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seraphim [82]
3 years ago
12

Are electrons found outside of an atom

Chemistry
1 answer:
sladkih [1.3K]3 years ago
8 0
Electrons are found outside of an atom.
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24 grams of CH4 was added to the above reaction. Calculate the theoretical yield of CO2. A. 66 grams B. 33 grams c. 132 grams. D
Sonbull [250]

Answer:

Option A. 66 g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CH₄ + 2O₂ —> CO₂ + 2H₂O

Next, we shall determine the mass of CH₄ that reacted and the mass of CO₂ produced from the balanced equation. This is illustrated below:

Molar mass of CH₄ = 12 + (4×1)

= 12 + 4 = 16 g/mol

Mass of CH₄ from the balanced equation = 1 × 16 = 16 g

Molar mass of CO₂ = 12 + (16×2)

= 12 + 32 = 44 g/mol

Mass of CO₂ from the balanced equation = 1 × 44 = 44 g

SUMMARY:

From the balanced equation above,

16 g of CH₄ reacted to produce 44 g of CO₂.

Finally, we shall determine the theoretical yield of CO₂. this can be obtained as follow:

From the balanced equation above,

16 g of CH₄ reacted to produce 44 g of CO₂.

Therefore, 24 g of CH₄ will react to produce = (24 × 44) /16 = 66 g of CO₂.

Thus, the theoretical yield of CO₂ 66 g

8 0
3 years ago
A 100.0-mL sample of 1.00 M NaOH is mixed with 50.0 mL of 1.00 M H2SO4 in a large Styrofoam coffee cup calorimeter fitted with a
worty [1.4K]

Answer:

THE ENTHALPY CHANGE IN KJ/MOLE IS +114 KJ/MOLE.

Explanation:

Heat = mass * specific heat capacity * temperature rise

Total volume = 100 + 50 = 150 mL

Total mass = density * volume

Total mass = 1 * 150 mL = 150 g

So therefore, the heat evolved during the reaction is:

Heat = 150 * 4.18 * ( 31.4 - 22.3)

Heat = 150 * 4.18 * 9.1

Heat = 5705.7 J

Equation for the reaction:

2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)  

From the equation, 2 moles of NaOH reacts with 1 mole of H2SO4 to produce 1 mole of Na2SO4 and 2 moles of water

50 mL of 1 M of H2SO4 contains

50 * 1 / 1000 mole of acid

= 0.05 mole of acid

The production of 1 mole of water evolved 5705.7 J of heat and hence the enthalpy changein kJ per mole will be:

0.05 mole of H2SO4 produces 5705.7 J of heat

1 mole of H2SO4 will produce 5705.7 / 0.05 J

= 114,114 J / mole

In kj/mole = 114 kJ/mole.

Hence, the enthalpy change of the reaction in kJ /mole is +114 kJ/mole.

5 0
3 years ago
Which of the following statements are true regarding the dissociation of Al(NO3)3? Select more than one
Gelneren [198K]

Answer:

There will be one Al3+ ion.

There will be 3 NO3- ions

Explanation:

Dissociation equation:

Al(NO₃)₃   →  Al³⁺ + 3NO₃¹⁻

When aluminium nitrate dissociate it produces one silver ion (Al³⁺) and three (NO₃¹⁻) ions.

Properties of Al(NO₃)₃:

It is inorganic compound having molecular mass 169.87 g/mol.

It is white odor less compound.

Its density is 4.35 g/mL.

Its melting and boiling points are 120°C and 440°C.

It is soluble in water.

It is sued to treat infections.

It is used in the photographic films.

It s toxic and must be handled with great care.

4 0
3 years ago
What substance can be used to electrolyze water? any electrolyte that is not easily reduced or oxidized hydrochloric acid only s
Alisiya [41]
<span>any electrolyte that is not easily reduced or oxidized</span>
7 0
3 years ago
A student prepares a aqueous solution of butanoic acid . Calculate the fraction of butanoic acid that is in the dissociated form
ella [17]

Answer:

15.4%

Explanation:

If Ka = 0.54 mM = 1.51x10⁻⁵

Then;

C₄H₈O₂               -------->            C₄H₇O₂⁻          +           H⁺

I                    0.54x10⁻³                             0                                0

E                   0.54x10⁻³(1-x)                      0.54x10⁻³x                0.54x10⁻³x

Recall that x is the percentage degree of dissociation

From the ICE table;

Ka = [C₄H₇O₂⁻] [ H⁺]/[C₄H₈O₂]

1.51x10⁻⁵=(0.54x10⁻³x) (0.54x10⁻³x)/ 0.54x10⁻³(1-x)  

1.51x10⁻⁵ = 0.54x10⁻³x^2/1-x

1.51x10⁻⁵(1-x) = 0.54x10⁻³x^2

1.51x10⁻⁵ - 1.51x10⁻⁵x = 0.54x10⁻³x^2

Hence;

0.54x10⁻³x^2 + 1.51x10⁻⁵x - 1.51x10⁻⁵=0

x^2 + 0.028x - 0.028 = 0

Solving the quadratic equation here;

x = 0.154 or −0.182

Ignoring the negative result, x = 0.154

Hence, fraction of butanoic acid that is in the dissociated form in this solution = 15.4%

3 0
3 years ago
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