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julsineya [31]
3 years ago
11

Which two elements make up the compound named butane? What is the ratio of these two hours elements in butane? How would you use

chemical symbols to represent butane?
Chemistry
1 answer:
otez555 [7]3 years ago
7 0

Answer:

1. Carbon and hydrogen

2. The ratio is C : H => 4 : 10

3. C4H10

Explanation:

1. Butane is made up carbon (C) and hydrogen (H)

2. Butane is the 4th member of the alkane series

The alkanes has generated formula CnH2n+2

Butane is number 4 in the family i.e n = 4

Butane => CnH2n+2

Butane => C4H2x4 +2

Butane => C4H8+2

Butane => C4H10

Therefore, the ratio of carbon and hydrogen in butane is:

C : H => 4 : 10

3. Butane contain carbon and hydrogen in the ratio 4 : 10. Therefore, butane is represented as C4H10

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Explanation:

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3 years ago
Find the solubility of agi in 2.5 m nh3 [ksp of agi = 8.3 × 10−17; kf of ag(nh3)2+ = 1.7 × 107].
vovangra [49]
When the first reaction equation is:

AgI(S) ↔ Ag+(Aq)  +  I-(Aq)

So, the Ksp expression = [Ag+][I-]

∴Ksp = [Ag+][I-] = 8.3 x 10^-17

Then the second reaction equation is:

 Ag+(aq)  + 2NH3(aq) ↔  Ag(NH3)2+  

So, Kf expression = [Ag(NH3)2+] / [Ag+] [NH3]^2

∴Kf = [Ag(NH3)2+] /[Ag+] [NH3]^2 = 1.7 x 10^7

by combining the two equations and solve for Ag+:

and by using ICE table:

               AgI(aq) + 2NH3  ↔  Ag(NH3)2+  + I-
initial                        2.5                    0            0 

change                    -2X                    +X             +X

Equ                     (2.5-2X)                   X               X

so K = [Ag(NH3)2+] [I-] / [NH3]^2

Kf * Ksp = X^2 / (2.5-2X)

8.3 x 10^-17 * 1.7 x10^7 = X^2 / (2.5-2X) by solving for X

∴ X = 5.9 x 10^-5 

∴ the solubility of AgI = X = 5.9 x 10^-5 M

4 0
3 years ago
HEllp no links no i dont knows and no hfdifewbhelr
Dahasolnce [82]

Answer:

second answer

Explanation:

The melting point would occur more quickly but stay the same.

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5 0
3 years ago
Read 2 more answers
A sucrose solution is prepared to a final concentration of 0.210 MM . Convert this value into terms of g/Lg/L, molality, and mas
vova2212 [387]

Answer:

1) 71.9 g/L

2) 0.221 m olal

3)  7.05% by mass

Explanation:

Step 1: Data given

Concentration of sucrose = 0.210 M

Molar weight of sucrose = 342.3 g/mol

Density of solution = 1.02 g/mL

Mass of water = 948.1 grams

Step 2: Convert this value into terms of g/L

(0.210 mol/L) * (342.3 g/mol) = 71.9 g/L

Calculate the molality

Step 1: Calculate mass water

Suppose we have a volume of 1.00L

Mass of the solution = 1000 mL * 1.02 g/mL = 1020 g solution

We know that there are 71.9 g of solute in a liter of solution from the first calculation. This means

(1020 grams solution) - (71.9 g solute) = 948.1 g = 0.9481 kg water

Step 2: Calculate molality

Molality = moles sucrose / mass water

(0.210 mol) / (0.9481 kg) = 0.221 mol/kg = 0.221 m olal

Mass %

% MAss = (mass solute / mass solution)*100%

(71.9 g) / (1020 g) *100% = 7.05% by mass

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arlik [135]
Hydrosphere because it is water and water is where the dolphins swim. 
4 0
4 years ago
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