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tekilochka [14]
3 years ago
10

Why is Jupiter so much larger than Earth? Check all that apply.

Chemistry
2 answers:
Debora [2.8K]3 years ago
6 0

Answer:

The answer is A, C and E

also

1, 3 and 5

Explanation:

Why is Jupiter so much larger than Earth? Check all that apply.

Inner planets were created at higher velocities.

Outer planets have less gravity than inner ones.

Outer planets attracted helium, hydrogen, and other gasses.

Inner planets formed where temperatures were cooler.

Gases escaped from inner planets and not from outer ones.

SIZIF [17.4K]3 years ago
3 0

Answer:

The answer is A, C and E

also

1, 3 and 5

Explanation:

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In which of the three acid-base concepts can water be a product of an acid-base reaction? In which is it the only product?
Archy [21]

Salt and water is formed as product, when Arrhenius acid and Arrhenius base reacts and the reaction is known as neutralization reaction.

<h3>What is Arrhenius acid-base reaction?</h3>
  • The concept of acid and base based on the theory of ionization which was first proposed in 1884 by Svante Arrhenius
  • According to Arrhenius, on dissociation in water the hydrogen-containing compounds which give H+ ions or protons are the acids and  the hydroxide compounds which give OH− ions on dissociation in water are bases.
  • This theory is applicable only to compounds which are dissolved in aqueous solution.

To learn more about Arrhenius acid-base reaction: brainly.com/question/15196401

#SPJ4                                                                                    

4 0
1 year ago
Which of the following examples depict(s) a physical property?
dimaraw [331]

Answer:

D

Explanation:

All explain physical no chemical examples were given

6 0
3 years ago
Read 2 more answers
- How many moles of H2O are required to react to form 2.5 grams of CH ?
abruzzese [7]

Answer:

0.312 moles of H2O

Explanation:

no. of moles of ch4= mass ÷ molar mass

                               =2.5 ÷ 16.04

                               =0.156 moles of ch4

According to balanced chemical equation

CH4        :        H2O

1 mole     :        2 moles

0.156 moles :       x moles  

by cross multiplication

x=  (0.156x2) ÷ 1

 = 0.312 moles of H2O

7 0
3 years ago
If 8.500 g CH is burned and the heat produced from the burning is added to 5691 g of water at 21 °C, what is the final
mojhsa [17]

The final temperature = 36 °C

<h3>Further explanation</h3>

The balanced combustion reaction for C₆H₆

2C₆H₆(l)+15O₂(g)⇒ 12CO₂(g)+6H₂O(l)  +6542 kJ

MW C₆H₆ : 78.11 g/mol

mol C₆H₆ :

\tt \dfrac{8.5}{78.11}=0.109

Heat released for 2 mol C₆H₆ =6542 kJ, so for 1 mol

\tt \dfrac{0.109}{2}\times 6542=356.539~kJ/mol

Heat transferred to water :

Q=m.c.ΔT

\tt 356.539=5.691~kg\times 4.18~kj/kg^oC\times (t_2-21)\\\\t_2-21=15\rightarrow t_2=36^oC

3 0
2 years ago
In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].
Karo-lina-s [1.5K]

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of  [H^+] is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and[H^+] both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose +  H^+\rightarrow  fructose+ glucose

The rate law of the reaction is given as:

R=k[H^+][sucrose]

[H^+]=0.01M

[sucrose]= 1.0 M

R=k[0.01M][1.0 M]..[1]

a)

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

R'=[0.01 M][2.5 M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}

R'=2.5\times R

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

R'=[0.01 M][0.5 M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}

R'=2.5\times R

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)

The rate of the reaction when [H^+] is changed to 0.001 M = R'

R'=[0.0001 M][1.0 M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}

R'=0.01\times R

If concentration of  [H^+] is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d)

The rate of the reaction when [sucrose] and[H^+] both are changed to 0.1 M = R'

R'=[0.1M][0.1M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}

R'=1\times R

If concentration when [sucrose] and[H^+] both are changed to 0.1 M than rate will be increased by the factor of 1.

5 0
3 years ago
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