If 260 ml of o2 gas is collected at 21 degrees celsius and 101.3 kpa, what volume would this gas occupy at stp?

241.1 mL.

Explanation:

From the general law of ideal gases:

PV = nRT.

where, P is the pressure of the gas.

V is the volume of the container.

n is the no. of moles of the gas.

R is the general gas constant.

T is the temperature of the gas (K).

For the same no. of moles of the gas at two different (P, V, and T):

P₁V₁/T₁ = P₂V₂/T₂.

P₁ = 101.3 kPa = 1.0 atm, V₁ = 260.0 mL, T₁ = 21°C + 273 = 294.0 K.P₂ = 1.0 atm (standard P), V₂ = ??? mL, T₂ = 0.0°C + 273 = 273.0 K (standard T).

∴ V₂ = (P₁V₁T₂)/(T₁P₂) = (1.0 atm)(260.0 mL)(273.0 K)/(294.0 K)(1.0 atm) = 241.1 mL.

0.12 & of NaCl

Explanation:

Determine the percentage of NaCl in the broth

First of all, we convert the mass from mg to g

340 mg . 1g/ 1000 mg = 0.340 g

Now  we can calculate the %

(0.340 g / 286.7 g) . 100 = 0.12 %