A straight handle for easy use and a wide tip for scooping food
You can consider the density
of the water. Thus, in order to properly measure the mass of a liquid, we can
first get the volume and density of the liquid material or substance.
We can firstly utilize the
formula to get the mass from deriving the set formula of density to mass:
Since density is mass over the
volume,
<span><span>
1. </span>D=m/v</span>
<span><span>
2. </span>We can transmute the formula to m = dv</span>
<span><span>3. </span>Mass is density times the volume</span>
Answer:
0.3229 M HBr(aq)
0.08436M H₂SO₄(aq)
Explanation:
<em>Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.</em>
<em />
Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).
NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)
When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.
![18.45 \times 10^{-3} L NaOH.\frac{0.3500molNaOH}{1LNaOH} .\frac{1molHBr}{1molNaOH} .\frac{1}{20.00 \times 10^{-3} LHBr} =\frac{0.3229molHBr}{1LHBr} =0.3229M](https://tex.z-dn.net/?f=18.45%20%5Ctimes%2010%5E%7B-3%7D%20L%20NaOH.%5Cfrac%7B0.3500molNaOH%7D%7B1LNaOH%7D%20.%5Cfrac%7B1molHBr%7D%7B1molNaOH%7D%20.%5Cfrac%7B1%7D%7B20.00%20%5Ctimes%2010%5E%7B-3%7D%20LHBr%7D%20%3D%5Cfrac%7B0.3229molHBr%7D%7B1LHBr%7D%20%3D0.3229M)
<em>Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?</em>
<em />
Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).
2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)
When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.
![42.18 \times 10^{-3} LNaOH.\frac{0.1000molNaOH}{1LNaOH} .\frac{1molH_{2}SO_{4}}{2molNaOH} .\frac{1}{25.00\times 10^{-3}LH_{2}SO_{4}} =\frac{0.08436molH_{2}SO_{4}}{1LH_{2}SO_{4}} =0.08436M](https://tex.z-dn.net/?f=42.18%20%5Ctimes%2010%5E%7B-3%7D%20LNaOH.%5Cfrac%7B0.1000molNaOH%7D%7B1LNaOH%7D%20.%5Cfrac%7B1molH_%7B2%7DSO_%7B4%7D%7D%7B2molNaOH%7D%20.%5Cfrac%7B1%7D%7B25.00%5Ctimes%2010%5E%7B-3%7DLH_%7B2%7DSO_%7B4%7D%7D%20%3D%5Cfrac%7B0.08436molH_%7B2%7DSO_%7B4%7D%7D%7B1LH_%7B2%7DSO_%7B4%7D%7D%20%3D0.08436M)
Answer:
Mass = 73.73 g
Explanation:
Given data:
Mass of Mg used = 24.48 g
Mass of HCl used = ?
Mass of hydrogen gas produced = 2.04 g
Mass of Magnesium chloride produced = 96.90 g
Solution:
Chemical equation:
Mg + 2HCl → MgCl₂ + H₂
Number of moles of Mg:
Number of moles = mass/ molar mass
Number of moles = 24.48 g/ 24.305 g/mol
Number of moles = 1.01 mol
Now we will compare the moles of Mg with HCl from balance chemical equation.
Mg : HCl
1 ; 2
1.01 : 2/1× 1.01 = 2.02 mol
Mass of HCl react:
Mass = number of moles × molar mass
Mass = 2.02 × 36.5 g/mol
Mass = 73.73 g
Ni = Nickel
SO4 = Sulphate
2 and 3 in the sub-script are the charges.
So , its name is NICKEL SULPHATE