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3 years ago

C12H22011+1202-->12CO2+11H20

Chemistry

1 answer:

kogti [31]3 years ago

4 0

Answer:

0.185moles

Explanation:

Given parameters:

Volume of O₂ = 49.8L

Unknown:

Number of moles of sucrose required = ?

Solution:

We can assume that the reaction takes place at standard temperature and pressure.

From this, we can find the number of moles of oxygen that reacted and extrapolate to that of sucrose.

Chemical equation;

C₁₂H₂₂0₁₁ + 120₂ → 12CO₂ + 11H₂0

Number moles = $\frac{volume of gas}{22.4}$ at STP

Number of moles of oxygen gas = $\frac{49.8}{22.4}$ = 2.22moles

12 moles of oxygen gas combines with 1 mole of sucrose

2.22 moles of oxygen gas will combine with $\frac{2.22}{12}$ = 0.185moles

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If the copper is drawn into wire whose diameter is 9.50 mm, how many feet of copper can be obtained from the ingot? The density

ankoles [38]

Answer:

The length of the wire = 352.66 feet.

Explanation:

A copper refinery produces a copper ingot weighing 150 lb. If the copper is drawn into wire whose diameter is 9.50 mm, how many feet of copper can be obtained from the ingot? The density of copper is 8.94 g/cm3. (Assume that the wire is a cylinder whose volume is V = πr2h, where r is the radius and h is its height or length.)

Step 1: Convert lb to kg

150 lb = 68.0389 kg

Step 2: Calculate volume of copper

Volume = mass / density

Volume = 68038.9 grams / 8.94 g/cm³

Volume = 7610.6 cm³ Cu

Step 3: Calculate length of wire

The diameter of the wire is 9.50 mm, so the radius is half of that (4.75 mm), or 0.475 cm.

The total "volume" of the wire is πr²h = (π)*(0.475 cm)²(h) = 0.708h = 7610 cm^3

7610 = 0.708h

h = 10749 cm = length of wire

The length of the wire = 352.66 feet.

7 0

3 years ago

How many grams of ammonia (NH3) can be produced by the synthesis of excess hydrogen gas (H2) and 253.8 grams of nitrogen gas (N2

kogti [31]

Answer:

308.2 g of NH₃.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3H₂ + N₂ —> 2NH₃

Next, we shall determine the mass of N₂ that reacted and the mass of NH₃ produced from the balanced equation. This can be obtained as follow:

Molar mass of N₂ = 2 × 14 = 28 g/mol

Mass of N₂ from the balanced equation = 1 × 28 = 28 g

Molar mass of NH₃ = 14 + (3×1)

= 14 + 3 = 17 g/mol

Mass of NH₃ from the balanced equation = 2 × 17 = 34 g

Summary:

From the balanced equation above,

28 g of N₂ reacted to produce 34 g of NH₃.

Finally, we shall determine the mass of NH₃ produced by the reaction of 253.8 g of N₂. This can be obtained as illustrated below:

From the balanced equation above,

28 g of N₂ reacted to produce 34 g of NH₃.

Therefore, 253.8 g of N₂ will react to produce = (253.8 × 34)/28 = 308.2 g of NH₃.

Thus, 308.2 g of NH₃ were obtained from the reaction.

8 0

2 years ago

How many molecules are in 2.50 moles of carbon dioxide?

podryga [215]

Answer:

1.5055×10²⁴ molecules

Explanation:

From the question given above, the following data were obtained:

Number of mole CO₂ = 2.5 moles

Number of molecules CO₂ =?

The number of molecules present in 2.5 moles CO₂ can be obtained as:

From Avogadro's hypothesis,

1 mole of CO₂ = 6.022×10²³ molecules

Therefore,

2.5 mole of CO₂ = 2.5 × 6.022×10²³

2.5 mole of CO₂ = 1.5055×10²⁴ molecules

Thus, 1.5055×10²⁴ molecules are present in 2.5 moles CO₂

7 0

2 years ago

The mole fraction of NaCl in an

posledela

The weight/weight percent(%w/w) of NaCl in solution : 33.03%

<h3>Further explanation</h3>

Given

Mole fraction of NaCl : 0.132

Required

The weight/weight percent of NaCl

Solution

1 mol solution : 0.132 mol NaCl + 0.868 mol H₂O

mass NaCl :

$\tt 0.132\times 58.44 g/mol=7.714~g$

mass H₂O :

$\tt 0.868\times 18.016~g/mol=15.64~g$

Total mass = 7.714 + 15.64 = 23.354 g

$\tt \%NaCl=\dfrac{7.714}{23.354}\times 100\%=33.03\%$

6 0

3 years ago

1. A flexible container filled with gas has a volume of 5.8 L and a pressure of 6.4 atm. What will be the

tino4ka555 [31]

Answer:

Explanation:

ygkoyul

7 0

2 years ago