Answer: 60 grams
Explanation: (60 ml)*(1g/ml) = 60g
Answer:
c. The atoms of one element can be identical to the atoms of another element.
Explanation:
<em>Which of the following is not a statement of Dalton's atomic theory of matter?</em>
<em>a. Elements are made of atoms.</em> TRUE. An atom is the smallest particle of a chemical element that can exist.
<em>b. Atoms of a given element are identical.</em> TRUE. The only slight difference is in the mass of isotopes.
<em>c. The atoms of one element can be identical to the atoms of another element.</em> FALSE. The atoms of different elements are different from one to another.
<em>d. A given compound always has the same number and kinds of atoms. </em>TRUE. This is known as Dalton's law of constant composition.
Answer:
5.158 mol/L
Explanation:
To find the molarity, you need to use the formula:
Molarity (M) = moles / volume (L)
You have been grams sodium carbonate. You need to (1) convert grams Na₂CO₃ to moles (via molar mass), then (2) convert moles Na₂CO₃ to moles HCl (via mole-to-mole ratio from equation), then (3) convert mL to L (by dividing by 1,000), and then (4) use the molarity equation.
<u>Steps 1 - 2:</u>
2 HCl + 1 Na₂CO₃ ----> 2 NaCl + H₂O + CO₂
6.5287 g Na₂CO₃ 1 mole 2 moles HCl
-------------------------- x ------------- x ------------------------- = 0.12318 mole HCl
106 g 1 mole Na₂CO₃
<u>Step 3:</u>
23.88 mL / 1,000 = 0.02388 L
<u>Step 4:</u>
Molarity = moles / volume
Molarity = 0.12318 mole / 0.02388 L
Molarity = 5.158 mole/L
**mole/L is equal to M**
Answer:
The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Explanation:
Given;
CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol
From the combustion reaction above, it can be observed that;
1 mole of methane (CH₄) released 890 kilojoules of energy.
Now, we convert 59.7 grams of methane to moles
CH₄ = 12 + (1x4) = 16 g/mol
59.7 g of CH₄ 
1 mole of methane (CH₄) released 890 kilojoules of energy
3.73125 moles of methane (CH₄) will release ?
= 3.73125 moles x -890 kJ/mol
= -3320.81 kJ
Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Answer:
Pb3O4
Explanation:
According to this question, 3.425g of lead oxide was reduced to form 3.105g of lead in an experiment. Since lead oxide contains both lead (Pb) and oxygen (O) element,
Mass of lead oxide = 3.425g
Mass of lead = 3.105g
Mass of oxygen = (3.425g - 3.105g) = 0.320g
Next, we convert each mass value to mole by dividing by respective molar mass
Pb = 3.105g ÷ 207.2 = 0.0149mol
O = 0.320g ÷ 16 = 0.02mol
Next, we divide each mole value by the smallest (0.0149)
Pb = 0.0149mol ÷ 0.0149mol = 1
O = 0.02mol ÷ 0.0149mol = 1.342
Multiply each ratio value by 3 to get:
Pb = 1 × 3 = 3
O = 1.342 × 3 = 4.026
The whole number ratio, approximately, of Pb and O is 3:4, hence, their empirical formula is Pb3O4.
