Which of the following is a common agricultural use of a mineral? (2 points)

Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation:

nignag [31]

Answer: 1) Maximum mass of ammonia 198.57g

2) The element that would be completely consumed is the N2

3) Mass that would keep unremained, is the one of the excess Reactant, that means the H2 with 3,44g

Explanation:

In order to calculate the Mass of ammonia , we first check the Equation is actually Balance:

N2(g) + 3H2(g) ⟶2NH3(g)

Both equal amount of atoms side to side.

Now we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant. ( Keep in mind that we use the molecular weight of 28.01 g/mol N2; 2.02 g/mol H2; 17.03g/mol NH3)

Moles of ammonia produced with 163.3g N2(g) ⟶ 163.3g N2(g) x (1mol N2(g)/ 28.01 g N2(g) )x (2 mol NH3(g) /1 mol N2(g)) = 11.66 mol NH3

Moles of ammonia produced with 38.77 g H2⟶ 38.77 g H2 x ( 1mol H2/ 2.02 g H2 ) x (2 mol NH3 /3 mol H2 ) = 12.79 mol NH3

As we can see the amount of NH3 formed with the N2 is the lowest one , therefore the limiting reactant is the N2 that means, N2 is the element that would be completey consumed, and the maximum mass of ammonia will be produced from it.
We proceed calculating the maximum mass of NH3 from the 163.3g of N2.

11.66 mol NH3 x (17.03 g NH3 /1mol NH3) = 198.57 g NH3

In order to estimate the mass of excess reagent, we start by calculating how much H2 reacts with the giving N2:

163.3g N2 x (1mol N2/28.01 g N2) x ( 3 mol H2 / 1 mol N2)x (2.02 g H2/ 1 mol H2) = 35.33 g H2

That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of H2, thus, 38.77g - 35.33 g = 3.44g H2 is left

3 0

3 years ago

A hydrate is compound that included water molecules within its crystal structure. During an experiment to determine the percent

stellarik [79]

Answer:

The percent by mass of water in this crystal is:

21.4%

Explanation:

This exercise can be easily solved using a simple rule of three where the initial weight of the hydrated crystal (6,235 g) is taken into account as 100% of the mass, and the percentage to which the mass of 4.90 g corresponds (after getting warm). First, the values and unknown variable are established:

6,235 g = 100%
4.90 g = X

And the value of the variable X is found:

X = (4.90 g * 100%) / 6,235 g
X = approximately 78.6%.

The calculated value is not yet the percentage of the water, since the water after heating the glass has evaporated, therefore, the remaining percentage must be taken, which can be calculated by subtraction:

Water percentage = Total percentage - Percentage after heating.
Water percentage = 100% - 78.6% = 21.4%

4 0

3 years ago

Read 2 more answers

If 5.0 kJ of energy is added to a 15.5-g sample of water at 10.°C, the water is Group of answer choices

BaLLatris [955]

Answer:

The water is completely vaporized at this stage.

Explanation:

The complete question is

If 5.0 kJ of energy is added to a 15.5-g sample of water at 10.°C, the water is

-boiling

-completely vaporized

-frozen solid

-decomposed

-still a liquid

Energy added = 50 kJ = 50000 J

mass of water = 15.5 g = 0.0155 kg

temperature of water = 10 °C

We know that the energy posses by a mass of water at a given temperature is given as

H = mcT

where H is the energy possessed by the mass of water

m is the mass of the water

c is the specific heat capacity of water = 4200 J/ kg- °C

T is the temperature of the water

substituting values, the energy of this amount of water is

H = 0.0155 x 4200 x 10 = 651 J

If 50 kJ is added to the water, the energy increases to

50000 J + 651 J = 50651 J

Temperature of this water at this stage will be gotten from

H = mcT

we solve for the new temperature

50651 = 0.0155 x 4200 x T

50651 = 65.1 x T

T = 50651/65.1 = 778.05 °C

This temperature is well over 100 °C, which is the vaporization temperature of water, but less than 3000 °C for its molecules to decompose.

3 0

3 years ago

Match these items.

mars1129 [50]

Here are the possible answers for the following questions above:
1. H-CC-H (name) - C. ethyne
2. cyclic compound with both saturated and unsaturated characteristics - G. benzene
3. CnH2n - E. general formula for alkenes
4. reaction typical of unsaturated hydrocarbons - A. addition
5. CnH2n-2 - F. general formula for alkynes
6. series name of hydrocarbons with triple bond - D. alkyne
7. CnH2n+2 - B. general formula of alkanes

6 0

3 years ago

Read 2 more answers

Thorium-234 has a half-life of 24.1 days. How many grams of a 150 g sample would you have after 120.5 days?

chubhunter [2.5K]

Answer:

8.625 grams of a 150 g sample of Thorium-234 would be left after 120.5 days

Explanation:

The nuclear half life represents the time taken for the initial amount of sample to reduce into half of its mass.

We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.

Initial amount of Thorium-234 available as per the question is 150 grams

So now we start with 150 grams of Thorium-234

$150 \times \frac{1}{2}=24.1$