Answer:
45.3°C
Explanation:
Step 1:
Data obtained from the question.
Initial pressure (P1) = 82KPa
Initial temperature (T1) = 26°C
Final pressure (P2) = 87.3KPa.
Final temperature (T2) =.?
Step 2:
Conversion of celsius temperature to Kelvin temperature.
This is illustrated below:
T(K) = T(°C) + 273
Initial temperature (T1) = 26°C
Initial temperature (T1) = 26°C + 273 = 299K.
Step 3:
Determination of the new temperature of the gas. This can be obtained as follow:
P1/T1 = P2/T2
82/299 = 87.3/T2
Cross multiply to express in linear form
82 x T2 = 299 x 87.3
Divide both side by 82
T2 = (299 x 87.3) /82
T2 = 318.3K
Step 4:
Conversion of 318.3K to celsius temperature. This is illustrated below:
T(°C) = T(K) – 273
T(K) = 318.3K
T(°C) = 318.3 – 273
T(°C) = 45.3°C.
Therefore, the new temperature of the gas in th tire is 45.3°C
Answer:
Opticians typically have a high school diploma or equivalent and receive some form of on-the-job training. Some opticians enter the occupation with an associate's degree or a certificate from a community college or technical school. About half of the states require opticians to be licensed.
During the experiment, scientists noted that several of the reaction beakers became hot to the touch. All of the following reactions could cause this result except endothermic and positive ∆H experiments.
<u>Explanation:</u>
If the beakers are becoming hot during experimentation, then that means the energy is being released from the reactants during this experiment. As the energy is being released that enthalpy change will also be negative as the enthalpy change is calculated as the difference of enthalpy of reactants from products.
So in these cases, heat is released making the beakers hot. So for the exceptional case, the experiment should be endothermic in nature and positive enthalpy change should be there in the experiment. Such that the heat will not be released leading to no heating of beakers.
Answer;
12 carbon atoms, 22 hydrogen atoms and 11 oxygen atoms.
Explanation:
Using a balanced chemical equation we can identify the number of carbon, hydrogen, and oxygen atoms in sugar.
CxHyOn + 12O₂ → 11 H₂O + 12CO₂
When an equation is completely balanced, then the number of each atom of an element is equal on the reactant side and the product side.
Therefore;
For carbon; x = 12
For Hydrogen; y = (11×2) = 22
For Oxygen; n + (12×2) = 11 + (12×2)
= n + 24 = 11 + 24
n = 11
Therefore the sugar has, 12 carbon atoms, 22 hydrogen atoms and 11 oxygen atoms.
Thus the balanced equation would be;
C₁₂H₂₂O₁₁ + 12O₂ → 11 H₂O + 12CO₂
