All categories

3 years ago

HELP me, this question not even really hard actually...

Chemistry

1 answer:

zheka24 [161]3 years ago

8 0

This is a hypothesis question. This is an educated guess, basically. In my opinion, I think the tails side would hold more water.

You might be interested in

The half-life of cesium-137 is 30 years. Suppose we have a 180 mg sample. Find the mass that remains after t years. Let y(t) be

Stels [109]

Explanation:

1) Initial mass of the Cesium-137= $N_o$ = 180 mg

Mass of Cesium after time t = N

Formula used :

$N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

Half life of the cesium-137 = $t_{1/2}=30 years[p/tex]where,
[tex]N_o$ = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

$t_{\frac{1}{2}}$ = half life of the isotope

$\lambda$ = rate constant

$N=N_o\times e^{-(\frac{0.693}{t_{1/2}})\times t}$

Now put all the given values in this formula, we get

$N=180mg\times e^{-\frac{0.693}{30 years}\times t}$

Mass that remains after t years.

$N=180 mg\times e^{0.0231 year^{-1}\times t}$

Therefore, the parent isotope remain after one half life will be, 100 grams.

t = 70 years

$N_o=180 mg$

$t_{1/2}= 30 yeras$

$N=180mg\times e^{-\frac{0.693}{30 years}\times 70 years}$

N = 35.73 mg

35.73 mg of cesium-137 will remain after 70 years.

$N_o=180 mg$

$t_{1/2}= 30 yeras$

N = 1 mg

t = ?

$1 mg =180mg\times e^{-\frac{0.693}{30 years}\times t}$

$\frac{-30 year}{0.693}\times \ln \frac{1 mg}{180 mg}=t$

t = 224.80 years ≈ 225 years

After 225 years only 1 mg of cesium-137 will remain.

7 0

3 years ago

What type of reaction is represented by the following? A + B ⟶ AB

Elden [556K]

Answer:

Synthesis Reaction

Explanation:

It is a synthesis Reaction because it is taking little reactants and forming a big product!

6 0

2 years ago

Propane gas reacts with oxygen according to this balanced equation: C subscript 3 H subscript 8 space (g )space plus space 5 spa

Misha Larkins [42]

Explanation:

The balanced chemical equation of the reaction is:

$C_3H_8(g)+ 5O_2 (g)->3CO_2(g)+4H_2O(g)$

From the balanced chemical equation,

1 mole of propane forms ------ 3 mol. of $CO_2$ gas.

The molar mass of propane is 44.1 g/mol.

One mole of any gas at STP occupies --- 22.4 L.

Hence, 44 g of propane forms (3x22.4 L=) 67.2 L of CO2 gas at STP.

Answer:

Thus, 67.2 L of CO2 is formed at STP.

4 0

3 years ago

In the combustion of ethane, what is/are the reactant(s)?

aleksley [76]

Reactant are those that combine or reacts to give products !!

so in combustion of ethane ; ethane and o2 are reactants so ... your answer is B !!

5 0

3 years ago

Read 2 more answers

Explain how are elements classified and provide an example

swat32

By nonmetals, metals, and gases

4 0

3 years ago