Answer:
independent variable - the thing you change
dependent variable - the thing you measure
control variables - the things you keep the same
According to the law of conservation of mass, the amount of BARIUM present of the reactants is the same as the amount present in the products (the precipitate).
(11.21 g BaSO4) / (233.4 g/mol BaSO4) = 0.0480 mol BaSO4 and original barium salt
(10.0 g) / (0.0480 mol) = 208.3 g/mol
So it must have been BaCl2, because the molar mass of Barium is 137 which leave 71 grams left. Since Barium is a +2 charge, it means the atom next to it must be twice. Chlorine mass is 35, which twice is 71
Answer:
Kp = 0.022
Explanation:
<em>Full question: ...With 2.3 atm of ammonia gas at 32. °C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of hydrogen gas to be 0.69 atm. </em>
<em />
The equilibrium of ammonia occurs as follows:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
Where Kp is defined as:

<em>Where P represents partial pressure of each gas.</em>
<em />
As initial pressure of ammonia is 2.3atm, its equilibrium concentration will be:
P(NH₃) = 2.3atm - 2X
<em>Where X represents reaction coordinate</em>
<em />
Thus, pressure of hydrogen and nitrogen is:
P(N₂) = X
P(H₂) = 3X.
As partial pressure of hydrogen is 0.69atm:
3X = 0.69
X = 0.23atm:
P(NH₃) = 2.3atm - 2(0.23atm) = 1.84atm
P(N₂) = 0.23atm
P(H₂) = 0.69atm

<h3>Kp = 0.022</h3>
Answer:
The calcium concentration must be greater outside the cell than inside the cell.
Explanation:
My previous answer was deleted from the explanation I provided from another website.
Density of the gas is 3.05 × 10⁻³ g / cm³.
<u>Explanation:</u>
Volume of the cylinder = π r² h
where r is the radius and h is the height of the height or the length of the glass tube.
Here r = 4 cm and h = 27.4 cm
Volume of the cylinder = 3.14 × 4 × 4 × 27.4 = 1376.6 cm³
We have to find the mass of the gas by subtracting the mass of the tube filled with the substance from the mass of the empty tube.
Mass of the substance = 258.5 - 254.3 = 4.2 g
We have to find the density using the formula as,

Plugin the values as,
= 3.05 × 10⁻³ g / cm³
So the Density of the gas is 3.05 × 10⁻³ g / cm³.