Answer:
Gamma
Explanation:
I'm not sure how to do it without calculations but:
E=hv
7*10^7 J/mol=6.626*10^34 Js * v
v=1*10^41
Gamma rays.
More here: https://www.hasd.org/faculty/AndrewSchweitzer/spectroscopy.pdf
It means “the study of.” Think of it:
Those are subjects you study. It makes more sense than the other choices.
The easiest way to do this is to assume that you have 100g of compound. Then you will have 69.9g of iron and 30.1g of oxygen.
You will need to find how many moles there are of each. Divide these masses by the molar mass of each element (for oxygen, just use the molar mass of O, not O2, as the calculation is easier with atoms than dioxygen):
69.9g / 55.845g/mol = 1.25mol Fe
30.1g / 15.999g/mol = 1.88mol O
Next, find the ratio of these 2 molar amounts. If we have 1 mole of Fe, we will have 1.88 / 1.25 = 1.50 mol O. Make everything whole numbers now: 1.50 is half of 3, so multiply both by 2. We get 2 moles of Fe to every 3 moles of O, so the empirical formula is Fe2O3.
Answer:
HCOOH(aq) + OH-(aq) —> HCOO-(aq) + H2O(l)
Explanation:
HCOOH is a weak acid and so will not ionised completely in solution.
KOH is a strong base and will ionised completely as shown below
KOH(aq) –> K+(aq) + OH-(aq)
The overall reaction can be written as follow:
HCOOH(aq) + K+(aq) + OH-(aq) —> HCOO-(aq) + K+(aq) + H2O(l)
Cancel out the K+ to obtain the net ionic equation as shown below
HCOOH(aq) + OH-(aq) —> HCOO-(aq) + H2O(l)
Answer:
<span>ρ≅13.0⋅g⋅m<span>L<span>−1</span></span></span> = <span>13.0⋅g⋅c<span>m<span>−3</span></span></span>
Explanation:
<span>Density=<span>MassPer unit Volume</span></span> = <span><span>75.0⋅g</span><span><span>(36.5−31.4)</span>⋅mL</span></span> <span>=??g⋅m<span>L<span>−1</span></span></span>
Note that <span>1⋅mL</span> = <span>1⋅c<span>m<span>−3</span></span></span>; these are equivalent units of volume;
i.e. <span>1⋅c<span>m3</span></span> = <span>1×<span><span>(<span>10<span>−2</span></span>⋅m)</span>3</span>=1×<span>10<span>−6</span></span>⋅<span>m3</span>=<span>10<span>−3</span></span>⋅L=1⋅mL</span>.