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Alchen [17]
3 years ago
12

2H2 + O2 -> 2H20

Chemistry
1 answer:
irina [24]3 years ago
6 0

Answer:

                    144.08 g of H₂O

Explanation:

                     The balance chemical equation for the synthesis of water is as follow;

                                           2 H₂ + O₂ → 2 H₂O

Step 1: <u>Find out moles of H₂O produced as;</u>

According to balance chemical equation,

                    1 mole of O₂ produced  =  2 moles of H₂O

So,

                 4 moles of O₂ will produce  =  X moles of H₂O

Solving for X,

                    X =  4 mol × 2 mol / 1 mol

                    X  =  8 moles of H₂O

Step 2: <u>Calculate Mass of H₂O as;</u>

                 Mass  =  Moles × M.Mass

                 Mass  =  8 mol × 18.01 g/mol

                 Mass  = 144.08 g of H₂O

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4 0
3 years ago
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SIZIF [17.4K]

1. NaF, Na₂S, Na₃P, Na₂O

2. MgF₂, MgS, Mg₃P₂, MgO

3. AlF₃, Al₂S₃, AlP, Al₂O₃

<h3>Further explanation</h3>

Given

Ionic charge

Required

The formula of binary ionic compounds

Solution

Ionic compounds consisting of cations (ions +) and anions (ions -)

Ionic compounds usually consist of metal cations and non-metal anions

Metal: cation, positively charged.

Nonmetal: negatively charged

The anion cation's charge is crossed

The ionic compounds :

1. NaF, Na₂S, Na₃P, Na₂O

2. MgF₂, MgS, Mg₃P₂, MgO

3. AlF₃, Al₂S₃, AlP, Al₂O₃

4 0
3 years ago
A flexible container at an initial volume of 6.13 L contains 2.51 mol of gas. More gas is then added to the container until it r
aleksley [76]

Answer:

2.12 moles of gas were added.

Explanation:

We can solve this problem by using<em> Avogadro's law</em>, which states that at constant temperature and pressure:

  • V₁n₂=V₂n₁

Where in this case:

  • V₁ = 6.13 L
  • n₂ = ?
  • V₂ = 11.3 L
  • n₁ = 2.51 mol

We <u>input the data</u>:

  • 6.13 L * n₂ = 11.3 L * 2.51 mol
  • n₂ = 4.63

As <em>4.63 moles is the final number of moles</em>, the number of moles added is:

  • 4.63 - 2.51 = 2.12 moles
5 0
2 years ago
A sample of gas at a pressure of 121.59 kPa, a volume of 31 L, and a temperature of 360 K contains how many moles of gas? A. 0.7
natali 33 [55]

Answer:

C. 1.3 mol

Explanation:

PV = nRT

where P is absolute pressure,

V is volume,

n is number of moles,

R is universal gas constant,

and T is absolute temperature.

Given:

P = 121.59 kPa

V = 31 L

T = 360 K

R = 8.3145 L kPa / mol / K

Find: n

n = PV / (RT)

n = (121.59 kPa × 31 L) / (8.3145 L kPa / mol / K × 360 K)

n = (3769.29 L kPa) / (2993.22 L kPa / mol)

n = 1.26 mol

Round to two significant figures, there are 1.3 moles of gas.

4 0
3 years ago
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