<h3>
Answer:</h3>
2.624 g
<h3>
Explanation:</h3>
The equation for the reaction is given as;
- CuSO₄(aq) + 2NaOH(aq) → Cu(OH)₂(s) + Na₂SO₄(aq)
- Volume of CuSO₄ as 46.0 mL;
- Molarity of CuSO₄ as 0.584 M
We are required to calculate the mass of Cu(OH)₂ precipitated
- We are going to use the following steps;
<h3>Step 1: Calculate the number of moles of CuSO₄ used</h3>
Molarity = Number of moles ÷ Volume
To get the number of moles;
Moles = Molarity × volume
= 0.584 M × 0.046 L
= 0.0269 moles
<h3>
Step 2: Calculate the number of moles of Cu(OH)₂ produced </h3>
- From the equation 1 mole of CuSO₄ reacts to give out 1 mole of Cu(OH)₂
- Therefore; Mole ratio of CuSO₄ to Cu(OH)₂ is 1 : 1.
Thus, Moles of CuSO₄ = Moles of Cu(OH)₂
Hence, moles of Cu(OH)₂ = 0.0269 moles
<h3>
Step 3: Calculate the mass of Cu(OH)₂</h3>
To get mass we multiply the number of moles with the molar mass.
Mass = Moles × Molar mass
Molar mass of Cu(OH)₂ is 97.561 g/mol
Therefore;
Mass of Cu(OH)₂ = 0.0269 moles × 97.561 g/mol
= 2.624 g
Thus, the mass of Cu(OH)₂ that will precipitate is 2.624 g
PV = n RT
P: pressure =10atm
V volume
n number of mole = 35.8 moles
R universal gas constant = 0.082
T: The temperature= 70°C= 343.15 Kelvin
V= (n RT) / P = 35.8 x 0.082 x 343.15 / 10 = 100.7 ≈ 101 L
V = 101L
Answer:
<h3>The answer is 9.0 kg/L</h3>
Explanation:
The density of a substance can be found by using the formula
From the question
mass = 1.8 kg
volume = 0.2 L
We have
We have the final answer as
<h3>9.0 kg/L</h3>
Hope this helps you
Answer:
i am not sure tho
Explanation:
Cd ²+ + F¹- = CdF2
F fluor is a non metal so he takes the electons that Cd gives so if Vd has to give 2 electrons and F can take only 1, there has to be 2 F atoms so all the electrona can be neutral and in ionisation
Answer:
Friction and Automobile Tires. ... On dry surfaces you might get as high as 0.9 as a coefficient of friction, but driving them on wet roads would be dangerous since the wet road coefficient might be as low as 0.1
Explanation:
hope this helps
