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3 years ago

What is biodiversity?

Chemistry

1 answer:

kifflom [539]3 years ago

7 0

Different amount of species in an area.

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Calculate the number of total atoms in 195 grams of Ni(OH)2.

Anni [7]

1.267 x 10 ^ 24 is the total number of atoms

6 0

3 years ago

You are studying the equilibirum between S8 and S2 gases at 0 celcius, you place a smaple of s2 in an otherwise empty, rigid con

Rainbow [258]

Answer:

Kp and Kc are 0.01266 and 145.17, respectively.

Explanation:

Please check document attached.

Download docx

5 0

3 years ago

The correct answer for the addition of 5.8 g + 2.26g + 1.311g + 2 g is?

jeka94

Answer:

= 11.371g

Explanation:

Hope This Helps! :]

4 0

3 years ago

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

3 years ago

What is the volume in liters of 5.25 moles of He gas?

serious [3.7K]

Answer:

$\boxed {\boxed {\sf 117.6 \ L \ He}}$

Explanation:

Regardless of the type of gas, 1 mole at standard temperature and pressure (STP) occupies a volume of 22.4 liters. In this case the gas is helium (He).

We can set up a ratio.

$\frac { 22.4 \ L \ He}{ 1 \ mol \ He}$

Multiply by the given number of moles.

$5.25 \ mol \ He *\frac { 22.4 \ L \ He}{ 1 \ mol \ He}$

The moles of helium will cancel.

$5.25 *\frac { 22.4 \ L \ He}{ 1 }$

$5.25 * { 22.4 \ L \ He}$

Multiply.

$117.6 \ L \ He$

5.25 moles of helium gas at STP is 117.6 liters of helium.

5 0

3 years ago