; (g)$ : $2 \; c = -1$

Thus

$a = -1/2\\b = 1/2\\c = -1/2$ and

$-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)$

Verify this conclusion against a fourth species involved- $\text{N}_2 \; (g)$ for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

$a + b = -1/2 + 1/2 = 0$

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

$\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

3 0

3 years ago

Salts are which part of an aqueous solution?

masha68 [24]

Explanation:

Salts are the solutes in an aqueous solution. An aqueous solution is solution whose solvent water.

To form a solution,a solute must be dissolved in a solvent.
For aqueous solutions, the solvent which is the dissolving medium is made up of water.
The solute is the substance that is dissolved in it.
Salts for example can be a solute in an aqueous solution.
A salt is generally an ionic compound consisting of positive ions such as metallic, ammonium ans complex ions and negative ions such as acid radicals and complex ions.

Learn more:

Aqueous solution brainly.com/question/8426727

#learnwithBrainly

3 0

3 years ago

Which diagram best represents reciprocal disclosure?

kramer

Answer:I think its B

Explanation:

6 0

3 years ago

Read 2 more answers

Help pls again HAHAHAH

Alenkasestr [34]

I believe the answer you are looking for is position 4 , because the northern face of the hemisphere is facing away from the sun not getting

to much heat nor daylight therefore its cold making it winter fully in option 4.

Just a tip option 3 looks like its facing back but half of it is still shown to the sun.

5 0

3 years ago

A student pours 44.3 g of water at 10Â°C into a beaker containing 115.2 g of water at 10Â°C. What are the final mass, temperatur

Agata [3.3K]

Answer:

Final mass = 159.5 g

Final temperature = 10 C

Final density = 1.00 g/ml

Explanation:

Given:

Beaker 1:

Mass of water = 44.3 g

Temperature = 10 C

Beaker 2:

Mass of water = 115.2 g

Temperature = 10 C

Density of water at 10C = 1.00 g/ml

To determine:

The final mass, temperature and density of water

Calculation:

$Final\ mass\ of \ water = Beaker\ 1 + Beaker\ 2 = 44.3 + 115.2 = 159.5 g$

Since there is no change in temperature, the final temperature will be 10 C

Density of a substance is an intensive property i.e. it is independent of the mass. Hence the density of water will remain constant i.e. 1.00 g/ml

3 0

3 years ago