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2 years ago

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A gas sample enclosed in a rigid metal container at room temperature (20.0∘C∘C) has an absolute pressure p1p1p_1. The container

is immersed in hot water until it warms to 40.0∘C∘C. What is the new absolute pressure p2p2p_2? Express your answer in terms of p1p1p_1.

1 answer:

choli [55]2 years ago

3 0

Answer: p2 = 1.06p1

Explanation: pressure increases with temperature increase.

According to Gass law

P1/T1 = P2/T2

T1 = 20°c = 20 +273 = 293k

T2 = 40°c = 40 +373 = 313k

Therefore

P2 = P1T2/T1 = 313P2/293

P2 = 1.06P1

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When backpacking in the wilderness, hikers often boil water to sterilize it for drinking. Suppose that you are planning a backpa

Pavlova-9 [17]

Answer:

2.104 L fuel

Explanation:

Given that:

Volume of water = 35 L = 35 × 10³ mL

initial temperature of water = 25.0 ° C

The amount of heat needed to boil water at this temperature can be calculated by using the formula:

$q_{boiling} = mc \Delta T$

where

specific heat of water c= 4.18 J/g° C

$q_{boiling} = 35 \times 10^{3} \times \dfrac{1.00 \ g}{1 \ mL} \times 4.18 \ J/g^0 C \times (100 - 25)^0 C$

$q_{boiling} = 10.9725 \times 10^6 \ J$

Also; Assume that the fuel has an average formula of C7 H16 and 15% of the heat generated from combustion goes to heat the water;

thus the heat of combustion can be determined via the expression

$q_{combustion} =- \dfrac{q_{boiling}}{0.15}$

$q_{combustion} =- \dfrac{10.9725 \times 10^6 J}{0.15}$

$q_{combustion} = -7.315 \times 10^{7} \ J$

$q_{combustion} = -7.315 \times 10^{4} \ kJ$

For heptane; the equation for its combustion reaction can be written as:

$C_7H_{16} + 11O_{2(g)} -----> 7CO_{2(g)}+ 8H_2O_{(g)}$

The standard enthalpies of the products and the reactants are:

$\Delta H _f \ CO_{2(g)} = -393.5 kJ/mol$

$\Delta H _f \ H_2O_{(g)} = -242 kJ/mol$

$\Delta H _f \ C_7H_{16 }_{(g)} = -224.4 kJ/mol$

$\Delta H _f \ O_{2{(g)}} = 0 kJ/mol$

Therefore; the standard enthalpy for this combustion reaction is:

$\Delta H ^0= \sum n_p\Delta H^0_{f(products)}- \sum n_r\Delta H^0_{f(reactants)}$

$\Delta H^0 =( 7 \ mol ( -393.5 \ kJ/mol) + 8 \ mol (-242 \ kJ/mol) -1 \ mol( -224.4 \ kJ/mol) - 11 \ mol (0 \ kJ/mol))$

$\Delta H^0 = (-2754.5 \ \ kJ - 1936 \ \ kJ+224.4 \ \ kJ+0 \ \ kJ)$

$\Delta H^0 = -4466.1 \ kJ$

This simply implies that the amount of heat released from 1 mol of C7H16 = 4466.1 kJ

However the number of moles of fuel required to burn $7.315 \times 10^{4} \ kJ$ heat released is:

$n_{fuel} = \dfrac{q}{\Delta \ H^0}$

$n_{fuel} = \dfrac{-7.315 \times 10^{4} \ kJ}{-4466.1 \ kJ}$

$n_{fuel} = 16.38 \ mol \ of \ C_7 H_{16$

Since number of moles = mass/molar mass

The mass of the fuel is:

$m_{fuel } = 16.38 mol \times 100.198 \ g/mol}$

$m_{fuel } = 1.641 \times 10^{3} \ g$

Given that the density of the fuel is = 0.78 g/mL

and we know that :

density = mass/volume

therefore making volume the subject of the formula in order to determine the volume of the fuel ; we have

volume of the fuel = mass of the fuel / density of the fuel

volume of the fuel = $\dfrac{1.641 \times 10^3 \ g }{0.78 g/mL} \times \dfrac{L}{10^3 \ mL}$

volume of the fuel = 2.104 L fuel

3 0

2 years ago

KClKCl has a lattice energy of −701 kJ/mol.−701 kJ/mol. Consider a generic salt, ABAB , where A2+A2+ has the same radius as K+,K

elena-14-01-66 [18.8K]

Explanation:

It is given that lattice energy is -701 kJ/mol.

Whereas it is known that realtion between lattice energy and radius is as follows.

Lattice energy $\propto \frac{Z_{+}Z_{-}}{r}$

where, $Z_{+}$ = +2, and $Z_{-}$ = -2

Therefore, lattice energy of AB = $4 \times \text{lattice energy of KCl}$

= $4 \times -701 kJ/mol$

= -2804 kJ/mol

Thus, we can conclude that lattice energy of the salt ABAB is -2804 kJ/mol.

6 0

1 year ago

Surface currents are mainly caused by prevailing winds. What is the best synonym for "prevailing?"

77julia77 [94]

Answer:

gentle

Explanation:

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2 years ago

What volume will 50.2 grams of co2 (g) occupy at stp?

Genrish500 [490]

Number of moles of CO2 =
Mass /Ar
= 50.2 / (12 + 32)
1.14 mols

For every 1 mol of gas, there will be
24000 cm^3 of gas

Vol. = 1.14 x 24 dm^3
= 27.36 dm^3

8 0

1 year ago

"The elementary reaction 2 NO2(g) → 2 NO(g) + O2(g) is second order in NO2 and the rate constant at 600 K is 6.77 × 10-1 M-1s-1.

blsea [12.9K]

Answer : The half-life at this temperature is, 3.28 s

Explanation :

To calculate the half-life for second order the expression will be:

$t_{1/2}=\frac{1}{k\times [A_o]}$

When,

$t_{1/2}$ = half-life = ?

$[A_o]$ = initial concentration = 0.45 M

k = rate constant = $6.77\times 10^{-1}M^{-1}s^{-1}$

Now put all the given values in the above formula, we get:

$t_{1/2}=\frac{1}{6.77\times 10^{-1}M^{-1}s^{-1}\times 0.45M}$

$t_{1/2}=3.28s$

Therefore, the half-life at this temperature is, 3.28 s

7 0

1 year ago