Answer:
The concentration the student should write down in her lab is 2.2 mol/L
Explanation:
Atomic mass of the elements are:
Na: 22.989 u
S: 32.065 u
O: 15.999 u
Molar mass of sodium thiosulfate, Na2S2O3 = (2*22.989 + 2*32.065 + 3*15.999) g/mol = 158.105 g/mol.
Mass of Na2S2O3 taken = (19.440 - 2.2) g = 17.240 g.
For mole(s) of Na2S2O3 = (mass taken)/(molar mass)
= (17.240 g)/(158.105 g/mol) = 0.1090 mole.
Volume of the solution = 50.29 mL = (50.29 mL)*(1 L)/(1000 mL)
= 0.05029 L.
To find the molar concentration of the sodium thiosulfate solution prepared we use the formula:
= (moles of sodium thiosulfate)/(volume of solution in L)
= (0.1090 mole)/(0.05029 L)
= 2.1674 mol/L
Element which possess different number of "Neutrons" in different situations could be an isotope of the atom
Ex. - 17 Cl 34, 17 Cl 35, 17 Cl 36
They all the isotopes.
Hope this helps!
Answer:
3.568 g of H₂O
Solution:
The Balance Chemical Equation is as follow,
2 H₂ + O₂ → 2 H₂O
Step 1: Calculate the Limiting Reagent,
According to Balance equation,
4.04 g (2 mol) H₂ reacts with = 32 g (1 mol) of O₂
So,
0.40 g of H₂ will react with = X g of O₂
Solving for X,
X = (0.40 g × 32 g) ÷ 4.04 g
X = 3.17 g of O₂
It means 0.4 g of H₂ requires 3.17 g of O₂, while we are provided with 3.2 g of O₂ which is in excess. Therefore, H₂ is the limiting reagent and will control the yield of products.
Step 2: Calculate amount of Water produced,
According to equation,
4.04 g (2 mol) of H₂ produces = 36.04 g (2 mol) of H₂O
So,
0.40 g of H₂ will produce = X moles of H₂O
Solving for X,
X = (0.40 g × 36.04 mol) ÷ 4.04 g
X = 3.568 g of H₂O
I believe the correct answers from the choices listed above are options two and five. It would be metals and hydroxyl ions that are common to bases. <span>Any of a class of compounds that form hydroxyl ions (OH) when dissolved in water, and whose aqueous solutions react with acids to form salts. Hope this answers the question.</span>
<u>Answer:</u> The partial pressure of hydrogen is 705.9 mmHg
<u>Explanation:</u>
Dalton's law of partial pressure states that the total pressure of the system is equal to the sum of partial pressure of each component present in it.
To calculate the partial pressure of hydrogen gas, we use the law given by Dalton, which is:
![P_T=p_{H_2}+p_{H_2O}](https://tex.z-dn.net/?f=P_T%3Dp_%7BH_2%7D%2Bp_%7BH_2O%7D)
We are given:
Total pressure of the collection tube,
= 729.8 mmHg
Vapor pressure of water,
= 23.8 mmHg
Putting values in above equation, we get:
![729.7=p_{H_2}+23.8\\\\p_{H_2}=729.7-23.8=705.9mmHg](https://tex.z-dn.net/?f=729.7%3Dp_%7BH_2%7D%2B23.8%5C%5C%5C%5Cp_%7BH_2%7D%3D729.7-23.8%3D705.9mmHg)
Hence, the partial pressure of hydrogen is 705.9 mmHg