In a bag of peas that weighs 454 grams, there are between 1261 and 4540 peas.
The average pea weighs between 0.1 and 0.36 grams.
If we take the lower value (0.1 g/pea), the number of peas in 454 g is:
If we take the higher value (0.36 g/pea), the number of peas in 454 g is:
In a bag of peas that weighs 454 grams, there are between 1261 and 4540 peas.
You can learn more about conversion factors here: brainly.com/question/1844638
One separation technique to be used is the paper chromatography. This works by separating the components of the mixture through the difference of their concentrations. There is a stationary phase and the mobile phase, which flows through the stationary phase. The components travel at different rates and is usually signified by the colors. If more than one color would appear, that means that the dye is a mixture.
Answer:
A) FeO
Explanation:
Options:
A) FeO
B) FeO2
C) Fe2O
D) Fe7O2
<u>The correct answer is A) FeO.</u>
(Hope this helps can I pls have brainlist (crown)☺️)
Answer:
The given statement- Aldehydes and ketones are converted into alkenes by means of a direct nucleophilic addition called the Wittig reaction, is<u> True.</u>
Explanation:
The Wittig reaction converts aldehydes and ketones into alkenes through a simple nucleophilic addition. A triphenylphosphorine ylide, also known as a phosphorane, reacts with an aldehyde/ketone to produce an oxaphosphetane, a four-membered cyclic intermediate. Instead of being isolated, the oxaphosphetane decomposes spontaneously to release triphenylphosphine oxide and an alkene.
In an SN2 reaction, triphenylphosphine, a good nucleophile, reacts with a primary alkyl halide, followed by deprotonation of the carbon with a solid base, such as butyllithium, to form the ylide. In the product alkene, the carbonyl carbon and the carbon initially bound to the halogen become two carbons with a double bond.
The Wittig reaction's true worth lies in its ability to produce an alkene with a predictable structure, as the C=C bond forms exactly where the C=O bond did in the reactant, with no isomers (other than E/Z isomers) forming.
<u>Hence , the correct option is (A) True.</u>
To take the percent by mass of this element, we use the
formula:
% mass = (mass of element / mass of ore) * 100%
% mass = (47.5 g / (660 kg * 1000 g / kg)) * 100*
<span>% mass = 7.20 x 10^-3 %</span>
