You did not include the options but I can tell you the product ratio.
The product ratio is the mole ratio of the products of the reaction.
From the balanced chemical equation you have all the mole ratios:
The given equation is: 2 C6H5COOH + 15O2 --> 14 CO2 + 6H2O
The mole ratios are: 2 C6H5COOH: 15 O2: 14 CO2 : 6 H2O
The products are CO2 and H2O
Their mole ratio = 14 CO2 : 6 H2O
That can be expressed as:
14 mol CO2 7 mol CO2
----------------- = -----------------
6 mol H2O 3 mol H2O
It is also the same that:
6 mol H2O : 14 mol CO2
6 mol H2O 3 mol H2O
------------------ = -------------------
14 mol CO2 7 mol CO2
So, compare your options to the ratios show above and pick the proper ratio.
The balanced equation given is:
4NH3 + 3O2 .....> 2N2 + 6H2O
From this equation, we can note that 4 moles of NH3 are required to produce 2 moles of N2.
Therefore, the mole ratio of NH3 to N2 is 4:2 which can be simplified into 2:1
M=11.20 g
m(H₂)=0.6854 g
M(H₂)=2.016 g/mol
M(Mg)=24.305 g/mol
M(Zn)=65.39 g/mol
w-?
m(Mg)=wm
m(Zn)=(1-w)m
Zn + 2HCl = ZnCl₂ + H₂
m₁(H₂)=M(H₂)m(Zn)/M(Zn)=M(H₂)(1-w)m/M(Zn)
Mg + 2HCl = MgCl₂ + H₂
m₂(H₂)=M(H₂)m(Mg)/M(Mg)=M(H₂)wm/M(Mg)
m(H₂)=m₁(H₂)+m₂(H₂)
m(H₂)=M(H₂)(1-w)m/M(Zn)+M(H₂)wm/M(Mg)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}
m(H₂)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}
(1-w)/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}
1/M(Zn)-w/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}
w(1/M(Mg)-1/M(Zn))=m(H₂)/{M(H₂)m}-1/M(Zn)
w=[m(H₂)/{M(H₂)m}-1/M(Zn)]/(1/M(Mg)-1/M(Zn))
w=0.583 (58.3%)
Answer:
C. It does not emit electromagnetic radiation.
Explanation:
Right now, Dark Matter is only a theory. Scientist proposed this to counter some of the strange phenomenon with matter in space.
Scientists know little about dark matter. Some say it's one of the driving forces of the universe. Currently, scientists have no way of measuring or identifying dark matter.
Answer:
<em>3.27·10²³ atoms of O</em>
Explanation:
To figure out the amount of oxygen atoms in this sample, we must first evaluate the sample.
The chemical formula for sodium sulfate is <em>Na₂SO₄, </em>and its molar mass is approximately 142.05
.
We will use stoichiometry to convert from our mass of <em>Na₂SO₄ </em>to moles of <em>Na₂SO₄</em>, and then from moles of <em>Na₂SO₄ </em>to moles of <em>O </em>using the mole ratio; then finally, we will convert from moles of <em>O </em>to atoms of <em>O </em>using Avogadro's constant.
19.3g <em>Na₂SO₄</em> · 
· 
·
After doing the math for this dimensional analysis, you should get a quantity of approximately <em>3.27·10²³ atoms of O</em>.
