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gogolik [260]
3 years ago
10

Each compound listed is soluble in water. For each compound, do you expect the resulting aqueous solution to conduct electrical

current? Why?
a) CsCl
b) CH3OH
c) Ca(NO2)2
d) C6H12O6
Chemistry
1 answer:
Strike441 [17]3 years ago
5 0

Answer:

See below in bold.

Explanation:

a) Yes  because it is an ionic compound. Forms ions in water solution.

b) This is an alcohol with covalent bonds it does not ionise in water - no.

c) Yes this forms ions in water solution.

d) This is a sugar and does not ionise so No.

You might be interested in
For a particular reaction, Δ=−111.4 kJ and Δ=−25.0 J/K.
vichka [17]

Answer:

\Delta G =-103.95kJ

Explanation:

Hello there!

In this case, since the thermodynamic definition of the Gibbs free energy for a change process is:

\Delta G =\Delta H-T\Delta S

It is possible to plug in the given H, T and S with consistent units, to obtain the correct G as shown below:

\Delta G =-111.4kJ-(298K)(-25.0\frac{J}{K}*\frac{1kJ}{1000J} )\\\\\Delta G =-103.95kJ

Best regards!

6 0
3 years ago
Sodas are made fizzy by the carbon dioxide ( CO ) dissolved in the liquid. An amount of carbon 2 dioxide equal to about 8.0 L of
eduard

Answer:

A) The number of moles of Carbon dioxide in 2.0 L bottle of soda = 0.325 mole

B) The mass of Carbon dioxide in 2.0 L bottle of soda = 14.3 g

Explanation:

Complete Question

Soda are made fizzy by the carbon dioxide (CO₂) dissolved int he liquid. An amount of carbon dioxide equal to about 8.0 L of carbon dioxide gas at atmospheric pressure and 300.0 K can be dissolved can be dissolved in a 2-L bottle of soda. The molar mass of CO₂ IS 44 g/mol.

A) How many moles of carbon dioxode are in the 2-L? (1L = 0.001 m³)

B) What is the mass of the carbon dioxide in the 2-L bottle of soda?

Solution

Assuming that CO₂ is an ideal gas.

A) The amount of carbon dioxide in 2.0 L of soda bottle is equal to the amount of Carbon dioxide in about 8.0 L of carbon dioxide gas at atmospheric pressure and 300.0 K

From the ideal gas equation, we know that

PV = nRT

P = pressure of the gas = atmospheric pressure = 101,325 Pa

V = volume of the gas = 8.0 L = 0.008 m³

n = number of moles of the ideal gas = ?

R = molar gas constant = 8.314 J/mol.K

T = absolute temperature of the gas in Kelvin = 300 K

101325 × 0.008 = n × 8.314 × 300

n = (101325×0.008) ÷ (8.314 × 300)

n = 0.324993986 = 0.325 mole of Carbon dioxide.

B) The mass of Carbon dioxide in 2.0 L bottle of soda

Mass = (Number of moles) × (Molar mass)

= 0.325 × 44.0

= 14.3 g

Hope this Helps!!!

7 0
2 years ago
Balance the following half reaction in basic conditions. Then, indicate the coefficients for H2O and OH– for the balanced half r
Ugo [173]

Answer:

The ballance half reactions are:

Mg²⁺  + 2e⁻ → Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

Coefficients for H2O and OH– are 3 for H₂O (in products side) and 6 for OH⁻ (in reactants side)

Explanation:

Si (s) + Mg(OH)₂ (s) → Mg (s) + SiO₃²⁻ (aq)

Let's see the oxidations number.

As any element in ground state, we know that oxidation state is 0, so Si in reactants and Mg in products, have 0.

Mg in reactants, acts with +2, so the oxidation number has decreased.

This is the reduction, so it has gained electrons.

Si in reactants acts with 0 so in products we find it with +4. The oxidation number increased it, so this is oxidation. The element has lost electrons.

Let's take a look to half reactions:

Mg²⁺  + 2e⁻ → Mg

Si  → SiO₃²⁻ + 4e⁻

In basic medium, we have to add water, as the same amount of oxygen we have, IN THE SAME SIDE. We have 3 oxygens in products, so we add 3 H₂O and in the opposite site we can add OH⁻, to balance the hydrogen. The half reaciton will be:

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

If we want to ballance the main reaction we have to multiply (x2) the half reaction of oxidation. So the electrons can be ballanced.

2Mg²⁺  + 4e⁻ → 2Mg

Now, that they are ballanced we can sum the half reactions:

2Mg²⁺  + 4e⁻ → 2Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

2Mg²⁺  + 4e⁻  + 6OH⁻ + Si  → 2Mg  +  SiO₃²⁻ + 4e⁻ + 3 H₂O

7 0
3 years ago
A sample of 0.370 mol of a metal oxide (m2o3) weighs 55.45 g. how many grams of o are in the sample?
Wewaii [24]
0.370 mol metal oxide = 55.45 g 

<span>1 mol = 55.45/0.370 = 149.86 g </span>

<span>in 1 mol there are 3 mol O = 16 * 3 = 48 g of O </span>

<span>there is 48/149.86 * 100% O in the sample </span>

<span>the sample has 48/149.86 * 0.370 = 0.119 g O</span>
6 0
3 years ago
EDTA EDTA is a hexaprotic system with the p K a pKa values: p K a1 = 0.00 pKa1=0.00 , p K a2 = 1.50 pKa2=1.50 , p K a3 = 2.00 pK
mihalych1998 [28]

Answer:

Check the explanation

Explanation:

When,

pH = -log[H+] = 3.30

[H+] = 5.0 X 10^{-4} M

Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 7.4 X 10^{-7} ; Ka6 = 4.3 X 10^{-11}

alpha[Y^-4] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6

= 1.56 X 10^{-20} + 3.12 X 10^{-17} + 2 X 10^{-15} + 4 X 10^{-14} + 1.6 X 10^{-13} + 2.34 X 10^{-16} + 2 X 10^{-23}

= 2.02 X 10^{-13}

When,

pH = -log[H+] = 10.15

[H+] = 7.08 X 10^{-11} M

Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 7.4 X 10^{-7} ; Ka6 = 4.3 X 10^-11

alpha[Y^{-4}] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6

= 1.26 X 10^{-61} + 1.8 X 10^{-51} + 8.1 X 10^{-43} + 1.12 X 10^{-34} + 3.17 X 10^{-27} + 3.3 X 10^{-23} + 1.83 X 10^{-23}

= 5.12 X 10^{-23}

4 0
3 years ago
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