3 years ago

What is the charge of the following polyatomic ions

Chemistry

1 answer:

Veronika [31]3 years ago

3 0

Answer:

Explanation:

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Suppose the concentration of the standard stock solution is 50.04 ppm and that you delivered 4.60 mL of this standard stock solu

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Answer:

9.21 ppm

Explanation:

Considering

$concentration_{working\ solution}\times Volume_{working\ solution}=concentration_{stock\ solution}\times Volume_{stock\ solution}$

Given that:

$concentration_{working\ solution}=?$

$Volume_{working\ solution}=25.00mL$

$Volume_{stock\ solution}=4.60mL$

$concentration_{stock\ solution}=50.04 ppm$

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$concentration_{working\ solution}\times 25.00mL=50.04 ppm\times 4.60mL$

$concentration_{working\ solution}=\frac{50.04\times 4.60}{25.00}\ ppm=9.21\ ppm$

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What isotope is the product of the alpha (*2He) nuclear decay of Radon-222 (222U)?

NikAS [45]

Answer:

ⁿₐX => ²¹⁸₈₄Po

Explanation:

Let ⁿₐX be the isotope.

Thus, the equation can be written as follow:

²²²₈₆Rn —> ⁴₂α + ⁿₐX

Next, we shall determine the value of 'n' and 'a'. This can be obtained as follow:

222 = 4 + n

Collect like terms

222 – 4 = n

218 = n

Thus,

n = 218

86 = 2 + a

Collect like terms

86 – 2 = a

84 = a

Thus,

a = 84

ⁿₐX => ²¹⁸₈₄Po

²²²₈₆Rn —> ⁴₂α + ⁿₐX

²²²₈₆Rn —> ⁴₂α + ²¹⁸₈₄Po

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Hey there!:

Sodium cations and phosphate anions .

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