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3 years ago

How much force is required to lift a rock a vertical distance of 8 m if 80 j of work were done

Chemistry

1 answer:

yanalaym [24]3 years ago

5 0

we have

work done (W)= force(F) × displacemen(s)

or, 80= F× 8

or, F= 10 N

therefore, 10 N force is required to lift the rock.

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An iron nail rusts when exposed to oxygen. According to the following reaction, how many moles of iron(III) oxide will be formed

Zigmanuir [339]

Answer:

0.453 moles

Explanation:

The balanced equation for the reaction is:

2Fe(s) + 3O2(g) ==> 2Fe2O3

From the equation, mass of O2 involved = 16 x 2 x 3 = 96g

mass of Fe2O3 involved = [(2x26) + 3 x 16] x 2

= 100g

Therefore 96g of O2 produced 100g of Fe2O3

32.2g of O2 Will produce 100x32.2/96

= 33.54g of Fe2O3

Converting it to mole using number of mole = mass/molar mass

but molar mass of Fe2O3 = 26 + (16 X 3)

= 74g/mole

Therefore number of mole of 33.54g of Fe2O3 = 33.54/74

= 0.453 moles

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3 years ago

The Earth's asthenosphere is a structural layer made up on what?

ehidna [41]

Gas is the answer ok

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3 years ago

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Which of the following represent Dinitrogen tricarbide <br>1) NC <br>2) NC2 <br>3) N2C2 <br>4) N203

Ganezh [65]

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3 years ago

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What mass of HCL, in grams, is required to react with 0.610 g of al(oh)3 ?

kompoz [17]

Answer: 0.8541 grams of HCl will be required.

Explanation: Moles can be calculated by using the formula:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of $Al(OH)_3$ = 0.610 g

Molar mass of $Al(OH)_3$ = 78 g/mol

$\text{Number of moles}=\frac{0.610g}{78g/mol}$

Number of moles of $Al(OH)_3$ = 0.0078 moles

The reaction between $Al(OH)_3$ and HCl is a type of neutralization reaction because here acid and base are reacting to form an salt and also releases water.

Chemical equation for the above reaction follows:

$Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O$

By Stoichiometry,

1 mole of $Al(OH)_3$ reacts with 3 moles of HCl

So, 0.0078 moles of $Al(OH)_3$ will react with $\frac{3}{1}\times 0.0078$ = 0.0234 moles

Mass of HCl is calculated by using the mole formula, we get

Molar mass of HCl = 36.5 g/mol

Putting values in the equation, we get

$0.0234moles=\frac{\text{Given mass}}{36.5g/mol}$

Mass of HCl required will be = 0.8541 grams

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3 years ago

The normal freezing point of water is 0.00 ⁰C. What is the freezing point of a solution containing450.0 mg of ethylene glycol (M

anyanavicka [17]

Answer:

Freezing T° of solution = - 8.98°C

Explanation:

We apply Freezing point depression to solve this problem, the colligative property that has this formula:

Freezing T° of pure solvent - Freezing T° of solution = Kf . m

Kf = 1.86°C/m, this is a constant which is unique for each solvent. In this case, we are using water

m = molality (moles of solute / 1kg of solvent)

We convert the mass of solvent from g to kg

1.5 g . 1kg/1000g = 0.0015 kg

We convert the mass of solute, to moles. Firstly we make this conversion, from mg to g → 450mg . 1g/1000mg = 0.450 g

0.450 g. 1mol / 62.07g = 7.25×10⁻³ moles

Molality → 7.25×10⁻³ mol / 0.0015 kg = 4.83 m

- Freezing T° of solution = 1.86°C /m . 4.83 m - Freezing T° of pure solvent

-Freezing T° of solution = 1.86°C /m . 4.83 m - 0°C

Freezing T° of solution = - 8.98°C

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3 years ago

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