All categories

3 years ago

Complete the following reaction. 14N+ on →?C+1H 18C 14 1 C 18C

Chemistry

1 answer:

CaHeK987 [17]3 years ago

6 0

Answer: The complete reaction is as follows.

$^{14}_{7}N + ^{1}_{0}n \rightarrow ^{14}_{6}C + ^{1}_{1}H$

Explanation:

When nucleus of two or more atoms are bombarded together then it leads to the formation of new particles with new identity. This type of reaction are called nuclear reaction.

For example, $^{14}_{7}N + ^{1}_{0}n \rightarrow ?C + ^{1}_{1}H$

Here, nitrogen atom when bombarded with a neutron then it is forming hydrogen and a carbon atom.

As total atomic mass on reactant side is (14 + 1) = 15

So, the atomic mass of carbon formed on product side is (15 - 1) = 14.

The number of protons holded by this carbon atom is (7 - 1) = 6.

Therefore, we can conclude that the complete reaction is as follows.

$^{14}_{7}N + ^{1}_{0}n \rightarrow ^{14}_{6}C + ^{1}_{1}H$

You might be interested in

A 50.6 grams sample of magnesium hydroxide (Mg(OH)2) is reacted with 45.0 grams of hydrochloric acid (HCl). What is the theoreti

boyakko [2]

The answer is C, 58.6

7 0

3 years ago

Read 2 more answers

What mass of ether will be produced is 221 g of ethanol reacts with a 66 percent yield?

Keith_Richards [23]

@pandamille help her please!!!!

5 0

3 years ago

Can someone help please

Serhud [2]

Let the mass be X g

percentage = X/ 6.50 * 100 =2.2%

X= 0.143 g

The mass is 0.143 g

8 0

3 years ago

Read 2 more answers

How many grams of water are theoretically produced for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca

leva [86]

Answer:

The answer to your question is: 6.8 g of water

Explanation:

Data

2.6 moles of HCl

1.4 moles of Ca(OH)2

2HCl + Ca(OH)2 → 2H2O + CaCl2

MW 2(36.5) 74 36 g 111 g

73g

1 mol of HCl ---------------- 36.5 g

2.6 mol -------------- x

x = (2.6 x 36.5) / 1 = 94.9 g

1 mol of Ca(OH)2 -------------- 74 g

1.4 mol --------------- x

x = (1.4 x 74) / 1 = 103.6 g

Grams of water

73 g of HCl ------------------ 36g of H2O

94.9 g ------------------- x

x = (94.9 x 36) / 73 = 46.8 g of water

6 0

3 years ago

The air bags in cars are inflated when a collision triggers the explosive, highly exothermic decomposition of sodium azide (NaN3

Oksanka [162]

Answer : The mass of $NaN_3$ required is, 166.4 grams.

Explanation :

First we have to calculate the moles of nitrogen gas.

Using ideal gas equation:

$PV=nRT$

where,

P = Pressure of $N_2$ gas = 1.00 atm

V = Volume of $N_2$ gas = 113 L

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $N_2$ gas = $85^oC=273+85=358K$

Putting values in above equation, we get:

$1.00atm\times 113L=n\times (0.0821L.atm/mol.K)\times 358K$

$n=3.84mol$

Now we have to calculate the moles of sodium azide.

The balanced chemical reaction is,

$2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)$

From the balanced reaction we conclude that

As, 3 mole of $N_2$ produced from 2 mole of $NaN_3$

So, 3.84 moles of $N_2$ produced from $\frac{2}{3}\times 3.84=2.56$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

Molar mass of $NaN_3$ = 65 g/mole

$\text{ Mass of }NaN_3=(2.56moles)\times (65g/mole)=166.4g$

Therefore, the mass of $NaN_3$ required is, 166.4 grams.

3 0

3 years ago