Both D and G options. Weak base and its conjugate acid or weak acid and its conjugate base are the possible components of a buffer solution.
Explanation:
Buffer solution is the solution which gets easily dissolved in water and so called as "Aqueous solution".
Buffer solution is essentially made up of two components Known as:
i.) Weak base and its conjugate acid
ii.) Weak acid and its conjugate base
This weak acid and base solution is used to maintain the pH value of the solution in a balanced way.
When the weak acid or base solution is added to strong acid or base solution that is the way pH gets balanced .
In one word buffer solution is the solution which resists for the pH change when strong acids or bases are added.
Answer: The questions looks unclear
Explanation: Periodic table is a table that contains elements arranged according to their increasing atomic number.
1. D belongs to group 4
E. Belongs to group 7
B belongs to group 1
A belongs to group 8. A noble gas.
R belongs to group 3. K belongs to group 6 C belongs to group 1. H belongs to group 8
Answer:
50000ppm and 0.855M.
Explanation:
ppm is an unit of chemistry defined as the ratio between mg of solute (NaCl) and Liters of solution. Molarity, M, is the ratio between moles of NaCl and liters
A 5% (w/v) NaCl contains 5g of NaCl in 100mL of solution.
To solve the ppm of this solution we need to find the mg of NaCl and the L of solution:
<em>mg NaCl:</em>
5g * (1000mg / 1g) = 5000mg
<em>L Solution:</em>
100mL * (1L / 1000mL) = 0.100L
ppm:
5000mg / 0.100L = 50000ppm
To find molarity we need to obtain the moles of NaCl in 5g using its molar mass:
5g * (1mol / 58.5g) = 0.0855moles NaCl
Molarity:
0.0855mol NaCl / 0.100L = 0.855M
Answer:
2.1056L or 2105.6mL
Explanation:
We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:
Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol
Mass of Na2CO3 = 10g
Mole of Na2CO3 =.?
Mole = mass /molar mass
Mole of Na2CO3 = 10/106
Mole of Na2CO3 = 0.094 mole
Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:
Na2CO3 + 2HCl —> 2NaCl + H2O + CO2
From the balanced equation above,
1 mole of Na2CO3 reacted to produce 1 mole of CO2.
Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.
Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:
1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.
Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L
Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL
Yes he was a pillar in the early church
