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faust18 [17]
2 years ago
10

An isotope of what element is used as the standard for the

Chemistry
1 answer:
balandron [24]2 years ago
7 0

Answer:

Since 1961 the standard unit of atomic mass has been one-twelfth the mass of an atom of the isotope carbon-12. An isotope is one of two or more species of atoms of the same chemical element that have different atomic mass numbers (protons + neutrons).

isotope carbon-12

Explanation:

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50.00 mL of 0.10 M HNO 2 (nitrous acid, K a = 4.5 × 10 −4) is titrated with a 0.10 M KOH solution. After 25.00 mL of the KOH sol
boyakko [2]

Answer:

b. 3.35

Explanation:

To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.

pH = pKa + log ([salt]/[acid])     (Eq. 01)

Where  

pKa = -log(Ka)        (Eq. 02)

[salt] = Molar concentration of salt produced as a result of titration

[acid] = Molar concentration of acid left in the solution after titration

Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:

HNO2 + KOH ⇆ H2O + KNO2    (Eq. 03)

This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.

Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles

Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles

As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.

Therefore

Amount of salt produced i.e [salt] = 0.0025 moles       (Eq. 04)

Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)

Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:

pH= -log(4.5x10 -4) + log (0.0025/0.0025)

Solving above we get  

pH = 3.35

5 0
3 years ago
What is the total number of Joules of heat absorbed by 65.0 grams of water when the temperature of the water is raised from 25°C
Norma-Jean [14]

Answer:

Total number of heat absorbed is 4.08kJ

Explanation:

Explanation is contained in the picture attached

5 0
2 years ago
The methylene chloride solution is filtered through sodium sulfate to:
Vlad [161]

Answer:

C. Dry the methylene chloride by removing water

Explanation:

Anhydrous sodium sulfate is known for its high capacity to absorb water, for this reason it is widely used in laboratories as a drying agent.

Sodium sulfate is a neutral molecule so it cannot be used to neutralize and is very stable, so it is difficult to precipitate organic molecules.

5 0
2 years ago
2. 27.8 mL of an unknown were added to a 50.0-mL flask that weighs 464.7 g. The total mass of the flask and the liquid is 552.4
agasfer [191]

Answer:

d=4.24x10^{-4}\frac{lb}{in^3}

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary for us to set the equation for the calculation of density and mass divided by volume:

d=\frac{m}{V}

Thus, we can find the mass of the unknown by subtracting the total mass of the liquid to the mass of the flask and the liquid:

m=552.4g-464.7g=87.7g

So that we are now able to calculate the density in g/mL first:

d=\frac{87.7g}{27.8mL}=3.15g/mL

Now, we proceed to the conversion to lb/in³ by using the following setup:

d=3.15\frac{g}{mL}*\frac{1lb}{453.6g}*\frac{1in^3}{16.3871mL}\\\\d=4.24x10^{-4}\frac{lb}{in^3}

Regards!

6 0
2 years ago
Help?<br> The question is down below.
Murrr4er [49]
The first three are T I don’t know about the next two and the last one is T
6 0
2 years ago
Read 2 more answers
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