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Scrat [10]
3 years ago
7

I HAVE TIME LIMIT I NEED CORRECT ANSWERS!

Chemistry
1 answer:
Basile [38]3 years ago
6 0

Answer:

5) oxygen

7)0.5480 M

9)5s

10) 2 σ and 2 π bonds

12) All of the carbon-oxygen bonds in a carbonate ion are weaker than the carbon-oxygen bonds in a carbon dioxide molecule

19) 0.400 g

Explanation:

For question 5

C2H6 + 7/2 O2 -----> 2CO2 + 3H2O

If 1 mole of ethane gives 2 moles of carbon dioxide

13 moles of ethane will give 13×2 = 26 moles of carbon dioxide

If 3.5 moles of oxygen gives 2 moles of carbon dioxide

42 moles of oxygen will give 42×2/3.5 = 24 moles of carbon dioxide

The reactant that gives the lower number of moles of product is the limiting reactant. This means that oxygen is the limiting reactant here.

7) concentration of acid CA =2.456 M

Concentration of base CB= ???

Volume of acid VA= 15.41 ml

Volume of base VB= 34.53 ml

Number of moles of acid NA= 2

Number of moles of base NB= 1

Ca(OH)2(aq) + 2HCl(aq) ----> CaCl2(aq) + 2H2O (l)

CB= CA VA NB/VB NA

CB= 2.456 × 15.41 ×1/34.53×2

CB= 0.548 M

9) For question 9, we have to look at the electronic configuration of Rb. We have [Kr]5s1. The outermost 5s1 level will have the least effective nuclear charge and is easily lost.

10) There are two sigma bonds and two pi bonds in CO2

12) The C-O bond length in the carbonate ion is 136 pm while the C-O bond length in CO2 is 116 pm. The longer the bond, the weaker the bond hence the C-O bonds in the carbonate ion are weaker than those in the carbon dioxide molecule.

19) From

m= mass of solute

M= molar mass of solute

C= concentration of solute

V= volume of solution

m/M = CV

m= MCV

m= 40.0gmol-1 × 50.0/1000 × 0.200

m= 0.4 g

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If 75 grams of oxygen react, how many grams of aluminum are required?
german

Answer:

84.24 g

Explanation:

Given data:

Mass of oxygen = 75 g

Mass of Al required to react = ?

Solution:

Chemical equation:

4Al + 3O₂     →   2Al₂O₃

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 75 g/ 32 g/mol

Number of moles = 2.34 mol

Now we will compare the moles of oxygen with Al.

                          O₂         :          Al

                           3          :             4

                        2.34        :         4/3×2.34 = 3.12 mol

Mass of Al required:

Mass = number of moles × molar mass

Mass = 3.12 mol × 27 g/mol

Mass = 84.24 g

5 0
3 years ago
The standard enthalpy of formation (ΔHf°) of calcium carbonate is –1207 kJ/mol. Which ONE of the equations below has ΔH° = –1207
Art [367]

Answer:

A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)

Explanation:

Standard enthalpy of formation of a chemical is defined as the change in enthalpy durin the formation of 1 mole of the substance from its constituent elements in their standard states.

The consituent elements of calcium carbonate, CaCO₃, in their standard states (States you will find this pure elements in nature), are:

Ca(s), C(s) and O₂(g)

That means, the equation that represents standard enthalpy of CaCO₃ is:

<h3>A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)</h3><h3 />

<em>Is the equation that has ΔH° = -1207kJ/mol</em>

3 0
3 years ago
At 0 degrees Celsius, a gas occupies 22.4L. How hot must the gas be in celcius to reach a volume of 25.0L
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Answer:

31.7 °C

Explanation:

Charles law states that for volume of a gas is directly proportional to the absolute temperature for a fixed amount of gas at constant pressure

we can use the following equation

V1/T1 = V2/T2

where V1 is volume and T1 is temperature at first instance

V2 is volume and T2 is temperature at second instance

temperature should be in kelvin scale

T1 - 0 °C + 273 = 273 K

substituting the values in the equation

22.4 L / 273 K = 25.0 L / T2

T2 = 304.7 K

temperature in celcius is - 304.7 K - 273 = 31.7 °C

the gas must be 31.7 °C to reach a volume of 25.0 L

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ash QC ok

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