<u>Answer:</u> The final temperature of the mixture is 51.49°C
<u>Explanation:</u>
When two samples of water are mixed, the heat released by the water at high temperature will be equal to the amount of heat absorbed by water at low temperature
The equation used to calculate heat released or absorbed follows:
![m_1\times c\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
......(1)
where,
q = heat absorbed or released
= mass of water at high temperature = 140 g (Density of water = 1.00 g/mL)
= mass of water at low temperature = 230 g
= final temperature = ?°C
= initial temperature of water at high temperature = 95.00°C
= initial temperature of water at low temperature = 25.00°C
c = specific heat of water= 4.186 J/g°C
Putting values in equation 1, we get:
![140\times 4.186\times (T_{final}-95)=-[230\times 4.186\times (T_{final}-25)]](https://tex.z-dn.net/?f=140%5Ctimes%204.186%5Ctimes%20%28T_%7Bfinal%7D-95%29%3D-%5B230%5Ctimes%204.186%5Ctimes%20%28T_%7Bfinal%7D-25%29%5D)
Hence, the final temperature of the mixture is 51.49°C
Mass number is equal to the sum of number of neutrons (neutral subatomic particle) and number of protons (positively charge particle).
Mass number =
Now, atomic number is same as the number of protons i.e. atomic of nitrogen is 7 . Thus, number of protons is also equal to 7.
Therefore,
Mass number =
= 
Hence, mass number of nitrogen is equal to .
Answer:
1. 15.71 g CO2
2. 38.19 % of efficiency
Explanation:
According to the balanced reaction (2 CO(g) + O2(g) → 2 CO2(g)), it is clear that the CO is the limitant reagent, because for every 2 moles of CO we are using only 1 mole of O2, so even if we have the same quantity for both reagents, not all of the O2 will be consumed. This means that we can just use the stoichiometric ratios of the CO and the CO2 to solve this question, and for that we need to convert the gram units into moles:
For CO:
C = 12.01 g/mol
O = 16 g/mol
CO = 28.01 g/mol
(10.0g CO) x (1 mol CO/28.01 g) = 0.3570 mol CO
For CO2:
C = 12.01 g/mol
O = 16 x 2 = 32 g/mol
CO2 = 44.01 g/mol
We now that for every 2 moles of CO we are going to get 2 moles of CO2, so we resolve as follows:
(0.3570 mol CO) x (2 mol CO2/2 mol CO) = 0.3570 moles CO2
We are obtaining 0.3570 moles of CO2 with the 10g of CO, now lets convert the CO2 moles into grams:
(0.3570 moles CO2) x (44.01 g/1 mol CO2) = 15.71 g CO2
Now for the efficiency question:
From the previous result, we know that if we produce 15.71 CO2 with all the 10g of CO used, we would have an efficiency of 100%. So to know what would that efficiency be if we would only produce 6g of CO2, we resolve as follows,
(6g / 15.71g) x 100 = 38.19 % of efficiency
Covalent bonds do not involve electron transfer because in covalent bonds, the electrons are shared between the atoms to form the molecule rather than being transferred to form ions as it is in the case of ionic bonds.<span />
Your answer would be shorter wavelength and equal speed !
