C = vf
c stands for the speed of waves (which is a constant that is 3 x 10^8)
v stands for the wavelength (which is given)
f stands for frequency (what we are solving for)
3 x 10^8 = (1.08 x 10^-6)f
Divide both sides by the given wavelength
f = 2.78 * 10^14 seconds
Answer:
c. F1-
Explanation:
In this chemical reaction the expression is:
HF + NaF → NaHF2
The ion that always keep the negative charge is the fluorine with a -1, if in this mixture there is more positive ions (H1+) the negative ion (F1-) will join with them.
Remember that also the Cl1- will be free, but the fluorine is more reactive than the fluorine.
You may find bellow the balanced chemical equations.
Explanation:
Molecular equations:
3 Sr(NO₃)₂ (aq) + 2 K₃PO₄ (aq) → Sr₃(PO₄)₂ (s) + 6 KNO₃ (aq)
2 NaOH (aq) + Ni(NO₃)₂ (aq) = Ni(OH)₂ (s) + 2 NaNO₃ (aq)
Ionic equations:
3 Sr²⁺ (aq) + 6 NO₃⁻ (aq) + 6 K⁺ (aq) + 2 PO₄³⁻ (aq) → Sr₃(PO₄)₂ (s) + 6 K⁺ (aq) + 6 NO₃⁻ (aq)
2 Na⁺ (aq) + 2 OH⁻ (aq) + Ni²⁺ (aq) + 2 NO₃⁻ (aq) = Ni(OH)₂ (s) + 2 Na⁺ (aq) + 2 NO₃⁻ (aq)
To get the net ionic equation we remove the spectator ions:
3 Sr²⁺ (aq) + 2 PO₄³⁻ (aq) → Sr₃(PO₄)₂ (s)
2 OH⁻ (aq) + Ni²⁺ (aq) = Ni(OH)₂ (s)
Learn more about:
net ionic equations
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Answer:
1.99grams
Explanation:
- First, we need to calculate the molar mass of the compound: Ca(HCO3)2
Ca = 40g/mol, H = 1g/mol, C = 12g/mol, O = 16g/mol
Hence, Ca(HCO3)2
= 40 + {1 + 12 + 16(3)}2
= 40 + {13 + 48}2
= 40 + {61}2
= 40 + 122
= 162g/mol
Molar mass of Ca(HCO3)2 = 162g/mol
- Next, we calculate the mass of oxygen in one mole of the compound, Ca(HCO3)2.
Oxygen = {16(3)}2
= 48 × 2
= 96g of Oxygen
- Next, we calculate the percentage composition of oxygen by mass by dividing the mass of oxygen in the compound by the molar mass of the compound i.e.
% composition of O = 96/162 × 100
= 0.5926 × 100
= 59.26%.
- The number of moles of the compound, Ca(HCO3)2, must be converted to mass by using the formula; mole = mass/molar mass
0.0207 = mass/162
Mass = 162 × 0.0207
Mass = 3.353grams
However, in every gram of Ca(HCO3)2, there is 0.5926 g of oxygen
Hence, in 3.353grams of Ca(HCO3)2, there will be;
0.5926 × 3.353
= 1.986
= 1.99grams.
Therefore, there is 1.99grams of Oxygen in 0.0207 moles (3.353g) of Ca(HCO3)2.
Answer:
A
Explanation:
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