Viewed from afar, the wave function below may seem smooth and analog<span>, but when you look closely there are tiny discrete steps as the </span>signal<span> tries to approximate values: That's the big </span>difference between analog and digital<span> waves. </span>Analog<span> waves are smooth and continuous, </span>digital<span> waves are stepping, square, and discrete.</span>
To find the temperature in the problem, we apply the ideal gas law, PV=nRT where R=8.314 Pam3/mol K. Substituting the given, T= 153,000 Pa*1.5x10^-4 m3/ [(0.75 mol)(<span>8.314 Pam3/mol K)]. The temperature is equal to 3.68 kelvin. </span>
Since this is Argon it would be a noble gas
<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
I really do not want you to get it wrong but i will go with nitride ion, oxide ion, sodium ion, magnesium ion
