<span>He should use a scale and record the weight in N. The beaker is used for liquid measurement and the scale is used for measuring sold things. A dumbbell is not a liquid material that is why the need of the scale will give it the accurate measurement that Hunter wants to know.</span>
Cycloalkanes are those saturated organic compounds which exist in the form of Rings. Their Hydrogen Deficiency Index in one. The General formula for cycloalkanes is,
CnH2n
When number of Carbons = 8
Then
C₈H₂₍₈₎
C₈H₁₆
Result:
Cycloalkane containing 8 carbon atoms has
16 hydrogen atoms.
Answer:
C₆H₈O₇+ 3NaHCO₃ --› Na₃C₆H₅O₇ + 3CO2 + 3H₂O
Explanation:
The reaction occuring in lava lamp is acid base reaction.
When you drop tablet into water the citric acid reacts with sodium bicarbonate and forms water, a salt, and bubbles of carbon dioxide gas.
Answer:
-177.9 kJ.
Explanation:
Use Hess's law. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2Ca(s) + O2(g) → 2CaO(s) ΔH = -1269.8 kJ We need to get rid of the Ca and O2 in the equations, so we need to change the equations so that they're on both sides so they "cancel" out, similar to a system of equations. I changed the second equation. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ The sign changes in the second equation above since the reaction changed direction. Next, we need to multiply the first equation by two in order to get the coefficients of the Ca and O2 to match those in the second equation. We also multiply the enthalpy of the first equation by 2. 2Ca(s) + 2CO2(g) + O2(g) → 2CaCO3(s) ΔH = -1625.6 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ Now we add the two equations. The O2 and 2Ca "cancel" since they're on opposite sides of the arrow. Think of it more mathematically. We add the two enthalpies and get 2CaO(s) + 2CO2(g) → 2CaCO3(s) and ΔH = -355.8 kJ. Finally divide by two to get the given equation: CaO(s) + CO2(g) → CaCO3(s) and ΔH = -177.9 kJ.
So we know that the equation for density is:
where D is the density, m is the mass in grams, and V is the volume in mL.
So since we know two of the variables, mass and density, we can solve for the volume:
Therefore, the volume of this urine sample is 144.12mL.
