All categories

2 years ago

Suggest why sodium and hydrogen ions do not diffuse at the same rate

Chemistry

1 answer:

Troyanec [42]2 years ago

8 0

Answer:

sodium has got ionic bonds that are weak

compared to hydrogen covalent bonds that are strong

You might be interested in

A 63.5 g sample of an unidentified metal absorbs 355 ) of heat when its temperature changes

insens350 [35]

0.208 is the specific heat capacity of the metal.

Explanation:

Given:

mass (m) = 63.5 grams 0R 0.0635 kg

Heat absorbed (q) = 355 Joules

Δ T (change in temperature) = 4.56 degrees or 273.15+4.56 = 268.59 K

cp (specific heat capacity) = ?

the formula used for heat absorbed and to calculate specific heat capacity of a substance will be calculated by using the equation:

q = mc Δ T

c = $\frac{q}{mΔ T}$

c = $\frac{355}{63.5X 268.59}$

= 0.208 J/gm K

specific heat capacity of 0.208 J/gm K

The specific heat capacity is defined as the heat required to raise the temperature of a substance which is 1 gram. The temperature is in Kelvin and energy required is in joules.

5 0

2 years ago

In a hydrogen fuel cell, hydrogen gas and oxygen gas are combined to form water. Write the balanced chemical equation describing

xxTIMURxx [149]

<u>Answer:</u> The chemical reaction is given below.

<u>Explanation:</u>

A fuel cell is defined as the electrochemical cell which converts the chemical energy of a fuel (often used hydrogen) and an oxidizing agent (often used oxygen) into electrical energy via a pair of redox reactions.

The reactions which occur in hydrogen-oxygen fuel cell are:

At cathode: $H_2+2OH^-\rightarrow 2H_2O+2e^-$

At anode: $\frac{1}{2}O_2+H_2O+2e^-\rightarrow 2OH^-$

Net reaction: $H_2+\frac{1}{2}O_2\rightarrow H_2O$

Thus, the chemical reaction is given above.

5 0

2 years ago

Read 2 more answers

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

How many hydrogen atoms are in 11C2H6

tresset_1 [31]

Answer:

Explanation:

You will see H6 and the H stands for helium and the 6 is how many of that atom is there

5 0

2 years ago

Read 2 more answers

A 74.0-gram piece of metal at 94.0 °C is placed in 120.0 g of water in a calorimeter at 26.5 °C. The final temperature in the ca

disa [49]

Answer:

Explanation:

Heat lost by metal = mass x specific heat x fall in temperature

= 74 x S x ( 94 - 32 )

= 4588 S

heat gained by water = mass x specific heat x rise in temperature

= 120 x 1 x ( 32 - 26.5 ) ( specific heat of water is 1 cals / gm )

= 660

Heat lost = heat gained

4588S = 660

S = .14 cal /gm .

Specific heat of metal = .14 cal / gm

7 0

2 years ago