Answer:
B. Ca(NO2)2
Explanation:
Ions (charged atoms) combine with one another to form stable ionic compounds. In this case, calicium ion (Ca2+) is said to react with NO2, which has a charge of -1 i.e. +1-1(2) = 1-2 = -1.
This means that calcium has a charge of +2 while nitrite ion has charge of -1, hence, when they combine, they exchange their charges, which become their subscript as follows:
Ca2+ + NO2- → Ca(NO2)2
Ca(NO2)2 is a stable ionic compound called calcium nitrite. Notice that it takes two atoms of NO2- to react with one atom of Ca2+.
The question incomplete , the complete question is:
A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.
Answer:
The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
Explanation:
Moles of urea = 
Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)
Molarity of the urea solution ;
Mass of solvent = m
Volume of solvent = V = 200.0 mL
Density of the urea = d = 0.95 g/mL
(1 g = 0.001 kg)
Molality of the urea solution ;
The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
Answer:
m H2O = 56 g
Explanation:
∴ The heat ceded (-) by the Aluminum part is equal to the heat received (+) by the water:
⇒ - (mCΔT)Al = (mCΔT)H2O
∴ m Al = 25.0 g
∴ Mw Al = 26.981 g/mol
⇒ n Al = (25.0g)×(mol/26.981gAl) = 0.927 mol Al
⇒ Q Al = - (0.927 mol)(24.03 J/mol°C)(26.8 - 86.4)°C
⇒ Q Al = 1327.64 J
∴ mH2O = Q Al / ( C×ΔT) = 1327.64 J / (4.18 J/g.°C)(26.8 - 21.1)°C
⇒ mH2O = 55.722 g ≅ 56 g
Answer:
8.33 atm
Explanation:
Xe is 5 out of (4+5) or 5 / 9 ths of the gas present
5/9 * 15 atm = 8.33 atm
They are both made out of atoms!
