Answer is: D. It is not sodium bicarbonate.
Balanced chemical reaction of heating sodium bicarbonate: 2NaHCO₃ → Na₂CO₃ + CO₂ + H₂O.
This is chemical change (chemical reaction), because new substances are formed (sodium carbonate, carbon(IV) oxide and water), the atoms are rearranged, so there is no sodium bicarbonate (NaHCO₃) in the test tube.
Answer:
The concentration is 50,8 % w/v and radio strengths = 1,96.
Explanation:
Phenobarbital sodium is a medication that could treat insomnia, for example.
2,0 M of Phenobarbital sodium means 2 moles in 1L.
The concentration units in this case are %w/v that means 1g in 100 mL and ratio strengths that means 1g in <em>r</em> mL. Thus, 2 moles must be converted in grams with molar weight -254 g/mole- and liters to mililiters -1 L are 1000mL-. So:
2 moles ×
= 508 g of Phenobarbital sodium.
1 L ×
= 1000 mL of solution
Thus, % w/v is:
× 100 = 50,8 % w/v
And radio strengths:
= 1,96. Thus, you have 1 g in 1,96 mL
I hope it helps!
Answer:
The final pressure is approximately 0.78 atm
Explanation:
The original temperature of the gas, T₁ = 263.0 K
The final temperature of the gas, T₂ = 298.0 K
The original volume of the gas, V₁ = 24.0 liters
The final volume of the gas, V₂ = 35.0 liters
The original pressure of the gas, P₁ = 1.00 atm
Let P₂ represent the final pressure, we get;



∴ The final pressure P₂ ≈ 0.78 atm.
About 2,500 grams Ans balkfdoaks;
Answer: 91.73g of NaCl
Explanation:
First, we solve for the number of moles of F2 using the ideal gas equation
V = 12L
P = 1.5 atm
T = 280K
R = 0.082atm.L/mol/K
n =?
PV = nRT
n = PV /RT
n = (1.5x12)/(0.082x280)
n = 0.784mol
Next, we convert this mole ( i.e 0.784mol) of F2 to mass
MM of F2 = 19x2 = 38g/mol
Mass conc of F2 = n x MM
= 0.784 x 38 = 29.792g
Equation for the reaction is given below
F2 + 2NaCl —> 2NaF + Cl2
Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol
Mass conc. of NaCl from the equation = 2 x 58.5 = 117g
Next, we find the mass of NaCl that reacted with 29.792g of F2.
From the equation,
38g of F2 redacted with 117g of NaCl.
Therefore, 29.792g of F2 will react with Xg of NaCl i.e
Xg of NaCl = (29.792 x 117)/38
= 91.73g
Therefore, 91.73g of NaCl reacted with f2