What is the volume, in liters, of 0.500 mol of c3 h8 gas at stp?

Maru [420]

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3 0

2 years ago

You have a 1.8 M solution of HCI. You need to make a 0.70 M solution with a volume of 550 mL. How many mL of the stock solution

Arte-miy333 [17]

The volume of the stock solution needed is 213.88 mL to get new concentration.

<h3>What is molarity?</h3>

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution.

Determination of the volume of stock solution.

Volume of diluted solution (V₂) =550 mL

Molarity of diluted solution (M₂) =0.70 M

Molarity of stock solution (M₁) = 1.8 M

Volume of stock solution needed (V₁) =?

M₁V₁ = M₂V₂

1.8 M × V₁ = 0.70 M × 550 mL

V₁ = 213.88 mL

Thus, the volume of the stock solution needed is 213.88 mL.

Learn more about the molarity here:

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5 0

2 years ago

Na+ and Cl- __________ ___________________________ Na+ and PO4 3- __________ ___________________________ Na+ and SO4 2- ________

bagirrra123 [75]

Answer:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

The cations and anions being oppositely charged attract each other through strong coloumbic forces and form an ionic bond.

(1) Sodium is carrying +1 charge called as $Na^{+1}$ cation and chloride $Cl^{-1}$ is an anion carrying -1 charge. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $NaCl$ .

(2) Sodium is carrying +1 charge called as $Na^{+1}$ cation and phosphate $PO_4^{-3}$ is an anion carrying -3 charge. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $Na_3PO_4$ .

(3) Sodium is carrying +1 charge called as $Na^{+1}$ cation and sulfate $SO_4^{-2}$ is an anion carrying -2 charge. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $Na_2SO_4$ .

(4) Sodium is carrying +1 charge called as $Na^{+1}$ cation and carbonate $CO_3^{-2}$ is an anion carrying -2 charge. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $Na_2CO_3$ .

(5) Potassium is carrying +1 charge called as $K^{+1}$ cation and chloride $Cl^{-1}$ is an anion carrying -1 charge. They form $KCl$ .

(6) Potassium is carrying +1 charge called as $K^{+1}$ cation and phosphate $PO_4^{-3}$ is an anion carrying -3 charge. They form $K_3PO_4$ .

(7) Potassium is carrying +1 charge called as $K^{+1}$ cation and sulfate $SO_4^{-2}$ is an anion carrying -2 charge. They form $K_2SO_4$ .

(8) Potassium is carrying +1 charge called as $K^{+1}$ cation and carbonate $CO_3^{-2}$ is an anion carrying -2 charge. They form $K_2CO_3$ .

(9) Calcium is carrying +2 charge called as $Ca^{+2}$ cation and chloride $Cl^{-1}$ is an anion carrying -1 charge. They form $CaCl_2$ .

(10) Calcium is carrying +2 charge called as $Ca^{+2}$ cation and phosphate $PO_4^{-3}$ is an anion carrying -3 charge. They form $Ca_3(PO_4)_2$ .

(11) Calcium is carrying +2 charge called as $Ca^{+2}$ cation and sulfate $SO_4^{-2}$ is an anion carrying -2 charge. They form $CaSO_4$ .

(12) Calcium is carrying +2 charge called as $Ca^{+2}$ cation and carbonate $CO_3^{-2}$ is an anion carrying -2 charge. They form $CaCO_3$ .

(13) Ammonium ion is carrying +1 charge called as $NH_4^{+1}$ cation and chloride $Cl^{-1}$ is an anion carrying -1 charge. They form $NH_4Cl$ .

(14) Ammonium ion is carrying +1 charge called as $NH_4^{+1}$ cation and phosphate $PO_4^{-3}$ is an anion carrying -3 charge. They form $NH_4_3PO_4$ .

(15) Ammonium ion is carrying +1 charge called as $NH_4^{+1}$ cation and sulfate $SO_4^{-2}$ is an anion carrying -2 charge. They form $NH_4_2SO_4$ .

(16) Ammonium ion is carrying +1 charge called as $NH_4^{+1}$ cation and carbonate $CO_3^{-2}$ is an anion carrying -2 charge. They form $NH_4_2CO_3$ .