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2 years ago

Why is the formula for sodium thiosulfate Na2S2O3, not Na2SO3?

Chemistry

1 answer:

Tasya [4]2 years ago

6 0

Answer:

Sodium thiosulfate is an inorganic sodium salt composed of sodium and thiosulfate ions in a 2:1 ratio. It has a role as an antidote to cyanide poisoning, a nephroprotective agent and an antifungal drug. It contains a thiosulfate(2-).

Explanation:

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Which of these accurately describes Newton’s third law?

marshall27 [118]

newtons third law is.........

for every reaction there is an equal and oposite reaction force

ex. im pushing the squares on my keyboard and it pushing back on me

im siting on my chair, the chair pushes on the floor and my butt pushes on the chair

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2 years ago

The Ksp of calcium carbonate in water at 25 °C is 2.25 x 10-8. CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq) What is favored at

seraphim [82]

Answer:

the solubility of CaCO3 is 0.015g/l 25 °C

is favored at equilibrium

Explanation:

The Ksp of calcium carbonate in water at 25 °C is 2.25 x 10-8. CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq) What is favored at equilibrium?

solubility is the property of a solute to dissolve in a solvent(liquid, gas ) to form a solution(soution can be saturated ,unsaturated, or supersaturated)

CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq)

in partial dissociation , we can say

2.25x 10^-8= $Ca^{2+} +CO^{2-}_{3} }$

let Ca^2+=CO3^-2=S

2.25x10^-8=S*S

S^2=2.25x10^-8

S=0.00015mol/L

Converting that to g/l

the relative molecular mass of CaCO3=100g/mol

0.00015*100g/mol

0.015g/l

the solubility of CaCO3 is 0.015g/l @room temperature

is favored at equilibrium

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2 years ago

URGENT!! I will have brainliest!!!

Vlad [161]

Answer:

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Explanation:

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2 years ago

For the substituted cyclohexane compound shown, identify the atoms that will interact with the methyl group in a

zalisa [80]

Answer: choice: I & Choice E

Explanation:

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2 years ago

A buffer contains 0.18 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. What is the pH

Yuki888 [10]

Answer:

1) pH = 5.05

2) pH = 5.13

3) pH = 4.97

Explanation:

Step 1: Data given

Number of moles of propionic acid = 0.18 moles

Number of moles sodium propionate = 0.26 moles

Volume = 1.20 L

Ka = 1.3 * 10^-5 → pKa = 4.989

Step 2: Calculate concentrations

Concentration = moles / volume

[acid]= 0.18/ 1.2 =0.150 M

[salt]= 0.26/ 1.3 = 0.217 M

pH = 4.89 + log(0.217/0.150)=<u>5.05</u>

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What is the pH of the buffer after the addition of 0.02 mol of NaOH?

moles acid = 0.18 - 0.02 = 0.16

[acid]= 0.16/ 1.2=0.133 M

moles salt = 0.26 + 0.02 = 0.28

[salt]= 0.28/ 12=0.233

pH = 4.89 + log 0.233/ 0.133 = 5.13

What is the pH of the buffer after the addition of 0.02 mol of HI?

moles acid = 0.18+ 0.02 = 0.20 moles

[acid]= 0.20/ 1.2 = 0.167 M

[salt]= 0.26 - 0.02= 0.24 moles

[salt]= 0.24/ 1.2 = 0.20 M

pH = 4.89 + log 0.20/ 0.167= 4.97

8 0

2 years ago