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Using the given formula, the density of the material is 2.015 g/mL
<h3>Calculating Density </h3>
From the question, we are to determine the density of the material
From the given formula
Density = Mass / Volume
And from the given information,
Mass = 65.5 g
and volume = 32.5 mL
Putting the parameters into the equation,
Density = 65.5/32.5
Density = 2.015 g/mL
Hence, the density of the material is 2.015 g/mL.
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Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
The pH of a solution is 9.02.
c(HCN) = 1.25 M; concentration of the cyanide acid
n(NaCN) = 1.37 mol; amount of the salt
V = 1.699 l; volume of the solution
c(NaCN) = 1.37 mol ÷ 1.699 l
c(NaCN) = 0.806 M; concentration of the salt
Ka = 6.2 × 10⁻¹⁰; acid constant
pKa = -logKa
pKa = - log (6.2 × 10⁻¹⁰)
pKa = 9.21
Henderson–Hasselbalch equation for the buffer solution:
pH = pKa + log(cs/ck)
pH = pKa + log(cs/ck)
pH = 9.21 + log (0.806M/1.25M)
pH = 9.21 - 0.19
pH = 9.02; potential of hydrogen
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<u>Answer:</u>
<em>4.5 L water we have in litres (L).</em>
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<u>Explanation:</u>

where
= Final T - Initial T
Q is the heat energy in calories
c is the specific heat capacity (for water 1.0 cal/(g℃))
m is the mass of water
Plugging in the values

So,
Volume of water = mass/density

=4.5 L (Answer)