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3 years ago

Find the horizontal range of a projectile launched at 15 degrees to the horizontal with speed of 40m/s

Chemistry

1 answer:

vfiekz [6]3 years ago

4 0

To Find :

The horizontal range of a projectile launched at 15 degrees to the horizontal with speed of 40 m/s.

Solution :

The horizontal range of a projectile is given by :

$R = \dfrac{u^2 sin 2\theta}{g}$ ( Here, g is acceleration due to gravity = 10 m/s² )

Putting all value in above equation :

$R = \dfrac{40^2 \times sin (2 \times 15)}{10} \ m\\\\R = \dfrac{1600 \times 1}{2\times 10} \ m\\\\R = 80 \ m$

Therefore, the horizontal range of projectile is 80 m.

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Answer:

1. 9.4 grams of methane produce 25.85 grams of CO2

2.Grams of water produced = 11.81 grams

3.Mass of Methane produced by 10.1 gram of O2 = 2.52 grams

4.Amount of methane consumed = 46.9 grams

5. Grams of Co2 produced = 8.32 grams

Explanation:

Molar masses :

Methane = CH4 = mass of C + 4x (mass of H)

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1 mole of CH4 = 16 gram

Oxygen O2 = 2 x (mass of O) = 2x(16) = 32 gram (1 mole of O2 =32 gram)

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Water = H2O = 18 grams ( 1 mole of H2O = 18 gram)

1 mole of each molecule is equal to their molar masses

The balanced equation is :

$1CH_{4}(g)+2O_{2}\rightarrow 1CO_{2}+2H_{2}O(l)$

According to Stoichiometry :

1 mole of CH4 = 2 Mole of O2 = 1 mole of CO2 = 2 mole of H2O

1. From the equation ,

1 mole of methane produce =1 mole of CO2

16 gram of methane = 44 gram of CO2

1 gram of methane =

$\frac{44}{16}$ gram of CO2

9.4 gram of CH4 =

$\frac{44}{16}\times 9.4$ gram of CO2

= 25 .85 gram of CO2

2 mole of O2 produces = 2 mole of H2O(water)

1 mole of O2 produces = 1 mole of H2O

32 gram of O2 = 18 gram of water

1 gram of O2 =

$\frac{18}{32}$

21 gram of O2 =

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3. 1 mole of CH4 = 2 mole of O2

16 gram of CH4 = 2(32) = 64 grams of O2

64 gram of O2 needs = 16 grams of CH4

1 gram of O2 needs =

$\frac{16}{64}$

10.1 gram of O =

$\frac{16}{64}\times 10.1$ of CH4

= 2.52 gram

1 mole of CO2 is produced from = 1 mole of CH4

44 gram of CO2 is produced from 16 gram of CH4

1 gram CO2 =

$\frac{16}{44}$ gram of CH4

129 gram of CO2 =

$\frac{16}{44}\times 129$ gram of CH4

= 46.90 grams

2 mole of O2 produce = 1 mole of CO2

2x 32 gram of O2 = 44 gram of CO2

1 gram of O2 =

$\frac{44}{64}$ of CO2

12.1 gram of O2 produce=

$\frac{44}{64}\times 12.1$ of CO2

= 8.318 gram

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Given the equation Ca + H2O --> Ca(OH)2 + H2 how many grams of calcium will react completely with 10.0 grams of water? *

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Answer: 11.0 g of calcium will react with 10.0 grams of water.

Explanation:

To calculate the moles, we use the equation:

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$\text{Number of moles}=\frac{10.0g}{18g/mol}=0.55moles$

The balanced chemical equation is:

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According to stoichiometry :

2 moles of $H_2O$ require = 1 mole of $Ca$

Thus 0.55 moles of $H_2O$ require= $\frac{1}{2}\times 0.55=0.275moles$ of $Ca$

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3 years ago

Find the horizontal range of a projectile launched at 15 degrees to the horizontal with speed of 40m/s​

Find the horizontal range of a projectile launched at 15 degrees to the horizontal with speed of 40m/s