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3 years ago

2.5kg weighing cat climbing at the top of the tree has a potential energy of 986 J. Find the height.

Chemistry

1 answer:

Kruka [31]3 years ago

3 0

Answer:

The height = 39.44 m

Explanation:

Given:

Mass m = 2.5 kg

Potential energy = 986 J

Find:

The height

Computation:

Potential energy = mgh

g = Acceleration due to gravity = 10 m/s²

986 = (2.5)(10)(h)

h = 39.44 m

The height = 39.44 m

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What is Keq for the reaction N₂ + 3H2 = 2NH3 if the equilibrium

hram777 [196]

The Keq for the reaction N₂ + 3H2 = 2NH3 if the equilibrium concentrations are Keq = 1.5. The correct option is D.

Keq is the ratio of the concentration of reactant to the concentration of the product.

The balanced equation is

N₂ + 3H₂ = 2NH₃

The equilibrium constant is $\rm \dfrac{[NH_3]^2}{[N_2]\; [H_2]^3}$

The given concentrations of the compounds have been:

Ammonia = 3 M

Nitrogen = 1 M

Hydrogen = 2 M

$\rm \dfrac{9}{1\times 8} = 1.5$

Thus, the correct option is D. Keq = 1.5.

Learn more about Keq

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2 years ago

Glucose (C6H12O6) can be fermented to yield ethanol (CH3CH2OH) and carbon dioxide (CO2). C6H12O6⟶2CH3CH2OH+2CO2 The molar mass o

jekas [21]

Answer:

The % yield is 74.45 %

Explanation:

<u>Step 1:</u> The balanced equation

C6H12O6⟶2CH3CH2OH+2CO2

<u>Step 2</u>: Data given

Molar mass glucose = 180.15 g/mol

Molar mass of ethanol = 46.08 g/mol

Molar mass of carbon dioxide = 44.01 g/mol

Mass of glucose = 61.5 grams

Mass of ethanol = 23.4 grams

<u>Step 3:</u> Calculate moles of glucose

Moles glucose = Mass glucose / Molar mass of glucose

Moles glucose = 61.5 grams / 180.15 g/mol

Moles glucose = 0.341 moles

<u>Step 4:</u> Calculate moles of ethanol

1 mole of glucose consumed, produces 2 moles of ethanol and 2 moles of CO2

0.341 moles of glucose, will produce 2*0.341 = 0.682 moles of ethanol

<u>Step 5:</u> Calculate mass of ethanol

Mass ethanol = moles ethanol * Molar mass ethanol

Mass ethanol = 0.682 moles * 46.08 g/mol

Mass ethanol = 31.43 grams = theoretical mass

<u>Step 6:</u> Calculate % yield

% yield = actual mass / theoretical mass

% yield = (23.4 grams / 31.43 grams) * 100%

% yield = 74.45 %

The % yield is 74.45 %

7 0

3 years ago

Lab report on flock flock y flock z insects and fruits

n200080 [17]

Answer:

In this lab, you will learn the anatomy of an insect, how to identify an insect to order, how to collect and curate insects,

Explanation:

7 0

2 years ago

Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79

BARSIC [14]

Answer : The mole fraction and partial pressure of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of $CH_4$ = 1.79 mole

Moles of $CO_2$ = 1.20 mole

Moles of $He$ = 3.71 mole

Now we have to calculate the mole fraction of $CH_4,CO_2$ and $He$ gases.

$\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267$

and,

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179$

and,

$\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554$

Thus, the mole fraction of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of $CH_4,CO_2$ and $He$ gases.

According to the Raoult's law,

$p_i=X_i\times p_T$

where,

$p_i$ = partial pressure of gas

$p_T$ = total pressure of gas = 5.78 atm

$X_i$ = mole fraction of gas

$p_{CH_4}=X_{CH_4}\times p_T$

$p_{CH_4}=0.267\times 5.78atm=1.54atm$

and,

$p_{CO_2}=X_{CO_2}\times p_T$

$p_{CO_2}=0.179\times 5.78atm=1.03atm$

and,

$p_{He}=X_{He}\times p_T$

$p_{He}=0.554\times 5.78atm=3.20atm$

Thus, the partial pressure of $CH_4,CO_2$ and $He$ gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0

3 years ago

Titanium dioxide (TiO2) is used extensively as a white pigment. It is produced from an ore that contains ilmenite (FeTiO3) and f

STALIN [3.7K]

Answer:

2928kg of ore are required.

2585kg of the 80% H₂SO₄ solution are required.

Explanation:

To solve this question we need first to find the moles of titanium in 1000kg of TiO₂. Keeping in mind the 89% of descomposition we can find the mass of the ore and the mass of the 80% sulfuric acid required:

<em>Moles TiO₂ -Molar mass: 79.866g/mol-:</em>

1x10⁶g * (1mol / 79.866g) = 12521 moles Titanium

In mass -Molar mass Ti: 47.867g/mol-:

12521 moles Titanium * (47.867g / mol) = 599341.4g of Ti.

As the ore contains 24.3% of Ti:

599341.4g of Ti = 599.34kg Ti * (100 / 24.3) = 2606kg ore

As the descomposition is just of 89%:

2606kg ore * (100 / 89) =

<h3>2928kg of ore are required</h3><h3 />

<em>Mass 80% sulfuric acid:</em>

12521 moles Titanium = 12521 moles H₂SO₄ * (100/89) = 14068.5 moles of H₂SO₄ are required.

In an excess of 50% =

14068.5 moles of H₂SO₄ are required * 1.5 = 21102.8 moles of H₂SO₄.

The mass is:

21102.8 moles of H₂SO₄ * (98g / mol) = 2068075g = 2068kg of sulfuric acid

That is in the 80%:

2068kg of sulfuric acid * (100/ 80) =

<h3>2585kg of the 80% H₂SO₄ solution are required</h3>

3 0

3 years ago