Answer is #4 (neutrons)
Explanation:
Total mass before the reaction is 236 amu. The total mass of the two elements is 233 amu. This means that there are 3 amus carried by element X. Looking at the total atomic number of the two elements after reaction, it's 92, the same as the one at the beginning. This means that element X has zero atomic number, which could only be a neutron.
0.0006 in scientific notation is 6 × 10-4
<h2>Step by step Explanation:</h2>
All numbers in scientific notation or standard form are written in the form
m × 10n, where m is a number between 1 and 10 ( 1 ≤ |m| < 10 ) and the exponent n is a positive or negative integer.
To convert 0.0006 into scientific notation, follow these steps:
Move the decimal 4 times to right in the number so that the resulting number, m = 6, is greater than or equal to 1 but less than 10
Since we moved the decimal to the right the exponent n is negative
n = -4
Write in the scientific notation form m × 10n
= 6 × 10-4
Therefore,
6 × 10-4 is the scientific notation form of 0.0006 number and 6e-4 is the scientific e-notation form for 0.0006
<h2>HOPE IT HELPS ☺️</h2>
Answer:
1 and 2
Explanation:
Message me for explanation.
Answer:
Atoms are neither created or destroyed.
Explanation:
The end result of a chemical change does not create or destroy any atoms. Matter cannot be created or destroyed, meaning the same amount of atoms exist before and after the change.
Answer:
See explanation
Explanation:
The nucleophile here is CH3OH. We know that CH3OH is a good nucleophile that promotes SN2 reanction. However, (R)-6-bromo-2,6-dimethylnonane is a tertiary alkyl halide so the reaction proceeds by SN1 mechanism. This means that a racemic mixture is obtained at the end of the reaction because the attack occurs at the stereogenic carbon atom (6R) hence the product is optically inactive.
On the other hand, when (5R)-2-bromo-2,5-dimethylnonane is reacted with CH3OH, an optically active product is obtained because; though a tertiary alkyl halide and reaction occurs by SN1 mechanism, the attack does not occur at the stereogenic carbon atom (5R). Therefore, an optically active product is obtained in this case.
