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2 years ago

Which of the outer planets is the biggest A.jupiter B.saturn C.your anus D.neptune

Chemistry

2 answers:

Alina [70]2 years ago

5 0

A. Jupiter. is the correct answer. Mark as brainliest please.

masha68 [24]2 years ago

4 0

It’s spelt Uranus by the way. The answer is Jupiter

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What is the freezing point of an aqueous suger (C12H22O11) solution that boils at 110C°?

AURORKA [14]

Sucrose doesn't show appreciable dissociation in water, so i≈1 in this case. by extension, −7.3°C is the approximate freezing point of this solution.

<h3>What is the freezing point?</h3>

The temperature at which a liquid solidifies.

Sucrose doesn't show appreciable dissociation in water, so i≈1 in this case. by extension, −7.3°C is the approximate freezing point of this solution.

Learn more about the freezing point here:

brainly.com/question/3543605

#SPJ1

7 0

1 year ago

Calculate the ph of a 0.60 m h2so3, solution that has the stepwise dissociation constants ka1 = 1.5 × 10-2 and 1.82 1.06 1.02 2.

Vsevolod [243]

Missing in your question Ka2 =6.3x10^-8
From this reaction:
H2SO3 + H2O ↔ H3O+ + HSO3-
by using the ICE table :
H2SO3 ↔ H3O + HSO3-
intial 0.6 0 0
change -X +X +X
Equ (0.6-X) X X

when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
∴[H2SO3] = 0.6 - 0.088 = 0.512
[HSO3-] = [H3O+] = 0.088

by using the ICE table 2:
HSO3- ↔ H3O + SO3-
initial 0.088 0.088 0
change -X +X +X
Equ (0.088-X) (0.088+X) X

Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] = 0.088 as the value of Ka2 is very small
6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
= 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
= -㏒ 0.088 = 1.06

5 0

2 years ago

Read 2 more answers

Convert 3.82 inches to km

weeeeeb [17]

the answer is 0.000097 KM

6 0

3 years ago

Read 2 more answers

A student has two unopened 33 CL cans containing carbonated water. Can A has been stored in the garage 32 degrees and can B has

IgorC [24]

Answer:

A. Can A will make a louder and stronger fizz than can B.

Explanation:

The solubility of a gas in a liquid decreases as the temperature increases, so the warmer can will have more undissolved carbon dioxide.

The warmer can will be under greater pressure, so it will make a louder and stronger fizz.

7 0

2 years ago

Calculate %ic of the interatomic bonds for the intermetallic compound al6mn. on the basis of this result, what type of interatom

pashok25 [27]

Answer : % ionic character is 0.20%

Bonding between the two metals will be purely metallic.

Explanation: For the calculation of % ionic character, we use the formula

$\% \text{ ionic character}= [1-e^{\frac{-(X_A-X_B)^2)}{4}}]\times(100\%)$

where $X_A$ & $X_B$ are the Pauling's electronegativities.

The table attached has the values of electronegativities, by taking the values of Al and Mn from there,

$X_{Al}=1.61$

$X_{Mn}=1.55$

Putting the values in the electronegativity formula, we get

$\% \text{ ionic character}= [1-e^{\frac{-(1.61-1.55)^2)}{4}}]\times(100\%)$

= 0.20%

Now, there are 3 types of inter atomic bonding

1) Ionic Bonding: It refers to the chemical bond in which there is complete transfer of valence electrons between atoms

2) Covalent Bonding: It refers to the chemical bond involving the sharing of electron pairs between 2 atoms.

3) Metallic Bonding: It refers to the chemical bond in which there is an electrostatic force between the positively charged metal ions and delocalised electrons.

In $Al_6Mn$ compound, the % ionic character is minimal that is $\sim$ 0.20% and there are two metal ions present, therefore this compound will have metallic bonding.

8 0

2 years ago