Answer:
they could collide 241.66 molec / in² by increasing the volume to 40.2L
Explanation:
ideal gas:
<u>Boyle Law</u>: at constant temperature the pressure of a gas varies inversely with the volume
- V1 * P1 = V2 * P2
- P = F / A
∴ V1 = 6.70 L;
∴ P1 = 1450 molec / in²
∴ V2 = 40.2 L
⇒ P2 = (( 6.70 L ) * ( 1450 molec/in²)) / 40.2 L
⇒ P2 = 241.66 molec/in²
Answer:
1) 0 C2H4O2 + 0 O2 -> 0 CO2 + 0 H2O (balanced)
2) V2O5 + CaS -> CaO + V2S5
<em>just additional info: V2O5 </em><em>is</em><em> </em><em>divanadium</em><em> </em><em>pentaoxide</em>
LHS (Left hand side)
V: 2
O: 5
Ca: 1
S: 1 x 5 [to balance with the right hand side of the equation]
RHS (Right hand side)
V: 2
O: 1 x 5 [to balance with the left hand side of the equation]
Ca: 1
S: 5
When you balance any elements, you have to balance the whole chemical compound.
Thus,
V2O5 + <em><u>5</u></em> CaS -> <em><u>5</u> CaO</em> + V2S5
LHS CHECK:
V: 2
O: 5
Ca: 5
S: 5
RHS CHECK:
V: 2
O: 5
Ca: 5
S: 5
3) S8 + O2 -> SO2
LHS:
S: 8
O: 2
RHS:
S: 1 x 8 [to balance with LHS]
O: 2
When you balance any elements, you have to balance the whole chemical compound.
S8 + O2 -> <em><u>8</u></em><em><u> </u></em>SO2
When we add 8 to the RHS, it gives us 8S, 16 O.
In order to balance that into the RHS, I need to multiply the O2 by 8, which will give 8(O2) = 16 O particles.
Therefore, <em><u>S8</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>8</u></em><em><u> </u></em><em><u>O2</u></em><em><u> </u></em><em><u>-</u></em><em><u>></u></em><em><u> </u></em><em><u>8</u></em><em><u> </u></em><em><u>SO2</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>final</u></em><em><u> </u></em><em><u>answer</u></em><em><u> </u></em><em><u>for</u></em><em><u> </u></em><em><u>(</u></em><em><u>3</u></em><em><u>)</u></em><em><u>.</u></em>
D. A sour liquid that forms gas bubblee when mixed with copper
The concept that can be used in order to answer this item is that of the conservation of heat among the system. We let T be equal to the final temperature. The equation that would allow us to relate the initial and final conditions of both substances is as follows,
m₁cp₁(T - T₁) = m₂cp₂(T₂ - T)
The first entity, 1, is the milk and the second entity, 2, is the coffee. We are given that the specific heats of both substances are just equal so we can eliminate them from the equation. Substituting the known values,
(10 g)(T - 10°) = (1.60 x 10^2 g)(90° - T)
The value of T from the equation is 85.29°C.
Answer: 85.29°C