The answer is C ionic bonds
Answer:
they have the same molar solubility at 44.5 degrees celsius
<span> ca. 0.4 moles. if this helps plz medal!</span>
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
- 3.5
- 9
- 24.7
- 8.01
- 0.006
- 13(1.25×1012=12.65)
Explanation:
Above we have simply looked the condition that where we put decimal point that gives your needed answer after round off.
For example:
we have to make 2 s.f from 345 so we kept decimal after 3 and we rounded off to make 3.5 from 3.45
Hope you got it
