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Anit [1.1K]
3 years ago
14

Help meeeeee

Chemistry
1 answer:
kolbaska11 [484]3 years ago
3 0

Answer:

469.9K

Explanation:

Using Charles's law equation:

V1/T1 = V2/T2

Where;

V1 = initial volume (ml)

V2 = final volume (ml)

T1 = initial temperature (K)

T2 = final temperature (K)

Based on the information provided:

V1 = 280.0ml

V2 = 440.0 ml

T1 = 26.0°C = 26 + 273 = 299K

T2 = ?

Using V1/T1 = V2/T2

280/299 = 440/T2

0.936 = 440/T2

T2 = 440 ÷ 0.936

T2 = 469.9K

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Basile [38]

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

6 0
3 years ago
Which of the following is a product in the chemical equation N + O2 = NO2
zlopas [31]
The answer is 1. NO2
3 0
3 years ago
WILL GIVE MOST BRAINLY ( PLEASE HURRY )
sergiy2304 [10]

Answer:

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Explanation:

6 0
3 years ago
Read 2 more answers
Murcury is the only metal at room temperature. Its density is 13.6g/mL. How many grams of murcury will occupy a volume of 95.8mL
Hatshy [7]

Answer:

<h3>The answer is 1.30288 × 10³ g</h3>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume = 95.8mL

density = 13.6g/mL

We have

mass = 13.6 × 95.8 = 1302.88

We have the final answer as

<h3>1.30288 × 10³ g</h3>

Hope this helps you

3 0
3 years ago
9. Given the following reaction:CO (g) + 2 H2(g) CH3OH (g)In an experiment, 0.45 mol of CO and 0.57 mol of H2 were placed in a 1
WITCHER [35]

Answer:

Keq=11.5

Explanation:

Hello,

In this case, for the given reaction at equilibrium:

CO (g) + 2 H_2(g) \rightleftharpoons CH_3OH (g)

We can write the law of mass action as:

Keq=\frac{[CH_3OH]}{[CO][H_2]^2}

That in terms of the change x due to the reaction extent we can write:

Keq=\frac{x}{([CO]_0-x)([H_2]_0-2x)^2}

Nevertheless, for the carbon monoxide, we can directly compute x as shown below:

[CO]_0=\frac{0.45mol}{1.00L}=0.45M\\

[H_2]_0=\frac{0.57mol}{1.00L}=0.57M\\

[CO]_{eq}=\frac{0.28mol}{1.00L}=0.28M\\

x=[CO]_0-[CO]_{eq}=0.45M-0.28M=0.17M

Finally, we can compute the equilibrium constant:

Keq=\frac{0.17M}{(0.45M-0.17M)(0.57M-2*0.17M)^2}\\\\Keq=11.5

Best regards.

3 0
4 years ago
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