Ask question

Ask question

All categories

2 years ago

7

Harry uses a pulley to pull up a bucket of water from a well. He uses chemical energy in his body to apply mechanical energy to

the system. Which of the following energy transformations also occurs as the bucket rises?

1 answer:

Fudgin [204]2 years ago

6 0

The potential energy the bucket started with transforms into kinetic energy as it moves.

You might be interested in

For which of the following reactions is ΔH∘rxn equal to ΔH∘f of the product(s)?You do not need to look up any values to answer t

Nutka1998 [239]

Answer:

Reactions 1, 3 and 5

Explanation:

First thing's first, let's ensure that all the reactions given are balanced. This is given as;

CO(g) + 1/2 O2(g )→ CO2(g)

Li(s) + 1/2 F2(l) → LiF(s)

C(s) + O2(g) → CO2(g)

CaCO3(g) → CaO + CO2(g)

2Li(s) + F2(g) → 2LiF(s)

For the condition to be valid;

- There is by convention 1 mol of product made. This means we eliminate reactions with more than one mole of compound formed. This eliminates reaction 5.

- The lements haveto be in their state at room temperature. Fluorine is a gas, not a liquid, at room temperature ans pressure, so 2 is not a correct answer.

This leaves us with reactions 1, 3 and 5 as the correct reactions that satisify the condition.

3 0

2 years ago

The density of o2 gas at 16 degrees Celsius and 1.27atm is?

velikii [3]

Answer:

The density of O₂ gas is 1.71 $\frac{g}{L}$

Explanation:

Density is a quantity that allows you to measure the amount of mass in a given volume of a substance. So density is defined as the quotient between the mass of a body and the volume it occupies:

$density=\frac{mass}{volume}$

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

So, you can get:

$\frac{n}{V} =\frac{P}{R*T}$

The relationship between number of moles and mass is:

$n=\frac{mass}{molar mass}$

Replacing:

$\frac{\frac{mass}{molar mass} }{V} =\frac{P}{R*T}$

$\frac{mass}{V*Molar mass} =\frac{P}{R*T}$

So:

$\frac{mass}{V} =\frac{P*molar mass}{R*T}$

Knowing that 1 mol of O has 16 g, the molar mass of O₂ gas is 32 $\frac{g}{mol}$ .

Then:

$\frac{mass}{V} =\frac{P*molar mass of O_{2} }{R*T}$

In this case you know:

P=1.27 atm
molar mass of O₂= 32 $\frac{g}{mol}$ .

R= 0.0821 $\frac{atm*L}{mol*K}$

T= 16 °C= 289 °K (0°C= 273°K)

Replacing:

$density=\frac{mass}{V} =\frac{1.27atm*32\frac{g}{mol} }{0.0821\frac{atm*L}{mol*K} *289 K}$

Solving:

density= 1.71 $\frac{g}{L}$

<u><em>The density of O₂ gas is 1.71 </em></u> $\frac{g}{L}$ <u><em></em></u>

3 0

2 years ago

Consider the following reaction: CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at25 ∘C under these conditions: PCO2PCC

Salsk061 [2.6K]

<u>Answer:</u> The $\Delta G$ for the reaction is 54.6 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.760atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.180atm$

Putting values in above equation, we get:

$K_p=\frac{(0.760)^2}{0.140\times 0.180}\\\\K_p=22.92$

To calculate the Gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 22.92

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(22.92))\\\\\Delta G=54659.78J/mol=54.6kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.6 kJ/mol

7 0

2 years ago

What is the volume (in ml) of a 12.9 g piece of metal with a density of 7.25 g/cm3?

Leto [7]

Hey there:

1 cm³ = 1 mL

D = m / V

7.25 = 12.9 / V

V = 12.9 / 7.25

V = 1.779 cm³

6 0

2 years ago

Read 2 more answers

Suppose you take a piece of hard wax from an unlit candle. After you roll the wax between your fingers for a while, it becomes s

jolli1 [7]

Answer:

the candle is still solid..................in this case, yes!

Explanation:

it is still solid because the molecules are packed together tighter than the molecules in a liquid or gas.

6 0

2 years ago

Read 2 more answers