Answer:
Empirical formula is CaSO₄.
Explanation:
Given data:
Percentage of calcium =29.44%
Percentage of sulfur = 23.55%
Percentage of oxygen = 47.01%
Empirical formula = ?
Solution:
Number of gram atoms of Ca = 29.44 / 40 = 0.74
Number of gram atoms of S = 23.55 / 32 = 0.74
Number of gram atoms of O = 47.01 / 16 = 3
Atomic ratio:
Ca : S : O
0.74/0.74 : 0.74/0.74 : 3/0.74
1 : 1 : 4
Ca : S : O = 1 : 1 : 4
Empirical formula is CaSO₄.
Answer:
they both gain one electron
Explanation:
they are both halogens and they each have one unpaired electron so that unpaired electron join together to make them form a bond
Answers:
(a) 1s² 2s²2p³; (b) 1s² 2s²2p⁶ 3s²3p⁶ 4s²3d²; (c) 1s² 2s²2p⁶ 3s²3p⁵
Step-by-step explanation:
One way to solve this problem is to add electrons to the orbitals one-by-one until you have added the required amount.
Fill the subshells in the order listed in the diagram below. Remember that an s subshell can hold two electrons, while a p subshell can hold six, and a d subshell can hold ten.
(a) <em>Seven electrons
</em>
1s² 2s²2p³
There are two electrons in the 2s subshell and three in the 2p subshell. The remaining two electrons are in the inner 1s subshell.
(b) <em>22 electrons
</em>
1s² 2s²2p⁶ 3s²3p⁶ 4s²3d²
There are two electrons in the 4s subshell and two in the 2p subshell. The remaining 18 electrons are in the inner subshells.
(c) <em>17 electrons</em>
1s² 2s²2p⁶ 3s²3p⁵
There are two electrons in the 3s subshell and five in the 2p subshell. The remaining 10 electrons are in the inner subshells.
Answer:
The ΔHrxn for the above equation = 179 kJ/mol
Explanation:
The reaction bond enthalpies are for the reactant;
3 × N-H = 3 × 390 = 1,170 kJ/mol
2 × O=O = 2 × 502 = 1004 kJ/mol
The reaction bond enthalpies are for the product;
3 × N-O = 3 × 201 = 603 kJ/mol
3 × O-H = 3 × 464 = 1,392 kJ/mol
The ΔHrxn for the above equation is therefore;
ΔHrxn = 1,170 + 1,004 - (603 + 1,392) = 179 kJ/mol
Answer: The molarity of the malonic acid solution is 0.08335 M
Explanation:
To calculate the molarity of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
Thus the molarity of the malonic acid solution is 0.08335 M
