Answer:
Double replacement reaction.
Explanation:
The Na and Ag atoms both (double) trade places (replacement) with each other.
Answer:
The correct answer is 8.79 × 10⁻² M.
Explanation:
Based on the given information, the mass of NaI given is 4.11 grams. The molecular mass of NaI is 149.89 gram per mole. The moles of NaI can be determined by using the formula,
No. of moles of NaI = Weight of NaI/ Molecular mass
= 4.11 / 149.89
= 0.027420
The vol. of the solution given is 312 ml or 0.312 L
The molarity can be determined by using the formula,
Molarity = No. of moles/ Volume of the solution in L
= 0.027420/0.312
= 0.0879 M or 8.79 × 10⁻² M
Answer:
13mL
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
HNO3 + KOH —> KNO3 + H2O
From the balanced equation above, we obtained the following data:
Mole ratio of the acid (nA) = 1
Mole ratio of the base (nB) = 1
Step 2:
Data obtained from the question.
This includes the following:
Molarity of the acid (Ma) = 6M
Volume of the acid (Va) =?
Volume of the base (Vb) = 39mL
Molarity of the base (Mb) = 2M
Step 3:
Determination of the volume of the acid.
Using the equation:
MaVa/MbVb = nA/nB, the volume of the acid can be obtained as follow:
MaVa/MbVb = nA/nB
6 x Va / 2 x 39 = 1/1
Cross multiply to express in linear form
6 x Va = 2 x 39
Divide both side by 6
Va = (2 x 39)/6
Va = 13mL
Therefore, the volume of the acid (HNO3) needed for the reaction is 13mL
CH3CH2OH + 3O2 = 2CO2 + 3H2O
Basically you do trial and error on both sides so they can be equal
Answer:
3.861x10⁻⁹ mol Pb⁺²
Explanation:
We can <u>define ppm as mg of Pb²⁺ per liter of water</u>.
We<u> calculate the mass of lead ion in 100 mL of water</u>:
- 100.0 mL ⇒ 100.0 / 1000 = 0.100 L
- 0.100 L * 0.0080 ppm = 8x10⁻⁴ mg Pb⁺²
Now we <u>convert mass of lead to moles</u>, using its molar mass:
- 8x10⁻⁴ mg ⇒ 8x10⁻⁴ / 1000 = 8x10⁻⁷ g
- 8x10⁻⁷ g Pb²⁺ ÷ 207.2 g/mol = 3.861x10⁻⁹ mol Pb⁺²
