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1 year ago

Please help!! Balancing Nuclear Equations

Chemistry

1 answer:

sineoko [7]1 year ago

8 0

The missing part of the equation is found to be 4/2He. Option A

<h3>What are nuclear equations?</h3>

The term nuclear equations have to do with the type of equation in which one type of nucleus is transformed into another sometimes by the bombardment or loss of a particle.

Now the full equation ought to be written as 7/3Li + 1/1H -----> 4/2He + 4/2He. This is because the total mass on the left is 8 and the total charge on the left is 4.

Learn more about nuclear equations:brainly.com/question/19752321

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Reactants are<br> What you start with<br> What you end with

Naya [18.7K]

Answer:

Reactants are what you start with

Explanation:

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3 years ago

At which temperature would a can of soda be most fizzy when it is opened? 30°C 40°C 20°C 10°C

Helga [31]

The answer to your question is 40° because freezing temp for a liquid is 32°

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2 years ago

Read 2 more answers

If 200. g of water at 20°C absorbs 41 840 J of energy, what will its final temperature be? (Specific Heat of water is 4.184 J/g*

Elena-2011 [213]

Answer: The final temperature will be $70^0C$

Explanation:

To calculate the specific heat of substance during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed =41840 J

c = specific heat = $4.184J/g^0C$

m = mass of water = 200 g

$T_{final}$ = final temperature =?

$T_{initial}$ = initial temperature = $20^0C$

Now put all the given values in the above formula, we get:

$41840J=200g\times 4.184J/g^0C\times (T_{final}-20)^0C$

$T_{final}=70^0C$

Thus the final temperature will be $70^0C$

3 0

2 years ago

Argon (Ar) and helium (He) are initially in separate compartments of a container at 25°C. The

love history [14]

Answer:

(a) $V_B=11.68L$

(b) $x_{He}=0.533$

Explanation:

Hello,

In this case, since the both gases behave ideally, with the given information we can compute the moles of He in A:

$n_A=\frac{0.082\frac{atm*L}{mol*K}*298K}{1.974 atm*6.00L}=2.063mol$

Thus, since the final pressure is 3.60 bar, we can write:

$P=x_{Ar}P_A+x_{He}P_B\\\\P=\frac{n_{Ar}}{n_{Ar}+n_{He}} P_A+\frac{n_{He}}{n_{Ar}+n_{He}} P_B\\\\3.60bar=\frac{2.063mol}{2.063mol+n_{He}} *2.00bar+\frac{n_{He}}{2.063mol+n_{He}} *5.00bar$

The moles of helium could be computed via solver as:

$n_{He}=2.358mol$

Or algebraically:

$3.60bar=\frac{1}{2.063mol+n_{He}} *(4.0126+5.00*n_{He})\\\\7.314+3.60n_{He}=4.013+5.00*n_{He}\\\\7.314-4.013=5.00*n_{He}-3.60n_{He}\\\\n_{He}=\frac{3.3}{1.4}=2.358mol$