Answer:
1- Yes, we can calculate the solubility of mineral compound X.
2- 0.012 g/mL.
Explanation:
<em>1- Using only the information above, can you calculate the solubility of X in water at 15.0 °C? </em>
The information available is:
The volume of water sample = 25.0 mL.
Weight of the mineral compound X after evaporation, drying, and washing = 0.30 g.
∴ Yes, we can calculate the solubility of mineral compound X.
<u><em>2- If you said yes, calculate it.</em></u>
∵ 25.0 mL of water sample contains → 0.30 g of the mineral compound X.
∴ 1.0 mL of water sample contains → ??? g of the mineral compound X.
1.0 ml of water sample will contain (0.3 g/25.0 mL) 0.012 g.
<em>∴ The solubility of the mineral compound X in the water sample is</em> <u><em>0.012 g/mL.</em></u>
<u><em></em></u>
Answer:
The least whole number coefficient for HNO₃ is 6
Explanation:
The chemical equation above is the reaction between calcium orthophosphate and nitric acid.
To balance a chemical equation, we have to consider law of conservation of matter which states that matter can neither be created nor destroyed.
What this law implies is that, whatever we have at the reactant side must be equal to whatever is obtainable at the product side.
The above equation is
Ca₃(PO₄)₂ + HNO₃ → Ca(NO₃)₂ + H₃PO₄
To balance the equation, we'll have to check the number of atoms at each side and possibly balance the equation with the number of moles.
The balanced equation is
Ca₃(PO₄)₂ + 6HNO₃ → 3Ca(NO₃)₂ + 2H₃PO₄
From the balanced equation above, we can see that the number of calcium (Ca), Phosphorus (P), Oxygen(O), Nitrogen(N) and hydrogen (H) are balanced at both sides of the equation.
The least number coefficient for HNO₃ is 6
D is the answer!!!!!!!!!!!!!!!!
The molarity of a solution equals to the mole number of the solute/the volume of the solution. For NH4Br, we know that the mole mass is 98. So the molarity is (14/98) mol /0.15 L=0.95 mol/L.
Answer:
65.2L
Explanation:
Using the general gas equation;
PV = nRT
Where;
P = pressure (atm)
V = volume (Litres)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (Kelvin)
According to the information provided in this question,
P = 1.631 atm
V = ?
n = 4.3 moles
T = 28°C = 28 + 273 = 301K
Using PV = nRT
V = nRT/P
V = 4.3 × 0.0821 × 301 ÷ 1.631
V = 106.26 ÷ 1.631
V = 65.15
Volume of the gas = 65.2L
