Answer:
d) The dilution equation works because the number of moles remains the same.
Explanation:
Let’s say that you have 1 mol of a solute in I L of solution. The concentration is 1 mol·L⁻¹. and <em>M</em>₁<em>V</em>₁ = 1 mol.
Now, you dilute the solution to a volume of 2 L. You still have 1 mol of solute, but in 2 L of solution. The new concentration is 0.5 mol·L⁻¹.
The volume has doubled, but the volume has halved, and <em>M</em>₂<em>V</em>₂ = 1 mol.
b) <em>Wrong</em>. The molar concentration changes on dilution.
c) <em>Wrong</em>. The volume changes on dilution.
a) <em>Wrong</em>, although technically correct, because if the moles don’t change, the mass doesn’t change either. However, the formula <em>M</em>₁<em>V</em>₁ has units mol·L⁻¹ × L = mol. Thus, in the formula, it is moles that are constant.
Answer is: mass oxygen is 129,28 grams.
Chemical reaction: 2C₁₈H₃₈ + 55O₂ → 36CO₂ + 38H₂O.
m(C₁₈H₃₈) = 37,5 g.
n(C₁₈H₃₈) = m(C₁₈H₃₈) ÷ M(C₁₈H₃₈).
n(C₁₈H₃₈) = 37,5 g ÷ 254 g/mol.
n(C₁₈H₃₈) = 0,147 mol.
From chemical reaction: n(C₁₈H₃₈) : n(O₂) = 2 : 55.
n(O₂) = 4,04 mol.
m(O₂) = 4,04 mol · 32 g/mol.
m(O₂) = 129,28 g.
Answer:
FALSE!!! Thermal energy can be transformed to heat.
Explanation:
There are 3.98 × 10^23 atoms of oxygen in the sample.
Given that;
1 mole of Mo(NO3)6 contains 6.02 × 10^23 atoms of Nitrogen
x moles of Mo(NO3)6 contains 2.22 x 10^22 atoms of nitrogen
x = 1 mole × 2.22 x 10^22 atoms/6.02 × 10^23 atoms
x = 0.0368 moles
The number of oxygen atoms in the sample is given by; 0.0368 × 6.02 × 10^23 × 18
Therefore, there are 3.98 × 10^23 atoms of oxygen in the sample.
Learn more: brainly.com/question/9743981
Answer:
Kindly check the explanation section.
Explanation:
From the description given in the question above, that is '' H subscript f to the power of degree of the reaction" we have that the description matches what is known as the heat of formation of the reaction, ∆fH° where the 'f' is a subscript.
In order to determine the heat of formation of any of the species in the reaction, the heat of formation of the other species must be known and the value for the heat of reaction, ∆H(rxn) must also be known. Thus, heat of formation can be calculated by using the formula below;
∆H(rxn) = ∆fH°( products) - ∆fH°(reactants).
That is the heat of formation of products minus the heat of formation of the reaction g specie(s).
Say heat of formation for the species is known as N(g) = 472.435kj/mol, O(g) = 0kj/mol and NO = unknown, ∆H°(rxn) = −382.185 kj/mol.
−382.185 = x - 472.435kj/mol = 90.25 kJ/mol
