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Reptile [31]
3 years ago
10

scandium47 has a half-life of 35s. suppose you have a 45g sample of scadium 47 how much of the sample remains unchanged after 14

0 seconds
Chemistry
1 answer:
Mars2501 [29]3 years ago
4 0

 The much  of the sample that would remain  unchanged  after 140 seconds is 2.813 g

Explanation

Half life  is time taken for the quantity  to reduce  to half its original value.

if the half life  for Scandium  is 35 sec, then the number  of half life in 140 seconds

=140 sec/ 35 s = 4 half life

Therefore 45 g after first half life = 45 x1/2 =22.5 g

               22.5 g after second half life = 22.5 x 1/2 =11.25 g

            11.25 g after third half life = 11.25 x 1/2 = 5.625 g

             5.625 after  fourth half life = 5.625 x 1/2 = 2.813

therefore 2.813 g  of Scandium 47 remains  unchanged.

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If the concentration of the HCl used in your titration was 1.00 M and you used 23.68 mL to reach the endpoint, calculate the con
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Answer: The concentration of the OH-, CB = 0.473 M.

Explanation:

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Using titration equation of formula

CAVA/CBVB = NA/NB

Where NA is the number of mole of acid = 2 (from the balanced equation of reaction)

NB is the number of mole of base = 1 (from the balanced equation of reaction)

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1×23.65/CB×25 = 2/1

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6 0
3 years ago
When methane ( CH4 ) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reac
slega [8]
<h2>Answer:</h2>1.33*10^{-2}grams

<h2>Explanations</h2>

The complete balanced equation for the given reaction is expressed as;

CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)

Given the following parameters

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Determine the moles of methane

\begin{gathered} moles\text{ of CH}_4=\frac{mass}{molar\text{ mass}} \\ moles\text{ of CH}_4=\frac{0.0059}{16.04} \\ moles\text{ of CH}_4=0.000368moles \end{gathered}

According to stoichimetry, 1 mole of methane produces 2 moles of water, hence the moles of water required will be:

\begin{gathered} moles\text{ of H}_2O=\frac{2}{1}\times0.000368 \\ moles\text{ of H}_2O=0.000736moles \end{gathered}

Determine the mass of water produced

\begin{gathered} Mass\text{ of H}_2O=moles\times molar\text{ mass} \\ Mass\text{ of H}_2O=0.000736\times18.02 \\ Mass\text{ of H}_2O=0.0133grams=1.33\times10^{-2}grams \end{gathered}

Therefore the mass of water produced from the complete combustion of 5.90×10−3 g of methane is 1.33 * 10^-2grams

5 0
1 year ago
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