The answer to the problem is 7/10
These 5 metals are: zinc,copper,tin,bronze,nickel
Answer:
Explanation:
We can calculate the volume of the oxygen molecule as the radius of oxygen molecule is given as 2×10⁻¹⁰m.
We know that volume=4/3×πr³
volume =4/3×π(2.0×10⁻¹⁰m)³
volume=33.40×10⁻³⁰m³
Volume of oxygen molecule=33.40×10⁻³⁰m³
we know the ideal gas equation as:
PV=nRT
k=R/Na
R=k×Na
PV=n×k×Na×T
n×Na=N
PV=Nkt
p is pressure of gas
v is volume of gas
T is temperature of gas
N is numbetr of molecules
Na is avagadros number
k is boltzmann constant =1.38×10⁻²³J/K
R is real gas constant
So to calculate pressure using the formula;
PV=NkT
P=NkT/V
Since there is only one molecule of oxygen so N=1
P=[1×1.38×10⁻²³J/K×300]/[33.40×10⁻³⁰m³
p=12.39×10⁷Pascal
6= Only the digits 1 and 6 are the actual measured values. Therefore we have only 2 significant figures.
0.3= Zeros used as placeholders are not significant. Zeros that come before non-zero integers are never significant. Example 5: The zeros in 098, 0.3, and 0.000000000389 are not significant because they are all in front of non-zero integers. c. If the zeros come after non-zero integers and are followed by a decimal point, the zeros are significant.
Answer:
A. 
B. 
Explanation:
Hello!
In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:
Thus we proceed as follows:
A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:
Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:
B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:
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