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Gnom [1K]
3 years ago
11

What is the stoichiometric ratio between BaCl2 and NaCl

Chemistry
1 answer:
bixtya [17]3 years ago
7 0
<span>BaCl2+Na2SO4---->BaSO4+2NaCl There is 1.0g of BaCl2 and 1.0g of Na2SO4, which is the limiting reagent? "First convert grams into moles" 1.0g BaCl2 * (1 mol BaCl2 / 208.2g BaCl2) = 4.8 x 10^-3 mol BaCl2 1.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 7.0 x 10^-3 mol Na2SO4 (7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2 ) = 1.5 mol Na2SO4 / mol BaCl2 "From this ratio compare it to the equation, BaCl2+Na2SO4---->BaSO4+2NaCl" The equation shows that for every mol of BaCl2 requires 1 mol of Na2SO4. But we found that there is 1.5 mol of Na2SO4 per mol of BaCl2. Therefore, BaCl2 is the limiting reagent.</span>
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Indicate the number of significant figures in the measured number 0.0020 <br> A. 4<br> B. 5<br> C. 2
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<u>Answer:</u>

The correct answer option is C. 2.

<u>Explanation:</u>

We are given the number '0.0020' and we are to indicate the number of significant figures in the given measured number.

According to the rules of significant figures, numbers that are non-zero, zeros between any two significant numbers and the ending zeros in the decimal position are categorized as significant figures.

Since there is one non-zero number and one ending zero in the decimal position, therefore 0.0020 has 2 significant figures.

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You can answer this question by only searching the element in the periodic table.

The atomic number of iodine, I, is 53. It is placed in the column 17 (this is the Group) and row 5 (this is the Period).

The conclusion is that the iodine is located in Period 5, Group 17, and is classified as a nonmetal.
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