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.
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.
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%
There are 0.35 grams of iron in 350 mg. To convert milligrams (mg) to grams (g), you can divide by 1000.
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.
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.
The molar mass of MgCl2 = 95.211 g/mol
1.39 moles
There are approximately 12.35 ounces in 350 grams.
350 g sample of CO contain 12,49 moles.
That is 12.346 ounces
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.
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.
So far I come up with: HBr (aq) + MgSO3 (s) --> H2SO3 (aq) + MgBr2 (??)
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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%
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.