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3 years ago

In the following reaction, which species is reduced? 2k+br2 yield 2k+ + 2br-

1 answer:

6 0

The species that is reduced in

2K + Br2 = 2K^+ + 2Br^-

is Br2 is reduced

This is because Br2 move from oxidation state zero to oxidation state of negative one. K is been oxidized since it move from oxidation state zero to oxidation state of positive one. K act as a reducing agent while Br2 act as a oxidizing agent

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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

4 years ago

How many atoms of nitrogen and oxygen are present in a molecule of dinitrogen pentoxide?

Kitty [74]

Answer: 7

Explanation: oxygen has 5 atoms and nitrogen has 2. Put them together and you get 7

6 0

3 years ago

Read 2 more answers

Oxygen gas is compressed in a piston–cylinder device from an initial state of 0.8 m3 /kg and 25°C to a final state of 0.1 m3 /kg

zhuklara [117]

Answer:

ΔS = -0.1076 kJ /kg*K

Explanation:

Step 1: Data given

Initial state = 0.8 m³/kg and 25 °C = 298.15 K

Final state = 0. 3³/kg and 287 °C = 560.15 K

Cv = 0.686 kJ/kg*K

Step 2: Calculate the average temperature

The average temperature = (25°C + 287 °C)/2 =156 °C ( = 429 K)

Step 3: Calculate the ΔS

ΔS =(Cv, average) * ln(T2/T1) + R*ln(V2/V1)

ΔS = 0.686 * ln(560.15/298.15) + 0.2598*ln( 0.1/0.8)

ΔS = -0.1076 kJ /kg*K

3 0

3 years ago

The decomposition of ammonia is: 2 NH3(g) = N2(g) + 3 H2(g). If Kp is 1.5 × 103 at 400°C, what is the partial pressure of ammoni

hjlf

Answer:

A) = 4.7 × 10⁻⁴atm

Explanation:

Given that,

Kp = 1.5*10³ at 400°C

partial pressure pN2 = 0.10 atm

partial pressure pH2 = 0.15 atm

To determine:

Partial pressure pNH3 at equilibrium

The decomposition reaction is:-

2NH3(g) ↔N2(g) + 3H2(g)

Kp = [pH2]³[pN2]/[pNH3]²

pNH3 =√ [(pH2)³(pN2)/Kp]

pNH3 = √(0.15)³(0.10)/1.5*10³ = 4.74*10⁻⁴ atm

$K_p = \frac{[pH_2] ^3[pN_2]}{[pNH_3]^2} \\pNH_3 = \sqrt{\frac{(pH_2)^3(pN_2)}{pNH_3} } \\pNH_3 = \sqrt{\frac{(0.15)^3(0.10)}{1.5 \times 10^3} } \\=4.74 \times 10^-^4atm$

= 4.7 × 10⁻⁴atm

4 0

3 years ago

Read 2 more answers

If a compound has a log kow value of 6.5, what would be its predicted concentration (in ppm) in the fat of fish that swim in wat

blsea [12.9K]

Easy ans I don't know .......

5 0

3 years ago