All categories

2 years ago

If each fission reaction in a nuclear reactor emits three neutrons, how many neutrons are produced from three fission reactions?

Chemistry

1 answer:

Andrej [43]2 years ago

3 0

answer:

I think nine

explanation:

Nuclear fission : It is defined as the splitting of bigger nuclei into two smaller nuclei. During this process neutrons and some energy also released.

You might be interested in

In the following equation, which element has been reduced?

Mila [183]

The answer is A. you sre correct!

4 0

3 years ago

Be sure to answer all parts.The combustion of a 40.0−g gaseous mixture of H2 and CH4 releases 3766 kJ of heat. Calculate the amo

Svetach [21]

Answer:H2=11.4g

CH4=28.6g

Explanation:The complete combustion of the two gases can be represented by a balanced reaction below

1. CH4 +2O2___CO2+2H2O

2.2H2+O2___2H2O

Combining the two we have CH4 +2H2+3O2___

CO2+4H2O

Since the mixture contains 40gof CH4 and 2, therefore 20g of CH4 and 8g of H2 combines.

Calculated from their molecular Mass i.e CH4 12+4×2)=20 and 2H2= 2×2×2=8g

Mass of CH4=20/28×40=28.6g

2H2=8/28×40=11.4g

7 0

3 years ago

How many moles of NaCl are needed to make 10.0L of a 5 M solution of<br> Salt water?*

vesna_86 [32]

Answer: 50 mol

Explanation:

$Molarity=\frac{Moles -of -solute}{liters -of -solution}$

$M=\frac{mol}{L}$

$mol=M*L$

$mol=5*10\\mol=50mol$

3 0

3 years ago

State the importance of oxygen for germination process

tamaranim1 [39]

Plants need oxygen to survive.

3 0

3 years ago

The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure

Yuki888 [10]

Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate $\Delta H_{vap}$ of the reaction, we use clausius claypron equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = vapor pressure at temperature $T_1$

$P_2$ = vapor pressure at temperature $T_2$

$\Delta H_{vap}$ = Enthalpy of vaporization

R = Gas constant = 8.314 J/mol K

1) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =363 K

$T_2$ = final temperature =373 K

$P_2=1 atm, P_1=?$

Putting values in above equation, we get:

$\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]$

$P_1=0.69671 atm \approx 0.6970 atm$

The vapor pressure at temperature 363 K is 0.6970 atm

2) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =373 K

$T_2$ = final temperature =383 K

$P_1=1 atm, P_2?$

Putting values in above equation, we get:

$\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]$

$P_2=1.4084 atm \approx 1.410 atm$

The vapor pressure at 383 K is 1.410 atm

8 0

3 years ago