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NISA [10]
3 years ago
6

Balanced nuclear equation forces the alpha decay Po-210

Chemistry
1 answer:
Serjik [45]3 years ago
8 0

Answer:

_{84}^{210}\text{Po} \longrightarrow \, _{82}^{206}\text{Pb} \, + \, _{2}^{4}\text{He}

Explanation:

Your unbalanced nuclear equation is:

_{84}^{210}\text{Po} \longrightarrow \, _{x}^{y}\text{Z} + \, _{2}^{4}\text{He}

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.  

Then

 84 = x + 2, so x =  84 - 2 =   82

210 = y + 4, so y = 210 - 4 = 206

Element 82 is lead, so the nuclear equation becomes

_{84}^{210}\text{Po} \longrightarrow \, _{82}^{206}\text{Pb} \, + \, _{2}^{4}\text{He}

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since mass and volume is known, density can be calculated
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0.1 dm³.

Explanation:

To obtain the value of 100 cm³ in dm³, do the following:

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3 years ago
A gas occupies 1.00cm^3 at STP. What volume does it occupy at 710.0 mm Hg and 55.0°C
Lera25 [3.4K]

Answer:

Final volume is 1.29cm³

Explanation:

Using combined gas law:

P₁V₁/T₁ = P₂V₂/T₂

<em>Where P is pressure, V volume and T absolute temperature of 1, initial state and 2, final state of a gas.</em>

<em />

We can solve the volume of the gas:

P₁ = 1atm at STP

V₁ = 1cm³

T₁ = 273.15K at STP

P₂ = 710mmHg * (1atm / 760mmHg) = 0.9342atm

V₂ = ?

T₂ = 55.0°C + 273.15K = 328.15K

Replacing:

1atm*1cm³/273.15K = 0.9342atm*V₂/328.15K

1.29cm³ = V₂

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What is the law of conservation of mass
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