Answer:
Explanation:says these bars are like co vid i get em right off teh bat
Answer:
A) = 4.7 × 10⁻⁴atm
Explanation:
Given that,
Kp = 1.5*10³ at 400°C
partial pressure pN2 = 0.10 atm
partial pressure pH2 = 0.15 atm
To determine:
Partial pressure pNH3 at equilibrium
The decomposition reaction is:-
2NH3(g) ↔N2(g) + 3H2(g)
Kp = [pH2]³[pN2]/[pNH3]²
pNH3 =√ [(pH2)³(pN2)/Kp]
pNH3 = √(0.15)³(0.10)/1.5*10³ = 4.74*10⁻⁴ atm
![K_p = \frac{[pH_2] ^3[pN_2]}{[pNH_3]^2} \\pNH_3 = \sqrt{\frac{(pH_2)^3(pN_2)}{pNH_3} } \\pNH_3 = \sqrt{\frac{(0.15)^3(0.10)}{1.5 \times 10^3} } \\=4.74 \times 10^-^4atm](https://tex.z-dn.net/?f=K_p%20%3D%20%5Cfrac%7B%5BpH_2%5D%20%5E3%5BpN_2%5D%7D%7B%5BpNH_3%5D%5E2%7D%20%5C%5CpNH_3%20%3D%20%5Csqrt%7B%5Cfrac%7B%28pH_2%29%5E3%28pN_2%29%7D%7BpNH_3%7D%20%7D%20%5C%5CpNH_3%20%3D%20%5Csqrt%7B%5Cfrac%7B%280.15%29%5E3%280.10%29%7D%7B1.5%20%5Ctimes%2010%5E3%7D%20%7D%20%5C%5C%3D4.74%20%5Ctimes%2010%5E-%5E4atm)
= 4.7 × 10⁻⁴atm
To solve this problem, we use Beer's Law: A= ε.l.c
A is the absorbance- 0,558
<span>ε is</span> the molar absorptivity- is <span>15000 </span><span><span>L⋅mol-1</span><span>cm-1</span></span>
<span>l is </span>the length of the cuvette- 1 cm
<span>c is</span> the molar concentration
Applying the formula,
0,558= 15000 x 1 x c
0,558/15000= c
c= <span>3.72×<span>10⁻⁵ </span> <span>mol⋅L<span>⁻¹</span></span></span>
<span />
I think it’s C but not certainly positive
Answer:
A. While inconsistent, his average of 1.0 kg made his results quite
accurate
Explanation:
Accuracy is how close the value you gain from test or observation compared with the real value. In this test, all tests relatively close to 1kg and the average of the test is 1kg so this tool can be considered quite accurate.
The precision determines how close the value of multiple repetitions of the test. In this case, the value varies from 0.8kg to 1.2kg. The highest value is about 50% higher than the lowest value, so the range is pretty far and the test can be considered not precise.
