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4 years ago

4.2g of sodium bicarbonate is equivalent to how many moles of sodium bicarbonate

2 answers:

7 0

The mass of one mole of sodium bicarbonate (aka NaHCO3) is equal to 1 * 22.99g/mol + 1 * 1.00g/mol + 1 * 12.01g/mol + 3 * 16.00g/mol = 83.91g/mol. From this, we can convert 4.2g of NaHCO3 to moles by dividing by 83.91g/mol, to get 0.050 moles of sodium bicarbonate.

babunello [35]4 years ago

7 0

Answer: The number of moles of sodium bicarbonate are 0.05 moles.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of sodium bicarbonate = 4.2 g

Molar mass of sodium bicarbonate = 84 g/mol

Putting values in above equation, we get:

$\text{Moles of sodium bicarbonate}=\frac{4.2g}{84g/mol}=0.05mol$

Hence, the number of moles of sodium bicarbonate are 0.05 moles.

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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

Elements in Group 4A

steposvetlana [31]

Answer:

helium hydrogen

Explanation:

lithium beryllium bottom carbon

6 0

3 years ago

A sample of water at 100°C is converted to steam after absorbing 820 kJ of heat. How grams of H2O are contained in the sample? A

Varvara68 [4.7K]

Since water is already at 100°C all the energy is used to evaporate it.
Now we can calculate how many mols of water are evaporated with 820kJ.
$N= \frac{820}{41} =20 mol$
We calculated that we got 20 mols of water evaporated. Now, all we have to do is find how many grams is a mol of water. Molar mass of water is 20.16 g/mol.
The final answer is:
$m=20*20.16=403.2g$

7 0

3 years ago

Answer both of these true or false correctly brainliest if correct

maxonik [38]

Answer:

There both true.

Explanation:

Hope this helps

3 0

3 years ago

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Which of the following statements about the greenhouse effect is false?