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Maksim231197 [3]
3 years ago
9

How many grams of CO2could be produced from the combustion of 23g of C4H10in the presence of an adequate amount of O2?

Chemistry
1 answer:
blsea [12.9K]3 years ago
8 0

Answer:

Mass = 70.4 g

Explanation:

Given data:

Mass of CO₂ produced = ?

Mass of C₄H₁₀ = 23 g

Solution:

Chemical equation:

2C₄H₁₀ + 13O₂  →   8CO₂ + 10H₂O

Number of moles of C₄H₁₀:

Number of moles = mass/ molar mass

Number of moles = 23 g/ 58.12 g/mol

Number of moles = 0.4 mol

Now we will compare the moles of C₄H₁₀ and  CO₂  from balance chemical equation.

                   C₄H₁₀          :           CO₂

                        2             :             8

                      0.4           :          8/2×0.4 = 1.6 mol

Mass of CO₂:

Mass = number of moles × molar mass

Mass = 1.6 mol × 44 g/mol

Mass = 70.4 g

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What is the engine piston displacement in liters of an engine whose displacement is listed as 490 in^3?
marishachu [46]

Answer:

490 in^3 = 8.03 L

Explanation:

Given:

The engine displacement = 490 in^3

= 490 in³

To determine the engine piston displacement in liters L;

(NOTE: Both in^3 (in³) and L are units of volume). Hence, to find the engine piston displacement in liters (L), we will convert in^3 to liters (L)

First, we will convert in³ to cm³

Since 1 in = 2.54 cm

∴ 1 in³ = 16.387 cm³

If 1 in³ = 16.387 cm³

Then 490 in³ =  (490 in³ × 16.387 cm³) / 1 in³ = 8029.63 cm³

∴ 490 in³ = 8029.63 cm³

Now will convert cm³ to dm³

(NOTE: 1 L = 1 dm³)

1 cm = 1 × 10⁻² m = 1 × 10⁻¹ dm

∴ 1 cm³ = 1 × 10⁻⁶ m³ = 1 × 10⁻³ dm³

If 1 cm³ = 1 × 10⁻³ dm³

Then, 8029.63 cm³ = (8029.63 cm³ × 1 × 10⁻³ dm³) / 1 cm³ = 8.02963 dm³

≅ 8.03 dm³

∴ 8029.63 cm³ = 8.03 dm³

Hence, 490 in³ = 8029.63 cm³ = 8.03 dm³

Since 1L = 1 dm³

∴ 8.03 dm³ = 8.03 L

Hence, 490 in³ = 8.03 L

3 0
2 years ago
The radioisotope phosphorus-32 is used in tracers for measuring phosphorus uptake by plants. The half-life of phosphorus-32 is 1
Harman [31]

Answer:

54 days

Explanation:

We have to use the formula;

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Where;

t1/2= half-life of phosphorus-32= 14.3 days

t= time taken for the activity to fall to 7.34% of its original value

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A= activity of phosphorus-32 after a time t

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Substituting values;

0.693/14.3 = 2.303/t log Ao/0.0734Ao

0.693/14.3 = 2.303/t log 1/0.0734

0.693/14.3 = 2.6/t

0.048=2.6/t

t= 2.6/0.048

t= 54 days

3 0
2 years ago
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