Answer:
The answer is
<h2>155 g</h2>
Explanation:
The mass of a substance when given the density and volume can be found by using the formula
<h3>mass = Density × volume</h3>
From the question
volume of bromine = 50 mL
density = 3.10 g/cm³
It's mass is
mass = 50 × 3.10
We have the final answer as
<h3>155 g</h3>
Hope this<u> </u>helps you
Answer:

Explanation:
Hello there!
In this case, for these problems about collecting a gas over water, we must keep in mind that once the gas has been collected, the total pressure of the system is given by the atmospheric pressure, in this case 1.01 atm. Next, since we also have water in the mixture, we can write the following equation:

Thus, by solving for the pressure of nitrogen and using consistent units, we obtain:

Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Explanation:
the molar mass of the compound is 1502g/mol
The reaction between the magnesium, Mg, and the hydrochloric acid, HCl is given in the equation below,
Mg + 2HCl --> H2 + MgCl2
The number of moles of HCl that is needed for the reaction is calculated below.
n = (0.4681 g Mg)(1 mol Mg/24.305 g Mg)(2 mol HCl/1 mol Mg)
n = 0.0385 mols HCl
From the given concentration, we calculate for the required volume.
V = 0.0385 mols HCl/(0.650 mols/L)
V = 0.05926 L or 59.26 mL
<em>Answer: 59.26 mL of HCl</em>