Answer:
50000ppm and 0.855M.
Explanation:
ppm is an unit of chemistry defined as the ratio between mg of solute (NaCl) and Liters of solution. Molarity, M, is the ratio between moles of NaCl and liters
A 5% (w/v) NaCl contains 5g of NaCl in 100mL of solution.
To solve the ppm of this solution we need to find the mg of NaCl and the L of solution:
<em>mg NaCl:</em>
5g * (1000mg / 1g) = 5000mg
<em>L Solution:</em>
100mL * (1L / 1000mL) = 0.100L
ppm:
5000mg / 0.100L = 50000ppm
To find molarity we need to obtain the moles of NaCl in 5g using its molar mass:
5g * (1mol / 58.5g) = 0.0855moles NaCl
Molarity:
0.0855mol NaCl / 0.100L = 0.855M
The bond angles a and b are 120° respectively. The bond angle c is 111.4° .while the bond angle d is 120°. The bond angles e and f are 120° respectively.
In the carbonate ion, all the bond angles and bond lengths are equal hence three equivalent resonance structures can be drawn for the ion. All the bond angles, ( a and b) in carbonate ion all have bond angle of 120°.
The bond angle marked c in OCCl2 has a bond angle 111.4°, the bond angle marked d in the compound has the bond angle, 120°.
There are three bond angles present in the nitrate (NO3-) ion. Three resonance structures contribute to this bond. Based on these structures, the bond angles e and f in the molecule is 120°.
Learn more: brainly.com/question/20339399
Answer:
Va = (MbVb)/Ma
Explanation:
Divide both sides by Ma and voila!
Answer:
-125.4
Explanation:
Target equation is 4C(s) + 5H2(g) = C4H10
These are the data equations for enthalpy of combustion
- C(s) + O2(g) =O2(g) -393.5 kJ/mol * 4
- H2(g) + ½O2(g) =H20(l) = 285.8 kJ/mol * 5
- 2CO2(g) + 3H2O(l) = 13/2O2 (g) + C4H10 - 2877.1 reverse
To get target equation multiply data equation 1 by 4; multiply equation 2 by 5; and reverse equation 3, so...
Calculate 4(-393.5) + 5(-285.8) + 2877.6 and you should get the answer.
Photosynthesis. Because the plants give off oxygen as a waste product. Carbon dioxide then moves from the air into the leaves of plants through tiny openings in the plants leaves.
