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2 years ago

How do I turn 2 s into one in the products

Chemistry

1 answer:

-BARSIC- [3]2 years ago

5 0

Make the equation equal on each side.

With number 1 it says the left side of the arrow has 1 Sn, 2 S, and 2 O.

The right has 1 Sn, 4 O, and 1 S.

So you have to balance the equation by putting numbers in front of the compounds.

SnS2 + O2 = SnO2 + SO2

SnS2 + 3 O2 = SnO2 + 2 SO2

Now there is 1 Sn, 2 S. and 6 O on each side. The large number in front of the compounds essentially multiplies it by that the subscript of that number. So with O2 it says there are 2 Oxygens. So they come in pairs. If you get 3 pairs of Oxygens there will be 6.

The same process goes with the rest of them too. Sometimes it takes a few guesses. Just figure them out like an algebra problem. Put the equal sign between them like I did. It maked it easier for me.

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mojhsa [17]

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3 years ago

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The initial activity of 37Ar is 8540 disintegrations per minute. After 10.0 days, the activity is 6990 disintegrations per minut

mylen [45]

Answer:

Approximately 3318 disintegrations per minute.

Explanation:

The activity $A$ of a radioactive decay at time $t$ can be found with the following equation:

$\displaystyle A(t) = A_0 \cdot \mathrm{e}^{-\lambda\cdot t}$ .

In this equation,

$\mathrm{e}$ is the natural base. $\mathrm{e}\approx 2.71828$ .
$A_0$ is the initial activity of the decay. For this question, $A_0 = \rm 8540\; min^{-1}$ .
The decay constant $\lambda$ of this sample needs to be found.

The decay constant here can be found using the activity after 10 days. As long as both times are in the same unit (days in this case,) conversion will not be necessary.

$A(10) = A_0\cdot \mathrm{e}^{\rm -\lambda \times 10\;day}= (\mathrm{8540\; min^{-1}})\cdot \mathrm{e}^{\rm -\lambda \times 10\;day}$ .

$A(10) = \rm 6690\; min^{-1}$ .

$\displaystyle \frac{\rm 6690\; min^{-1}}{\mathrm{8540\; min^{-1}}} = \mathrm{e}^{\rm -\lambda \times 10\;day}$

Apply the natural logarithm to both sides of this equation.

$\displaystyle \ln{\left(\frac{\rm 6690\; min^{-1}}{\mathrm{8540\; min^{-1}}}\right)} = \ln{\left(\mathrm{e}^{\rm -\lambda \times 10\;day}\right)}$ .

$\displaystyle \rm -\lambda \cdot (10\;day) = \ln{\left(\frac{\rm 6690\; min^{-1}}{\mathrm{8540\; min^{-1}}}\right)}$ .

$\displaystyle \rm \lambda= \rm \frac{\displaystyle \ln{\left(\frac{\rm 6690\; min^{-1}}{\mathrm{8540\; min^{-1}}}\right)}}{-10 \; day} \approx 0.0200280\; day^{-1}$ .

Note that the unit of the decay constant $\lambda$ is $\rm day^{-1}$ (the reciprocal of days.) The exponent $-\lambda \cdot t$ should be dimensionless. In other words, the unit of $t$ should also be days. This observation confirms that there's no need for unit conversion as long as the two times are in the same unit.

Apply the equation for decay activity at time $t$ to find the decay activity after 47.2 days.

$\displaystyle \begin{aligned}A(t)& = A_0 \cdot \mathrm{e}^{-\lambda\cdot t}\\&\approx \rm \left(8540\; min^{-1}\right)\cdot \mathrm{e}^{-0.0200280\; day^{-1}\times 47.2\;day}\\&\approx \rm 3318\; min^{-1}\end{aligned}$ .

By dimensional analysis, the unit of activity here should also be disintegrations per minute. The activity after 47.2 days will be approximately 3318 disintegrations per minute.

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3 years ago

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3 years ago

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LenaWriter [7]

Answer:

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Explanation:

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