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butalik [34]
3 years ago
13

Write the formulas for the ionic compounds formed by the following:

Chemistry
1 answer:
PSYCHO15rus [73]3 years ago
7 0

1. KI

2. AlBr₃

3. CsNO₃

4. Al₂(CO₃)₃

Explanation:

1. potassium (K⁺) iodine (I⁻) - KI

2. aluminium (Al³⁺) bromine (Br⁻) - AlBr₃

3. caesium (Cs⁺) nitrate (NO₃⁻) - CsNO₃

4. aluminum (Al³⁺) carbonate (CO₃²⁻)  - Al₂(CO₃)₃

Learn more about:

formulas for the ionic compounds

brainly.com/question/13954262

#learnwithBrainly

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How many moles of na2co3 are necessary to reach stoichiometric quantities with cacl2
lbvjy [14]

0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ is necessary  to reach stoichiometric quantities with cacl2.

<h3>Explanation:</h3>

Based on the reaction

CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

1 mole of CaCl₂ reacts per mole of Na₂CO₃

we have to calculate how many moles of CaCl2•2H2O are present in 1.50 g

  • We must calculate the moles of CaCl2•2H2O using its molar mass (147.0146g/mol) in order to answer this issue.
  • These moles, which are equal to moles of CaCl2 and moles of Na2CO3, are required to obtain stoichiometric amounts.
  • Then, we must use the molar mass of Na2CO3 (105.99g/mol) to determine the mass:

<h3>Moles CaCl₂.2H₂O:</h3>

1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂

Moles Na₂CO₃:

0.0102 moles Na₂CO₃

Mass Na₂CO₃:

0.0102 moles * (105.99g / mol) = 1.08g of Na₂CO₃ are present

Therefore, we can conclude that 0.0102 moles Na₂CO₃  is necessary.to reach stoichiometric quantities with cacl2.

To learn more about stoichiometric quantities visit:

<h3>brainly.com/question/28174111</h3>

#SPJ4

7 0
2 years ago
What is the mass of 3.35 mol Hg(IO3)2? 1,700 g 1,840 g 1,960 g 2,110 g
vichka [17]

Answer:- 1840 g.

Solution:- We have been given with 3.35 moles of  and asked to calculate it's mass.

To convert the moles to grams we multiply the moles by the molar mass of the compound. Molar mass of the compound is the sum of atomic masses of all the atoms present in it.

molar mass of  = atomic mass of Hg + 2(atomic mass of I) + 6(atomic mass of O)

= 200.59+2(126.90)+6(16.00)

= 200.59+253.80+96.00

= 550.39 gram per mol

Let's multiply the given moles by the molar mass:

3.35mol(\frac{550.39g}{1mol})

= 1843.8 g

Since, there are three sig figs in the given moles of compound, we need to round the calculated my to three sig figs also. So, on rounding off to three sig figs the mass becomes 1840 g.



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