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erastova [34]
3 years ago
5

I need help with number 17 pls someone pls ???

Chemistry
2 answers:
VMariaS [17]3 years ago
8 0

Answer:

glass cup as it can trap oxygen from getting in and out meaning that the fire can't grow

LenaWriter [7]3 years ago
5 0

Answer:

can you pls say what subject it is pls and thank u

Explanation:

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Write balanced equations for the following reactions
Arte-miy333 [17]

Answer:

See explanation

Explanation:

The shorthand nuclear reaction equations have been given; the first particle in the parentheses is a reactant particle while the second particle is a product particle. These can now be rewritten as the longhand equations as follows;

238/92U + 4/2 He -------> 241/94Pu + 1/0 n

238/92U + 4/2 He ------> 241/94Pu + 1/0 n

14/7N + 4/2 He------> 17/8O + 1/1 p

56/26Fe + 2 4/2 He----> 60/29Cu + 4/2 He

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3 years ago
Does the production of a new gas by a substance represent a chemical property. Explain?
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Yes, because you made a new substance to make a gas.

4 0
2 years ago
The two isotopes of chlorine are LaTeX: \begin{matrix}35\\17\end{matrix}Cl35 17 C l and LaTeX: \begin{matrix}37\\17\end{matrix}C
Kay [80]

<u>Answer:</u> The percentage abundance of _{17}^{35}\textrm{Cl} and _{17}^{37}\textrm{Cl} isotopes are 77.5% and 22.5% respectively.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

Let the fractional abundance of _{17}^{35}\textrm{Cl} isotope be 'x'. So, fractional abundance of _{17}^{37}\textrm{Cl} isotope will be '1 - x'

  • <u>For _{17}^{35}\textrm{Cl} isotope:</u>

Mass of _{17}^{35}\textrm{Cl} isotope = 35 amu

Fractional abundance of _{17}^{35}\textrm{Cl} isotope = x

  • <u>For _{17}^{37}\textrm{Cl} isotope:</u>

Mass of _{17}^{37}\textrm{Cl} isotope = 37 amu

Fractional abundance of _{17}^{37}\textrm{Cl} isotope = 1 - x

Average atomic mass of chlorine = 35.45 amu

Putting values in equation 1, we get:

35.45=[(35\times x)+(37\times (1-x))]\\\\x=0.775

Percentage abundance of _{17}^{35}\textrm{Cl} isotope = 0.775\times 100=77.5\%

Percentage abundance of _{17}^{37}\textrm{Cl} isotope = (1-0.775)=0.225\times 100=22.5\%

Hence, the percentage abundance of _{17}^{35}\textrm{Cl} and _{17}^{37}\textrm{Cl} isotopes are 77.5% and 22.5% respectively.

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