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There are 22,9.10e21 anions.

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7y ago
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5mo ago

There are 0.021 moles of MgBr2 in 3.50 g. Since MgBr2 contains 2 bromide ions per formula unit, there are 0.042 moles of Br- ions. So, there are 0.042 * 6.022 x 10^23 = 2.53 x 10^22 Br- ions in 3.50 g of MgBr2.

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Q: How many anions are there in 3.50 g of MgBr2?
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How many anions are there in 2.50 g of MgBr2?

There are a total of 2 moles of anions in 2.50 g of MgBr2. Each formula unit of MgBr2 contains 2 moles of anions (Br-). The molar mass of MgBr2 is 184.113 g/mol, so 2.50 g is equivalent to 0.0136 moles, and therefore 0.0136 moles * 2 moles = 0.0272 moles of anions.


How many moles of anions are in 37.4 g of AlF3?

To find the number of moles of anions in 37.4 g of AlF3, we first need to calculate the molar mass of AlF3. Aluminum (Al) has a molar mass of 26.98 g/mol, and fluorine (F) has a molar mass of 19.00 g/mol. So, the molar mass of AlF3 is 26.98 + (3 * 19.00) = 83.98 g/mol. Therefore, the number of moles of AlF3 in 37.4 g is 37.4 g / 83.98 g/mol = 0.445 moles. Since there are 3 fluoride ions (anions) in each formula unit of AlF3, the number of moles of anions in 37.4 g of AlF3 is 0.445 moles * 3 = 1.34 moles.


What is the percent composition of a compound containing 32.0 g of bromine and 4.9 g of magnesium?

To find the percent composition, first calculate the molar mass of the compound, then divide the mass of each element by the molar mass and multiply by 100 to get the percent composition. The molar mass of MgBr2 is 184.11 g/mol. Percent composition of bromine: (32.0 g Br / 184.11 g) x 100 = 17.38% Percent composition of magnesium: (4.9 g Mg / 184.11 g) x 100 = 2.67%


How many grams of iron are in 350 Mg of iron?

There are 0.35 grams of iron in 350 mg. To convert milligrams (mg) to grams (g), you can divide by 1000.


What is the mass of carbon found in 350 grams of C2H6?

To find the mass of carbon in C2H6, we first need to calculate the molar mass of C2H6. Carbon has a molar mass of 12 g/mol, and hydrogen has a molar mass of 1 g/mol. The molar mass of C2H6 is (212) + (61) = 30 g/mol. The mass of carbon in 350 grams of C2H6 is then (2*12)/30 * 350 = 140 grams.

Related questions

How many anions are there in 2.50 g of MgBr2?

There are a total of 2 moles of anions in 2.50 g of MgBr2. Each formula unit of MgBr2 contains 2 moles of anions (Br-). The molar mass of MgBr2 is 184.113 g/mol, so 2.50 g is equivalent to 0.0136 moles, and therefore 0.0136 moles * 2 moles = 0.0272 moles of anions.


What is the mass of 7 moles of MgBr2?

The molar mass of MgCl2 = 95.211 g/mol


How many moles of anions are in 38.8 g of AlF3?

1.39 moles


How many ounces in 350 gramms?

There are approximately 12.35 ounces in 350 grams.


How many moles of carbon monoxide are in 350 g sample?

350 g sample of CO contain 12,49 moles.


How many oz are 350 g?

That is 12.346 ounces


How many moles of anions are in 37.4 g of AlF3?

To find the number of moles of anions in 37.4 g of AlF3, we first need to calculate the molar mass of AlF3. Aluminum (Al) has a molar mass of 26.98 g/mol, and fluorine (F) has a molar mass of 19.00 g/mol. So, the molar mass of AlF3 is 26.98 + (3 * 19.00) = 83.98 g/mol. Therefore, the number of moles of AlF3 in 37.4 g is 37.4 g / 83.98 g/mol = 0.445 moles. Since there are 3 fluoride ions (anions) in each formula unit of AlF3, the number of moles of anions in 37.4 g of AlF3 is 0.445 moles * 3 = 1.34 moles.


How many moles of anions are in 33.4 g of AlF3?

To find the number of moles of anions in AlF3, you need to calculate the molar mass of AlF3 (83.98 g/mol) and then determine the moles of AlF3 in 33.4 g (0.398 moles). Since there are 3 F^- ions per formula unit of AlF3, there are 3 times the number of moles of AlF3 in moles of F^- ions, which is 1.194 moles.


What is the balance formula for Hydrobromic acid and magnesium sulfite?

So far I come up with: HBr (aq) + MgSO3 (s) --> H2SO3 (aq) + MgBr2 (??)


How many g of dextrose are found in 350 ml of D10W?

35


What is the percent composition of a compound containing 32.0 g of bromine and 4.9 g of magnesium?

To find the percent composition, first calculate the molar mass of the compound, then divide the mass of each element by the molar mass and multiply by 100 to get the percent composition. The molar mass of MgBr2 is 184.11 g/mol. Percent composition of bromine: (32.0 g Br / 184.11 g) x 100 = 17.38% Percent composition of magnesium: (4.9 g Mg / 184.11 g) x 100 = 2.67%


How many teaspoons in 350g?

This question is impossible to answer without knowing what product is to be measured. For example, you'll get many more TBL from 350 g of flour than you would from 350 g of salt.