Answer:
6× 10⁸
Explanation:
We need to find the multiplication of 2 x 10⁴ by 3 x 10⁴.
2 x 10⁴ × 3 x 10⁴
= (3 × 2) x 10⁴ x 10⁴
= 6 x 10⁴ x 10⁴
= 6 × 10⁴⁺⁴
= 6× 10⁸
Hence, the required answer is 6× 10⁸.
, a crystal structure with a short symmetrical hydrogen bond.
<h3>What is Classical bonding?</h3>
Classical models of the chemical bond. By classical, we mean models that do not take into account the quantum behaviour of small particles, notably the electron. These models generally assume that electrons and ions behave as point charges which attract and repel according to the laws of electrostatics.
Sodium dihydrogen phosphate is a derivative composed of glycerol derivatives formed by reacting mono and diglycerides that are derived from edible sources with phosphorus pentoxide followed by neutralization with sodium carbonate.
Bonding in
, a crystal structure with a short symmetrical hydrogen bond. Sodium dihydrogen phosphate () is monoclinic, space group P2,/c, with a= 6.808 (2), b= 13.491 (3), c=7.331 (2)/~, fl=92.88 (3) ; Z=8.
Learn more about the bond here:
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The balanced equation for the reaction is as follows
Cu₂O + 2HCl ---> 2CuCl + H₂O
Molar ratio of Cu₂O to CuCl is 1:2
mass of Cu₂O reacted - 73.5 g
Number of moles of Cu₂O reacted - 73.5 g / 143 g/mol = 0.51 mol
According to the molar ratio,
when 1 mol of Cu₂O reacts then 2 mol of CuCl is formed
therefore when 0.51 mol of Cu₂O reacts then - 2 x 0.51 mol of CuCl is formed
number of CuCl moles formed - 1.02 mol
mass of CuCl formed - 1.02 mol x 99 g/mol = 101 g
mass of CuCl formed is 101 g
Answer:
Atomic Number - 11. Sodium
Answer:
[HF]₀ = 0.125M
Explanation:
NaOH + HF => NaF + H₂O
Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3. This is 0.089M NaF and 0.001M HF remaining.
=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.
HF ⇄ H⁺ + F⁻
C(eq) [HF] 10⁻³M 0.089M (<= soln after adding 20ml 0.200M NaOH)
Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka
[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M
