Answer:
1) thiamine pyrophosphate -activation of aldehydes
2) coenzyme A -acyl group transfer
3) biotin -CO2 activation/transfer
4) NAD -oxidation/reduction
Explanation:
1. Thiamine pyrophosphate: This is a derivative of Vitamin B1 also known as thiamine. It contains a pyrimidine group linked to the thiazole ring. This connection is further linked to the pyrophosphate group. It functions as a coenzyme in all reactions involving alpha-keto acids. This produces activated aldehydes that could be subject to oxidation.
2. Coenzyme A: This cofactor is a thiol that reacts with carboxylic acids to form thioesters. In so doing, it carries the acyl group. In this condition, it can also be referred to as acyl CoA.
3. Biotin: Also known as Vitamin B7, biotin consists of an ureido ring merged with tetrahydrothiophene. The ureido ring contains the CO2 that can be transferred or activated. Five carboxylase enzymes use biotin as a cofactor in processes such as fat synthesis, glucose generation and the breakdown of sugar.
4. NAD: Nicotinamide adenine dinucleotide consists of two dinucleotides connected to each other at their phosphate groups. NAD exists in two states which are the NAD+ and NADH states. These two states serve as oxidizing and reducing agents respectively. The oxidizing agent becomes reduced to NADH after accepting electrons from other compounds. NADH donates an electron and becomes oxidized to NAD+.
An atom typically has a neutral charge, as in most cases an atom has the same amount of positively charged protons as it does negatively charged electrons.
Atoms that do not have the same amount of electrons as protons are known as isotopes.
Answer:
Explanation:
<u>1) Mass of carbon (C) in 2.190 g of carbon dioxide (CO₂)</u>
- atomic mass of C: 12.0107 g/mol
- molar mass of CO₂: 44.01 g/mol
- Set a proportion: 12.0107 g of C / 44.01 g of CO₂ = x / 2.190 g of CO₂
x = (12.0107 g of C / 44.01 g of CO₂ ) × 2.190 g of CO₂ = 0.59767 g of C
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<u>2) Mass of hydrogen (H) in 0.930 g of water (H₂O)</u>
- atomic mass of H: 1.00784 g/mol
- molar mass of H₂O: 18.01528 g/mol
- proportion: 2 × 1.00784 g of H / 18.01528 g of H₂O = x / 0.930 g of H₂O
x = ( 2 × 1.00784 g of H / 18.01528 g of H₂O) × 0.930 g of H₂O = 0.10406 g of H
<u>3) Mass of oxygen (O) in 1.0857 g of pure sample</u>
- Mass of O = mass of pure sample - mass of C - mass of H
- Mass of O = 1.0857 g - 0.59767 g - 0.10406 = 0.38397 g O
Round to four decimals: Mass of O = 0.3840 g
<u>4) Mole calculations</u>
Divide the mass in grams of each element by its atomic mass:
- C: 0.59767 g / 12.0107 g/mol = 0.04976 mol
- H: 0.10406 g / 1.00784 g/mol = 0.10325 mol
- O: 0.3840 g / 15.999 g/mol = 0.02400 mol
<u>5) Divide every amount by the smallest value (to find the mole ratios)</u>
- C: 0.04976 mol / 0.02400 mol = 2.07 ≈ 2
- H: 0.10325 mol / 0.02400 mol = 4.3 ≈ 4
- O: 0.02400 mol / 0.02400 mol = 1
Thus the mole ratio is 2 : 4 : 1, and the empirical formula is:
Answer:
1. Land 2. open 3. broken 4. Lava matter 5. Force of the plates
Density = Mass / Volume
V = 1.00 * 4.00 * 2.50 = 10 cm³
22.57 g/cm³ = Mass / 10 cm³
M = 22.57 g/cm³ * 10 cm³
M = 225.7 g
Answer: The mass of the block of osmium is 225.7 g.
