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3 years ago

A rock weighing 2 newtons was lifted 3 meters. how much work was done

Chemistry

1 answer:

goblinko [34]3 years ago

3 0

Answer: The work done to lift the 2 N rock was 6 Joules.

Explanation:

Weight of the Rock = 2 Newtons

So, the force required to lift this rock will be equal to its weight or greater than its weight

Minimum force required to lift the Rock = 2 N

Displacement of the Rock when force was applied = 3 m,

Work Done: It is defined as product of force applied to the object into displacement of the object.

$\text{Work done}=Force\times Displacement=2 N\times 3 m=6 Nm=6 Joules$

The work done to lift the 2 N rock was 6 Joules.

You might be interested in

A chemist prepares a solution of silver(I) nitrate by measuring out of silver(I) nitrate into a volumetric flask and filling the

leva [86]

Answer:

3.33 M

Explanation:

It seems your question is incomplete, however, that same fragment has been found somewhere else in the web:

" A chemist prepares a solution of silver nitrate (AgNO3) by measuring out 85.g of silver nitrate into a 150.mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in mol/L of the chemist's silver nitrate solution. Be sure your answer has the correct number of significant digits. "

In this case, first we calculate the moles of AgNO₃, using its molecular weight:

85.0 g AgNO₃ ÷ 169.87 g/mol = 0.500 mol AgNO₃

Then we convert the 150 mL of the volumetric flask into L:

150 / 1000 = 0.150 L

Finally we divide the moles by the volume:

0.500 mol AgNO₃ / 0.150 L = 3.33 M

4 0

2 years ago

How many moles of gas would occupy 22.4 L at 273K in one arm?

andrew-mc [135]

Answer:

1 mol

Explanation:

Using the general gas law equation as follows:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

According to the provided information in the question;

V = 22.4L

T = 273K

P = 1 atm

R = 0.0821 Latm/molK

n = ?

Using PV = nRT

n = PV/RT

n = (1 × 22.4) ÷ (0.0821 × 273)

n = 22.4 ÷ 22.4

n = 1mol

4 0

2 years ago

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

Is the following nuclear equation balanced? No Yes

fredd [130]

Answer:

Yes.

The nuclear equation {226/88 Ra → 222/26 Rn + 4/2 He} is balanced. As we know that an alpha particle is identical to a helium atom. This implies that if an alpha particle is eliminated from an atom's nucleus, an atomic number of 2 and a mass number of 4 is lost.

Therefore, the equation will be reduced to:

226 - 4 = 222

88 - 2 = 86

Hence, the equation is balanced.

Explanation:

7 0

2 years ago

Read 2 more answers

What is this thermodynamic?

VLD [36.1K]

Answer:

oh coooool

I have one Id it is on ace with maybe 5800 points

maybe this picture will help in something

6 0

2 years ago