Answer:
C. The mass of an electron is much less than the mass of a proton or
a neutron.
Explanation:
When we compare the mass of an electron to that of proton or neutron, the mass of an electron is much less than the mass of a proton or a neutron.
Electrons are negatively charged particles in an atom
Protons are positively charged particles
Neutrons do not carry any charges.
- The relative mass of an electron compared to that of a proton is
- This is a very small value
- Electrons generally have mass of 9.11 x 10⁻³¹kg
- Protons weigh 1.67 x 10⁻²⁷kg
- Neutrons weigh 1.68 x 10⁻²⁷kg
We can see that electrons have very small mass and this is why when calculating the mass of an atom, we use the sum of the number of protons and neutrons.
Answer:
A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)
B - Increase(P), Increase(q), Decrease (R)
C - Triple (P) and reduce (q) to one third
Explanation:
<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>
P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.
In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.
Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.
<span> aluminum is an element. All elements are pure substances, so that means they are homogenous.
please mark as brainliest
</span>
So it would be the complimentary base pairing, meaning that the codon must have been:
GAC
(Which is the codon for aspartic acid)
Answer:
(R)-but-3-en-2-ylbenzene
Explanation:
In this reaction, we have a very <u>strong base</u> (<em>sodium ethoxide</em>). This base, will remove a hydrogen producing a double bond. We know that the reaction occurs through an <u>E2 mechanism</u>, therefore, the hydrogen that is removed must have an <u>angle of 180º</u> with respect to the leaving group (the "OH"). This is known as the <u>anti-periplanar configuration</u>.
The hydrogen that has this configuration is the one that placed with the <u>dashed bond</u> (<em>red hydrogen</em>). In such a way, that the base will remove this hydrogen, the "OH" will leave the molecule and a double bond will be formed between the methyl and the carbon that was previously attached to the "OH", producing the molecule (R) -but-3- en-2-ylbenzene.
See figure 1
I hope it helps!
