Answer:
0.17 moles
Explanation:
In the elements of the periodic table, the atomic mass = molar mass. <u>Ex:</u> Atomic mass of Carbon is 12.01 amu which means molar mass of Carbon is also 12.01g/mol.
In order to find the # of moles in a 12 g sample of NiC-12, we will need to multiply the number of each atom by its molar mass and then add the masses of both Nickel and C-12 found in the periodic table:
- Molar Mass of Ni (Nickel): 58.69 g/mol
- Molar Mass of C (Carbon): 12.01 g/mol
Since there's just one atom of both Carbon and Nickel, we just add up the masses to find the molar mass of the whole compound of NiC-12.
- 58.69 g/mol of Nickel + 12.01 g/mol of Carbon = 70.7 g/mol of NiC-12
There's 12g of NiC-12, which is less than the molar mass of NiC-12, so the number of moles should be less than 1. In order to find the # of moles in NiC-12, we need to do some dimensional analysis:
- 12g NiC-12 (1 mol of NiC-12/70.7g NiC-12) = 0.17 mol of NiC-12
- The grams cancel, leaving us with moles of NiC-12, so the answer is 0.17 moles of NiC-12 in a 12 g sample.
<em>P.S. C-12 or C12 just means that the Carbon atom has an atomic mass of 12amu and a molar mass of 12g/mol, or just regular carbon.</em>
Answer: ORGANIC ACIDS
Explanation:
CAM PLANTS CARBOXYLATE ORGANICS ACIDS through the addition of CO2 to PEP Carboxylase( a phosphoenolpyruvate carboxylase enzyme present in the mesophyll cells of the cytoplasm in a green plant) to produce Oxaloacetate (organic compound).
CO2 + PEP ⇒ C4H4O5 (oxaloacetate)
Oxaloacetate is then converted to a similar molecule, Malate (C4H6O5, another form of organic compound) that can be transported in to the bundle-sheath cells. Malate enters the plasmodesmata and releases the CO2. The CO2 then fixed by rubisco and made into sugars via the Calvin cycle.
This is your answer I hope it is right.
Answer:
10.11 mL
Explanation:
Given that :
Total volume of decolorizer = 500 mL
A common recipe is equal parts 95% ethanol and acetone.
So;
The volume of 95% ethanol in the decolorizer = 500mL/2 = 250 mL
Let represent V(ml) for the 99% of ethanol needed to make 95% of the 250 mL of ethanol;
Therefore:
V(ml) × 0.99 = 250 × 0.95
V(ml) × 0.99 = 237.5
V(ml) = 237.5/0.99
V(ml) = 239.89 mL
Hence; the amount of water to be added to 95% of ethanol = ( 250 - 239.89 )mL = 10.11 mL
