Chemistry, 26.06.2019 23:00 Zayybabii

How many atoms of n are in 137.0 grams of n2o3? a. 1.085x10^23b. 1.802x10^23c. 5.985x10^23d. 2.171x10^24e. 3.604x10^24will mark as branliest

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Answer from: jkw1222p0ttvq

d 2.171x10^24

Explanation:

#platofam

Answer from: jtal

Option (d) is the correct answer.

Explanation:

It is given that mass of $N_{2}O_{3}$ is 137.0 g.

Molar mass of $N_{2}O_{3}$ is as follows.

Molar mass of $N_{2}O_{3}$ = 2 × 14 + 3 × 16

= 76 g

Thus, number of moles will be calculated as follows.

No. of moles = $\frac{mass}{molar mass}$

= $\frac{137.0 g}{76 g/mol}$

= 1.803 mol

There are 2 nitrogen (N) atoms present in the formula of dinitrogen trioxide. Also Avogadro's number equals $6.022 \times 10^{23}$ atoms or molecules per mol.

Therefore, number of N molecules or atoms in $N_{2}O_{3}$ is as follows.

Number of N atoms = $2 \times 1.803 mol \times 6.022 \times 10^{23} atoms/mol$

= $2 \times 10.855 \times 10^{23}$ atoms

= $2.171 \times 10^{24}$ atoms

Thus, we can conclude that there are $2.171 \times 10^{24}$ atoms of N in in 137.0 grams of $N_{2}O_{3}$ .

Answer from: Quest

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied

Answer from: Quest

C) the increase in atmospheric pressure pushes the walls of the balloon outward thus making it is the answer.hope i you.

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How many atoms of n are in 137.0 grams of n2o3? a. 1.085x10^23b. 1.802x10^23c. 5.985x10^23d. 2.171x1...