The removal of trees would most likely decrease the amount of carbon in the atmosphere.
Please note that it is useful to add the options provided with the question, in order to get a most accurate answer and have your question answered quicker.
Hope this helps!!
Hrxn = Q reaction / mol of reaction
mol of reaction = M * V = 10 * 1 = 10 mmol = 0.01 mol
Q water = m * C * (Tf - Ti)
= (10 + 10) (4.184) (26-20) = 502.08 J
Q reaction = - Q water = -502.08 J
Hrxn = -502.08 / (0.01) = - 50208 J = - 50.21 kJ/mol
A technician mixes 80 ml of a 5% solution with 10 ml of water. the final percentage strength of the solution prepared is 40 %.
given that :
8 ml of a 5 % solution mix with 10 ml . that means the 80 mL of 5 % solution is diluted with water of 10 mL
therefore, 80 × 5 = 10 × x %
x % = 40 %
Therefore, the final percentage strength of the solution is 40 %
Thus, A technician mixes 80 ml of a 5% solution with 10 ml of water. the final percentage strength of the solution prepared is 40 %.
To learn more about percentage strength here
brainly.com/question/17130362
#SPJ4
The masses of the components are obtained as;
- Sodium hydrogen carbonate = 3.51 g
- Sodium carbonate = 8.708 g
<h3>What is decomposition?</h3>
The term decomposition has to do with the breakdown of the given substance into its components. The components of sodium hydrogen carbonate could be identified as water vapor, carbon dioxide gas and sodium carbonate. Among these products that have been listed here, we can see that it is only the sodium carbonate that remains as a solid. The others are gases that move away from the system that is under study.
Now putting down the equation of the reaction, we have;
Now, the loss in mass must be due to the carbon dioxide and the water. Hence we obtain the loss in mass to be 10.000 g - 8.708 g = 1.292 g
Mass of sodium hydrogen carbonate = 2 * 88 g/mol * 1.292 g/62 g/mol
= 3.51 g
Learn more about anhydrous sodium carbonate :brainly.com/question/20479996
#SPJ1
Explanation:
1 literThe total of water is equal to 1000.0 g of water
we need to find the molality of a solution containing 10.0 g of dissolved in Na₂S0₄1000.0 g of water
1. For that find the molar mass
Na: 2 x 22.99= 45.98
S: 32.07
O: 4 x 16= 64
The total molar mass is 142.05
We have to find the number of moles, y
To find the number of moles divide 10.0g by 142.05 g/mol.
So the number of moles is 0.0704 moles.
For the molarity, you need the number of moles divided by the volume. So, 0.0704 mol/1 L.
The molarity would end up being 0.0704 M
The molality of a solution containing 10.0 g of Na2SO4 dissolved in 1000.0 g of water is 0.0704 Mliter
