Answer:
The correct answer is B) it helps to ensure the result are consistent and repeatable.
Explanation:
Scientist generally repeat an experiment if he or she did not make a mistake in the first one to compare the results of two experiment, if there is no difference in the result or values obtainted from the observation of two experiment. he or she become sure that experiment was done in a right way because if there is some error made when experiment was carried out then the result of two same experiment would be different.
First find the number of moles of sulfur using dimensional analysis with avogadro’s number as the conversion factor. 4.2*10^24 atoms * (1 mol/6.022*10^23 atoms) = 7.0 mol sulfur. The molar mass of sulfur is 32.06 g/mol, which is found on the periodic table as sulfur’s (S) atomic weight. Use dimensional analysis again with the molar mass of sulfur as the conversion factor. 7.0 mol * 32.06 g/mol = 224.42 g sulfur. Since the problems gives us two significant figures, round the mass of sulfur to 220 grams, or 2.2 * 10^2 g.
Answer:
The rate of disappearance of C₂H₆O = 2.46 mol/min
Explanation:
The equation of the reaction is given below:
2 K₂Cr₂O₇ + 8 H₂SO₄ + 3 C₂H₆O → 2 Cr₂(SO₄)₃ + 2 K₂SO₄ + 11 H₂O
From the equation of the reaction, 3 moles of C₂H₆O is used when 2 moles of Cr₂(SO₄)₃ are produced, therefore, the mole ratio of C₂H₆O to Cr₂(SO₄)₃ is 3:2.
The rate of appearance of Cr₂(SO₄)₃ in that particular moment is given 1.64 mol/min. This would than means that C₂H₆O must be used up at a rate which is approximately equal to their mole ratios. Thus, the rate of of the disappearance of C₂H₆O can be calculated from the mole ratio of Cr₂(SO₄)₃ and C₂H₆O.
Rate of disappearance of C₂H₆O = 1.64 mol/min of Cr₂(SO₄)₃ * 3 moles of C₂H₆O / 2 moles of Cr₂(SO₄)₃
Rate of disappearance of C₂H₆O = 2.46 mol/min of C₂H₆O
Therefore, the rate of disappearance of C₂H₆O = 2.46 mol/min
The partial stress of H2 is 737.47 mmHg Let's observe the Ideal Gas Law to find out the whole mols.
We count on that the closed vessel has 1L of volume
- P.V=n.R.T
- We must convert mmHg to atm. 760 mmHg.
- 1 atm
- 755 mmHg (755/760) = 0.993 atm
- 0.993 m.1L=n.0.082 L.atm/mol.K .
- 293 K(0.993 atm 1.1L)/(0.082mol.K /L.atm).
- 293K = n
- 0.0413mols = n
These are the whole moles. Now we are able to know the moles of water vapor, to discover the molar fraction of it.
- P.V=n.R.T
- 760 mmHg. 1 atm
- 17.5 mmHg (17.5 mmHg / 760 mmHg)=0.0230 atm
- 0.0230 m.1L=n.0.082 L.atm/mol.K.293 K(0.0230atm.1L)/(0.082mol.K/L.atm .293K)=n 9.58 × 10 ^ 4 mols = n.
- Molar fraction = mols )f gas/general mols.
- Molar fraction water vapor =9.58×10^ -four mols / 0.0413 mols
- Sum of molar fraction =1
- 1 - 9.58 × 10 ^ 4 × mols / 0.0413 ×mols = molar fraction H2
- 0.9767 = molar fraction H2
- H2 pressure / Total pressure =molar fraction H2
- H2 pressure / 55mmHg = =0.9767 0.9767 = h2 pressure =755 mmHg.
- 737,47 mmHg.
<h3>What is a mole fraction?</h3>
Mole fraction is a unit of concentration, described to be identical to the variety of moles of an issue divided through the whole variety of moles of a solution. Because it's miles a ratio, mole fraction is a unitless expression.
Thus it is clear that the partial pressure of H2 is 737,47 mmHg.
To learn more about partial pressure refer to the link :
brainly.com/question/19813237
<h3 />
Answer:
Kc = 50.5
Explanation:
We determine the reaction:
H₂ + I₂ ⇄ 2HI
Initially we have 0.001 molesof H₂
and 0.002 moles of I₂
If we have produced 0.00187 moles of HI in the equilibrium we have to know, how many moles of I₂ and H₂, have reacted.
H₂ + I₂ ⇄ 2HI
In: 0.001 0.002 -
R: x x 2x
Eq: 0.001-x 0.002-x 0.00187
x = 0.00187/2 = 9.35×10⁻⁴ moles that have reacted
So in the equilibrium we have:
0.001 - 9.35×10⁻⁴ = 6.5×10⁻⁵ moles of H₂
0.002 - 9.35×10⁻⁴ = 1.065×10⁻³ moles of I₂
Expression for Kc is = (HI)² / (H₂) . (I₂)
0.00187 ² / 6.5×10⁻⁵ . 1.065×10⁻³ = 50.5
