Is this the full question?
<span>Heat that flows by conduction is the transfer of thermal energy between substances in contact. For this to happen, what must occur?
A) The two systems must be the same temperature.
B) The two systems must not be touching each another.
C) One system must have higher kinetic energy than the other system.
D) The thermal energy of one system must be the same as the thermal energy of the other system.</span>
7<span> to 49 10 to 100. 30 Secs. 3. What is the </span>pH<span> value of pure </span>water<span>? 0 3 </span>7<span> 10 ... How do acids </span>taste<span>? </span>bitter sour<span> sweet salty. 30 Secs. </span>7<span>. How do </span>bases taste<span>? </span>bitter<span> ... 8. Which kind of solution would react with a metal? acidic basic </span>neutral water<span> ... cocoa </span>has<span> a </span>bitter taste<span>. It is most likely which of the following? acid </span><span>base neutral</span>
Answer:
1)
Explanation:
the answer to you question Is 1)
Answer:
The final volume will be 5.80 L
Explanation:
Step 1: Data given
Number of moles gas = 0.140 moles
Volume of gas = 2.78 L
Number of moles added = 0.152 moles
Step 2: Calculate the final volume
V1/n1 = V2/n2
⇒ with V1 = the initial volume = 2.78 L
⇒ with n1 = the initial number of moles = 0.140 moles
⇒ with V2 = The new volume = TO BE DETERMINED
⇒ with n2 = the new number of moles = 0.140 + 0.152 = 0.292 moles
2.78/0.140 = V2 /0.292
V2 = 5.80 L
The final volume will be 5.80 L
Explanation:
mass H2O2 = 55 mL(1.407 g/mL) = 80.85 g
molar mass H2O2 = 2(1.01 g/mol) + 2(16.00 g/mol) = 34.02 g/mol
moles H2O2 = 80.85 g/34.02 g/mol = 2.377 moles H2O2
For each mole of H2O2 you obtain 0.5 mole of O2 (see the equation).
moles O2 = 2.377 moles H2O2 (1 mole O2)/(2 moles H2O2) = 1.188 moles O2
Now, you need the temperature. If you are at STP (273 K, and 1.00 atm) then 1 mole of an ideal gas at STP has a volume of 22.4 L. Without temperature you are not really able to continue. I will assume you are at STP.
Volume O2 = 1.188 moles O2(22.4 L/mole) = 0.0530 L of O2.
which is 53 mL.