Answer:
0.24 g/mL
Explanation:
The density of an object is given by the ratio between its mass and its volume:
where
m is the mass of the object
V is its density
In this problem, we have:
m = 7.0 g is the mass of the object
The volume of an irregular shaped object can be measured by putting it into water, and by measuring the difference in water volume.
In this case,
is the initial volume of water
is the final volume of water
So the volume of the object is
Therefore, the density of the object is:
Answer:
<h2>8 moles</h2>
Explanation:
To find the number of moles in a substance given it's number of entities we use the formula
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have
We have the final answer as
<h3>8 moles</h3>
Hope this helps you
Answer:
A) 31.22
Explanation:
The reaction of sulfuric acid with NaOH is:
H₂SO₄ + 2 NaOH → Na₂SO₄ + 2H₂O
To solve this problem we need to determine the moles of acid that will react, and, using the chemical equation we can determine the moles of NaOH and the volume that a 0.2389M NaOH solution would require to neutralize it.
<em>Moles H₂SO₄ (Molar mass: 98.08g/mol):</em>
0.9368g * 39.04% = 0.3657g H₂SO₄ * (1mol / 98.08g) =
3.7289x10⁻³moles H₂SO₄
And moles of NaOH that you require to neutralize the acid are:
3.7289x10⁻³moles H₂SO₄ * (2 moles NaOH / 1 mole H₂SO₄) =
7.4578x10⁻³ moles NaOH
Using a 0.2389M NaOH solution:
7.4578x10⁻³ moles NaOH * (1L / 0.2389mol) = 0.03122L = 31.22mL
Right answer is:
<h3>A) 31.22
</h3>
Answer:
21.8 grams.
Explanation:
Molar mass data from a modern periodic table:
How many moles of MgO will be produced if Mg is the limiting reactant?
Number of moles of Mg:
.
The ratio between the coefficient of Mg and that of MgO is 2:2. Two moles of Mg will make two moles of MgO. 0.670644 moles of MgO will be produced if Mg is the limiting reactant.
How many moles of MgO will be produced if O₂ is the limiting reactant?
Number of moles of O₂:
.
The ratio between the coefficient of O₂ and that of MgO is 1:2. One mole of O₂ will make two moles of MgO. 
of MgO will be produced if O₂ is in excess.
How many moles of MgO will be produced?
0.541284 is smaller than 0.670644. Only 0.541284 moles of MgO will be produced since O₂ will run out before all 16.3 grams of Mg is consumed.
What's the mass of 0.541284 moles of MgO?
Formula mass of MgO:
.
Mass of 0.541284 moles of MgO:
.
