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3 years ago

Round to 4 significant figures. 0.007062

Chemistry

1 answer:

Monica [59]3 years ago

7 0

.0071 you round 0 to1 because the 6 is bigger than 5 and if anything is bigger than five you round up that last number

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If 28 grams of N reacts completely with 12 grams of H2, then how many

Bogdan [553]

Answer:

Mass of NH₃ produced = 34 g

Explanation:

Given data:

Mass of nitrogen = 28 g

Mass of Hydrogen = 12 g

Mass of NH₃ produced = ?

Solution:

Chemical equation:

N₂ + 3H₂ → 2NH₃

Moles of nitrogen:

Number of moles = mass/molar mass

Number of moles = 28 g/ 28 g/mol

Number of moles = 1 mol

Moles of hydrogen:

Number of moles = mass/molar mass

Number of moles = 12 g/ 2 g/mol

Number of moles = 6 mol

Now we will compare the moles of hydrogen and nitrogen with ammonia.

H₂ : NH₃

3 : 2

6 : 2/3×6 = 4 mol

N₂ : NH₃

1 : 2

Number of moles of ammonia produced by nitrogen are less thus it will act as limiting reactant.

Mass of ammonia produced:

Mass = number of moles × molar mass

Mass = 2 mol × 17 g/mol

Mass = 34 g

5 0

3 years ago

Read 2 more answers

25g of potassium nitrate are dissolved in 100g of water. What is the percent by mass of potassium nitrate?

trapecia [35]

The the exact mass is 24.740% (according to the internet) but I’d say that it’s 25

8 0

3 years ago

What happens when soluble fiber is mixed with water?

storchak [24]

Answer:

Soluble fiber attracts water and turns to gel during digestion. This slows digestion. Soluble fiber is found in oat bran, barley, nuts, seeds, beans, lentils, peas, and some fruits and vegetables. It is also found in psyllium, a common fiber supplement.:

6 0

3 years ago

What is the boiling point of 5 ml of alcohol

NeTakaya

173.1f is the answer I believe, please let me know if I'm wrong then I would try to make up for it

4 0

3 years ago

wo reactions and their equilibrium constants are given. A + 2 B − ⇀ ↽ − 2 C K 1 = 2.57 2 C − ⇀ ↽ − D K 2 = 0.226 A+2B↽−−⇀2CK1=2.

Papessa [141]

Answer: The value of equilibrium constant for the net reaction is 11.37

Explanation:

The given chemical equations follows:

Equation 1: $A+2B\xrightarrow[]{K_1} 2C$

Equation 2: $2C\xrightarrow[]{K_2} D$

The net equation follows:

$D\xrightarrow[]{K} A+2B$

As, the net reaction is the result of the addition of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K=K_1\times \frac{1}{K_2}$

We are given:

$K_1=2.57$

$K_2=0.226$

Putting values in above equation, we get:

$K=2.57\times \frac{1}{0.226}=11.37$

Hence, the value of equilibrium constant for the net reaction is 11.37

6 0

3 years ago