Answer:
the answer is 7.936X 10⁻⁷kg.... the unit for volume must be cm³
Explanation:
we know,
ρ =m/v
given,
ρ =19.84/1000X(100)²= 1.984X10⁻⁶kg/m²
V (volume of the container means the volume of the element)=2.5 m³
NOW putting in the equation
1.984X10⁻⁶/2.5=m
or, m=7.936X 10⁻⁷kg(ANS)
thank u(i am not 100% sure but think it shoud be the answer)
Answer:
643g of methane will there be in the room
Explanation:
To solve this question we must, as first, find the volume of methane after 1h = 3600s. With the volume we can find the moles of methane using PV = nRT -<em>Assuming STP-</em>. With the moles and the molar mass of methane (16g/mol) we can find the mass of methane gas after 1 hour as follows:
<em>Volume Methane:</em>
3600s * (0.25L / s) = 900L Methane
<em>Moles methane:</em>
PV = nRT; PV / RT = n
<em>Where P = 1atm at STP, V is volume = 900L; R is gas constant = 0.082atmL/molK; T is absolute temperature = 273.15K at sTP</em>
Replacing:
PV / RT = n
1atm*900L / 0.082atmL/molK*273.15 = n
n = 40.18mol methane
<em>Mass methane:</em>
40.18 moles * (16g/mol) =
<h3>643g of methane will there be in the room</h3>
Answer:
A
Explanation:
Both students are Pushing/Pulling towards the same direction meaning if enough force is applied the object will move into that direction
Let's apply the principle of conservation of energy.
Heat of metal = Heat of water
mCmetalΔT = mCwaterΔT
Applying the given values,
(45 g)(Cmetal)(100 - 28 °C) = (130 g)(4.18 J/g-°C)(28 - 25 °C)
Solving for Cmetal,
<em>Cmetal = 0.503 J/g°C
Therefore, the heat capacity of the unknown metal is 0.503 J/g</em>°C.