__Al+__S —>___Al S 8 2 3

Nitrogen (N) and hydrogen (H) react with each other to produce ammonia (NH3). This reaction is shown in an equation as N2 + 3H2

Anika [276]

There are 2 Nitrogen atoms (or parts) of Nitrogen on the left side of the equation, and 2 Hydrogen, and only one Nitrogen but three Hydrogen on the other side. Where did the extra Nitrogen go? Where did that Hydrogen come from? The answer is Stoichiometry.
N2 + H2 --> NH3 has to be balanced, so we add coefficients to the reactants and products, which indicate in what ratio they are consumed in the reaction. They effectively multiply the subscripts on the elements.
To balance Nitrogen, we have to add a 2 to the front of NH3, so we get 2NH3. Nitrogen is balanced, but Hydrogen isn't. There are now 6 Hydrogen being produced by the reaction, so we can add a 3 to the products side, making 3H2.
Now we have N2 + 3H2 --> 2NH3, and everything is balanced.
The ratio is 2:6, or 1:3

3 0

4 years ago

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The system co2(g) + h2(g) ⇀↽ h2o(g) + co(g) is at equilibrium at some temperature. at equilibrium a 4.00 l vessel contains 1.00

Marina CMI [18]

Answer: The moles of $CO_2$ added to the system is 7.13 moles

Explanation:

We are given:

Moles of $CO_2$ at equilibrium = 1.00 moles

Moles of $H_2$ at equilibrium = 1.00 moles

Moles of $H_2O$ at equilibrium = 2.40 moles

Moles of $CO$ at equilibrium = 2.40 moles

Volume of the container = 4.00 L

Concentration is written as:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

The given chemical equation follows:

$CO_2(g)+H_2(g)\rightleftharpoons H_2O(g)+CO(g)$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[CO][H_2O]}{[CO_2][H_2]}$

Putting values in above equation, we get:

$K_c=\frac{(\frac{2.40}{4.00})\times (\frac{2.40}{4.00})}{(\frac{1.00}{4.00})\times (\frac{1.00}{4.00})}\\\\K_c=5.76$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of CO = 0.791 mol/L

Volume of solution = 4.00 L

Putting values in above equation, we get:

$0.791M=\frac{\text{Moles of CO}}{4.00L}\\\\\text{Moles of CO}=(0.791mol/\times 4.00L)=3.164mol$

Extra moles of CO = (3.164 - 2.40) = 0.764 moles

Let the moles of $CO_2$ needed be 'x' moles.

Now, equilibrium gets re-established:

$CO_2(g)+H_2(g)\rightleftharpoons H_2O(g)+CO(g)$

Initial: 1.00 1.00 2.40 2.40

At eqllm: (0.236+x) 0.236 3.164 3.164

Again, putting the values in the expression of $K_c$ , we get:

$5.76=\frac{(\frac{3.164}{4.00})\times (\frac{3.164}{4.00})}{(\frac{0.236+x}{4.00})\times (\frac{0.236}{4.00})}\\\\5.76=\frac{10.011}{0.056+0.236x}\\\\x=7.13$

Hence, the moles of $CO_2$ added to the system is 7.13 moles

4 0

3 years ago

3. The graph below shows the temperature changes occurring in two hot packs that are made from different materials. The same amo

nika2105 [10]

Answer:

LDPE would be the best choice for a hand warmer designed to release heat quickly.

Explanation:

Judging from the graph, the LDPE plastic releases heat quicker than the Latex. (and I also did the gizmo-)

4 0

3 years ago

MULTIPLE CHOICE What is the molality (m) of 3.5 mol of sugar in 2.5 kg of water?

otez555 [7]

The molality of the solution is 1.4 moles per kilogram (molal).

The molality of the solution is equal to the number of moles of sugar (solute) divided by the number of kilograms of water (solvent). Hence, the molality is determined below:

$m = \frac{3.5\,mol}{2.5\,kg}$

$m = 1.4\,\frac{mol}{kg}$

The molality of the solution is 1.4 moles per kilogram (molal).

We kindly invite to check this question on molality: brainly.com/question/4580605

5 0

3 years ago

Why cant (NaCl) light the bulb until the NaCl is dissolved in water. *

Aneli [31]

Answer:

NaCl will only conduct electricity in solutions

Explanation:

For electrical conduction, free mobile electrons as seen in most metals must be present or ions which are charged particles must be available for solutions and molten substances.

Sodium chloride is an ionic compound without free mobile electrons or ions despite being ionic.
It will maintain a subtle and unique charge stability when in solid form.
In solid, the ions are not free to move and remain locked up in the solid mass.
When introduced into a solution, the ions becomes free to move and this will aid electrical conduction.

6 0

3 years ago