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Papessa [141]
2 years ago
8

Select the correct answer. A student places a sample of white sodium bicarbonate powder in a test tube and heats it. The student

observes that the sodium bicarbonate undergoes a chemical change. After the change, what must be true about the substance in the test tube? A. It is not white. B. It is a liquid. C. It is an element. D. It is not sodium bicarbonate.
Chemistry
1 answer:
inessss [21]2 years ago
8 0

Answer is: D. It is not sodium bicarbonate.

Balanced chemical reaction of heating sodium bicarbonate:                    2NaHCO₃ → Na₂CO₃ + CO₂ + H₂O.

This is chemical change (chemical reaction), because new substances are formed (sodium carbonate, carbon(IV) oxide and water), the atoms are rearranged, so there is no sodium bicarbonate (NaHCO₃) in the test tube.

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Suppose you wanted to make a buffer of exactly ph 7.00 using kh2po4 and na2hpo4. if the final solution was 0.10 m in kh2po4, wha
OleMash [197]

Answer:- 0.138 M

Solution:- The buffer pH is calculated using Handerson equation:

pH=pKa+log(\frac{base}{acid})

KH_2PO_4 acts as a weak acid and Na_2HPO_4 as a base which is pretty conjugate base of the weak acid we have.

The acid hase two protons(hydrogen) where as the base has only one proton. So, we could write the equation as:

H_2PO_4^-\rightleftharpoons H^++HPO_4^-^2

Phosphoric acid gives protons in three steps. So, the above equation is the second step as the acid has only two protons and the base has one proton.

So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.

Let's plug in the values in the equation:

7.00=6.86+log(\frac{base}{0.10})

7.00-6.86=log(\frac{base}{0.10})

0.14=log(\frac{base}{0.10})

Taking antilog:

10^0^.^1^4=\frac{base}{0.10}

1.38=\frac{base}{0.10}

On cross multiply:

[base] = 1.38(0.10)

[base] = 0.138

So, the concentration of the base that is Na_2HPO_4 required to make the buffer is 0.138M.

5 0
2 years ago
Calculate the percent activity of the radioactive isotope strontium-89 remaining after 5 half-lives.
maria [59]
The answer to this question would be: 3.125%

Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%
5 0
2 years ago
What volume of a 0.716 m kbr solution is needed to provide 30.5 g of kbr?
Jlenok [28]
Answer is: volume of KBr is 357 mL.
c(KBr) = 0,716 M = 0,716 mol/L.
m(KBr) = 30,5 g.
n(KBr) = m(KBr) ÷ M(KBr).
n(KBr) = 30,5 g ÷ 119 g/mol.
n(KBr) = 0,256 mol.
V(KBr) = n(KBr) ÷ c(KBr).
V(KBr) = 0,256 mol ÷ 0,716 mol/L.
V(KBr) = 0,357 L · 1000 mL/L = 357 mL.
7 0
2 years ago
Read 2 more answers
compare the mass and volume of each object what is true of the mass and volume of all the floating objects
V125BC [204]

Answer:

Volume is always more than the mass for floating objects. For sinking objects mass is always more than the volume

Explanation:

None

3 0
2 years ago
How many moles of fluorine gas would occupy a volume of 42.3 L at a pressure of 106.1 kPa and a temperature of 940C?
timofeeve [1]

Answer:

5.7

Explanation:

6 0
2 years ago
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