Answer:
it helps us with our blood (Little)
Answer:
d. 12.3 grams of Al2O3
Explanation:
Based on the reaction:
4Al + 3O2 → 2Al2O3
<em>Where 4 moles of Al reacts in excess of oxygen to produce 2 moles of aluminium oxide.</em>
<em />
To solve this question we must find the moles of Aluminium. With these moles we can find the moles of aluminium oxide using the reaction:
<em>Moles Al -Molar mass: 26.9815g/mol-</em>
6.50g * (1mol / 26.9815g) = 0.241 moles Al
<em>Mass Al₂O₃ -Molar mass: 101.96g/mol-</em>
0.241 moles Al * (2 mol Al2O3 / 4 mol Al) = 0.120 moles Al2O3
0.120 moles Al2O3 * (101.96g / mol) =
12.3g of Al2O3 are produced.
Right answer is:
<h3>d. 12.3 grams of Al2O3
</h3>
Answer:
4 moles of neon
Explanation:
Given data:
Number of moles of neon = ?
Number of atoms of neon = 2.4×10²⁴ atoms
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen
For given neon atoms:
1 mol = 6.022 × 10²³ atoms
2.4×10²⁴ atoms × 1 mol / 6.022 × 10²³ atoms
0.4×10¹ mol = 4 mol
The answer is 615.91 grams of <span>n2f4
Solution:
225g F2 x [(1molF2)/(38gramsF2)] x [</span>(1molF2)/(1molN2F4)] x [(104.02 grams N2F4)/(1molN2F4)]
=615.91 grams
Answer:
2HNO3 (aq) + Na2CO3 (aq) → 2NaNO3 (aq) + CO2 (g) + H2O (l)
Explanation:
This question is asking to write and balance an equation between between aqueous sodium carbonate (Na2CO3) and aqueous nitric acid (HNO3). The equation is as follows:
HNO3 (aq) + Na2CO3 (aq) → NaNO3 (aq) + CO2 (g) + H2O (l)
However, this equation is not balanced as the number of atoms of each element must be the same on both sides of the equation. To balance the equation, one will make use of coefficients as follows:
2HNO3 (aq) + Na2CO3 (aq) → 2NaNO3 (aq) + CO2 (g) + H2O (l)
