Answer:
Hello friends
Explanation:
<h3>For a given principal quantum number for or n, the corresponding angular quantum number or is equivalent to a range between 0 and( n-1)</h3>
<h3>This means that the angular quantum number for a principal quantum number of 2 is equivalent to.</h3>
<h3>1 = 0 - > (n - 1) = 0 - > (2 - 1) = 0 - > 1</h3>
<h3>Hope it's helpfully. </h3>
1. Decreases by 4. (B)
2. The atomic number changes. (B)
3. 56/26 Fe. (C)
4. Potassium-40;t1/2=25 days. (B)
5. Takes place in the upper atmosphere. (A)
Answer:
The volume of a sample of gas (2.49 g) was 752 mL at 1.98 atm and 62∘C 62 ∘ C .
Answer:
2.1056L or 2105.6mL
Explanation:
We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:
Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol
Mass of Na2CO3 = 10g
Mole of Na2CO3 =.?
Mole = mass /molar mass
Mole of Na2CO3 = 10/106
Mole of Na2CO3 = 0.094 mole
Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:
Na2CO3 + 2HCl —> 2NaCl + H2O + CO2
From the balanced equation above,
1 mole of Na2CO3 reacted to produce 1 mole of CO2.
Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.
Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:
1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.
Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L
Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL
