Answer:
concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm
Explanation:
ppm (parts per million) concentration is defined as the mass (in milligrams) of a substance dissolved in one liter of solution.
In our case we have:
mass of MgBr₂ = 12.41 g
volume of water (which is equal to the final solution volume) = 2.55 L
Now we devise the following reasoning:
if 12.41 g of MgBr₂ are dissolved in 2.55 L of water
then X g of MgBr₂ are dissolved in 1 L of water
X = (1 × 12.41) / 2.55 = 4.867 g of MgBr₂
if in 184 g (1 mole) of MgBr₂ we have 160 g of Br⁻
then in 4.867 g of MgBr₂ we have Y g of Br⁻
Y = (4.867 × 160) / 184 = 4.232 g of bromide (Br⁻)
4.232 g of bromide (Br⁻) = 4234 mg of bromide (Br⁻)
concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm
Answer:
Explanation:
We have to start with the <u>reaction</u>:
We have the same amount of atoms on both sides, so, we can continue. The next step is to find the <u>number of moles</u> that we have in the 110.0 g of carbon dioxide, to this, we have to know the <u>atomic mass of each atom</u>:
C: 12 g/mol
O: 16 g/mol
Mg: 23.3 g/mol
If we take into account the number of atoms in the formula, we can calculate the <u>molar mass</u> of carbon dioxide:
In other words: . With this in mind, we can calculate the moles:
Now, the <u>molar ratio</u> between carbon dioxide and magnesium carbonate is 1:1, so:
With the molar mass of (. With this in mind, we can calculate the <u>grams of magnesium carbonate</u>:
I hope it helps!
Answer:
carbon mass = 12.01g/mol
hydrogen mass = 1.01g/mol
4 carbon atoms and 10 hydrogen so
12.01 x 4 + 1.01 x 10
48.04g/mol + 10.10g/mol
= 58.14g/mol
Answer: weigh is m = n × M = 2.87 mol × 58.44 g/mol
Explanation: mass = amount of substance × molar mass
M((NaCl) = 22.99 +35.45
15396 g
tell me if its correct