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muminat
3 years ago
11

How to balance this equation

Chemistry
2 answers:
Svetlanka [38]3 years ago
7 0

Answer:

2Al + 3Ni(SO4) ------->   1Al2(SO4)3 +  2Ni

Explanation:

Afina-wow [57]3 years ago
7 0
2Al+3Ni(SO4)-> 1Al2(SO4)3+3Ni
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During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.
Monica [59]

Answer:

0.74 grams of methane

Explanation:

The balanced equation of the combustion reaction of methane with oxygen is:

  • CH₄ + 2 O₂ → CO₂ + 2 H₂O

it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.

firstly, we need to calculate the number of moles of both

for CH₄:

number of moles = mass / molar mass = (3.00 g) /  (16.00 g/mol) = 0.1875 mol.

for O₂:

number of moles = mass / molar mass = (9.00 g) /  (32.00 g/mol) = 0.2812 mol.

  • it is clear that O₂ is the limiting reactant and methane will leftover.

using cross multiplication

1 mol of  CH₄ needs → 2 mol of O₂

???  mol of  CH₄  needs → 0.2812 mol of O₂

∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol

so 0.14 mol will react and the remaining CH₄

mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol

now we convert moles into grams

mass of CH₄ left over = no. of mol of CH₄ left over *  molar mass

                                    = 0.0469 mol * 16 g/mol = 0.7504 g

So, the right choice is 0.74 grams of methane

3 0
3 years ago
Given the formulas for the ionic compounds, draw the correct ratio of cations to anions (a) BaSO4, (b) CaF2, (c) Mg3N2, (d) K2O.
Karolina [17]

Answer:

A. It formed by barium(Ba+2) ion and sulfate ( SO42- )

B. It is formed by calcium ion (Ca+2) and two fluoride ions (2F-)

C. It is formed by magnesium ion (Mg+2) and nitride ion (N3-)

D. It is formed by two potassium ions (2 K+) and oxide ion(O2-)

3 0
3 years ago
Help question in the pic below
Fed [463]

Answer: A

Explanation:

6 0
3 years ago
Read 2 more answers
Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of
Kaylis [27]

Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of H_2

\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles

b) moles of I_2

\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles

H_2(g)+I_2(g)\rightarrow 2HI(g)

According to stoichiometry :

1 mole of I_2 require 1 mole of H_2

Thus 0.0787 moles of l_2 require=\frac{1}{1}\times 0.0787=0.0787moles of H_2

Thus l_2 is the limiting reagent as it limits the formation of product and H_2 acts as the excess reagent. (10.0-0.0787)= 9.92 moles of H_2are left unreacted.

Mass of H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g

Thus 19.9 g of H_2 remains unreacted.

5 0
3 years ago
Carbon-14 is an isotope used in carbon dating. The nucleus becomes Nitrogen-14 through beta decay. Its half-life is 5370 years.
frosja888 [35]

Answer:

If the half-life of 14C is 5730 years, when this period of time has passed it will have been halved, it is called the exponential decay law of radioactive isotopes.

8 0
3 years ago
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