What is the percent by mass of sodium in Na2SO4? total mass of element in compound molar mass of compound Use %Element x 100
The equivalency point is at the point of the titration where the amount of titrant added neutralize the solution. When it’s a strong acid strong base titration, the equivalence point will be 7. When it is a weak acid strong base, the equivalence point it more basic (the exact number depends on what acid and base you use). And when it is a strong acid weak base, the equivalence number is more acid (the exact number depends on what acid and base you use). Hope this helps!
Answer:
3.676 L.
Explanation:
- We can use the general law of ideal gas: PV = nRT.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and P are constant, and have different values of V and T:
(V₁T₂) = (V₂T₁)
V₁ = 3.5 L, T₁ = 25°C + 273 = 298 K,
V₂ = ??? L, T₂ = 40°C + 273 = 313 K,
- Applying in the above equation
(V₁T₂) = (V₂T₁)
∴ V₂ = (V₁T₂)/(T₁) = (3.5 L)(313 K)/(298 K) = 3.676 L.
Answer:
0.718L of 0.81M HCl are required
Explanation:
Based on the reaction:
Cd(s)+2HCI(aq) → H2(g)+CdCl2(aq)
<em>1 mol of Cd reacts with 2 moles of HCl</em>
<em />
To solve this question we must, as first, find the moles of Cd. With the moles of Cd we can find the moles of HCl needed to react completely with the Cd. With the moles and the molarity we can find the volume:
<em>Moles Cd -Molar mass: 112.411g/mol-:</em>
32.71g * (1mol / 112.411g) = 0.2910 moles Cd
<em>Moles HCl:</em>
0.2910 moles Cd * (2 moles HCl / 1mol Cd) =
0.5820 moles HCl
<em>Volume:</em>
0.5820 moles HCl * (1L / 0.81moles) =
<h3>0.718L of 0.81M HCl are required</h3>
Answer:
The answer is
<h2>5.43 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
From the question
mass = 250 g
volume = 46 mL
The density is
We have the final answer as
<h3>5.43 g/mL</h3>
Hope this helps you
