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Sophie [7]
4 years ago
15

Gas bubbles form when baking soda and vinegar react

Chemistry
1 answer:
user100 [1]4 years ago
6 0
Yes! Becuz baking soda is strong and if you mix it with any liqued  it makes BUBBLES
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11. How did the solubility product constant Ksp of KHT in pure water compare to its solubility product constant Ksp of KHT in KC
Airida [17]

Answer:

Explanation:

KHT is a salt which ionises in water as follows

KHT ⇄ K⁺ + HT⁻

Solubility product Kw= [ K⁺ ] [ HT⁻ ]

product of concentration of K⁺ and HT⁻ in water

In KCl solution , the solubility product of KHT will be decreased .

In KCl solution , there is already presence of K⁺  ion in the solution . So

in the equation  

[ K⁺ ] [ HT⁻ ]  = constant

when K⁺ increases [ HT⁻ ] decreases . Hence less of KHT dissociates due to which its  solubility decreases . It is called common ion effect . It is so because here the presence of common ion that is K⁺ in both salt to be dissolved and in solvent , results in decrease of solubility of the salt .

7 0
3 years ago
I bet that outfit is made of Copper and Tellurium, because it is so ________.
Aleonysh [2.5K]

THE SYMBOLS!!!

CuTe

LOLOLOL

7 0
4 years ago
Read 2 more answers
A student would like to determine how heating a liquid changes its volume. The student hypothesizes that the liquid will increas
Zigmanuir [339]

Answer:

The volume of the liquid should be measured before it is heated.

Explanation:

3 0
3 years ago
Complete this reaction of a carboxylic acid with a strong base.
raketka [301]

Answer:

C6H5COOH + OH- —> C6H5COO- + H2O

Explanation:

C6H5COOH + OH- —> C6H5COO- + H2O

In the reaction above, C6H5COOH donate a proton(H+) to form the carboxylate ion C6H5COO-. The proton (H+) combines with the OH- to form H2O.

This can better be understood in the illustration below

C6H5COOH + NaOH —> C6H5COONa + H2O

6 0
3 years ago
How many grams of H2O are produced when 35.0 g of NaOH reacts with 17.5 g of CO,?
zhenek [66]

Answer:

2NaOH + CO2 -> Na2CO3 + H2O

1) Find the moles of each substance

\eq n(NaOH)=\frac{35.0}{22.99+16.00+1.008\\}\  =\frac{35.0}{39.998} \ = 0.8750437522 moles\\n(CO_{2} ) = \frac{17.5}{12.01+32.00} = \frac{17.5}{44.01} = 0.3976369007 moles\\

2) Determine the limitting reagent

\\NaOH = \frac{0.8750437522}{2} = 0.4375218761\\\\

∴ Carbon dioxide is limitting as it has a smaller value.

3) multiply the limiting reagent by the mole ratio of unknown over known

n(H2O ) = 0.3976369007 × 1/2

             = 0.1988184504 moles

4) Multiply the number of moles by the molar mass of the substance.

m = 0.1988184504 × (1.008 × 2 + 16.00)

   = 0.1988184504 × 18.016

   = 3.581913202 g

Explanation:

6 0
3 years ago
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