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:
4.5 kilograms of silicon dioxide is required to produce 3.00 kg of SiC.
Explanation:
The balanced equation for the reaction between silicon dioxide and carbon at high temperature is given as:
1 mole silicon dioxide reacts with 3 moles of carbon to give 1 moles of silicon carbide and 2 moles of carbon monoxide.
Mass of SiC = 3.00kg = 3000.00 g
1 kg = 1000 g
Molecular mass of SiC = 40 g/mol
Moles of SiC = 
According to reaction, 1 mole of SiC is produced from 1 mole of silicon dioxide.
Then 75 moles of SiC will be produce from:
of silicon dioxide.
mass of 75 moles of silicon dioxde:
4.5 kilograms of silicon dioxide is required to produce 3.00 kg of SiC.
Answer:
The equation shows balance.
Explanation:
You can easily count the number of elements on each side.
On the left side of the equation, you have 2 moles of Na and 2 moles of Cl.
On the right side of the equation, you also have 2 moles of Na and 2 moles of Cl because the two elements formed a compound meaning that whatever number is in front of them, both elements will receive the same number.
Answer:
Experimental: the dogs who get the toys
Control: the dogs who dont recieve a toy
Explanation:
expermental is the group that recieved the tested variable in this case that being the toy
control is the group that doesn't recieve the tested variable
Answer:
n C2H5OH = 0.2496 mol
Explanation:
- moles (n) = mass / molecular weight
∴ m C2H5OH = 11.5 g
∴ Mw C2H5OH = (2)(12.011) + (6)(1.00794) + 15.9994 = 46.069 g/mol
⇒ n C2H5OH = (11.5 g)/(46.069 g/mol) = 0.2496 mol C2H5OH
