A scientist is tracking an object orbiting the sun that is found between mars and Jupiter. Which additional feature can be used
1 answer:
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Answer:
<u>2.26 </u>
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Explanation:
First you should know what are significant figures,
1 . All non - zero digits are significant.
2 . The zero between two non- zero are significant.Ex 302 has 3 significant figures.
3. The zero before the decimal and before any non-zero digit is non significant. Example : 0.003 has only 1 significant figure.
4. The zero after non zero are non - significant . But the zeros after the decimal point are significant.
300 has only 1 significant figure
300.0 has 4 significant figure
The least number of significant figures present in the number should be there in the final answer of the calculation.
Look at the number having minimum significant digits :
It is 3 (every number has 3- significant figures ) So the answer should also contain 3 - significant figures.
First solve the numerator part
Round off this number to 3 significant figures.
Answer is =<u> 2.26</u>
It has 3 - significant digit since all the digits are non-zero.
<u>Answer:</u> The volume of solution required is 0.0275 L.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
Given mass of copper = 4.24 g
Molar mass of copper = 63.55 g/mol
Putting values in above equation, we get:
The chemical equation for the reaction of potassium dichromate and copper follows:
By Stoichiometry of the reaction:
3 moles of copper reacts with 1 mole of potassium dichromate.
So, 0.067 moles of copper will react with = of potassium dichromate
To calculate the volume of potassium dichromate, we use the equation:
We are given:
Moles of potassium dichromate = 0.022 mol
Molarity of solution = 0.800 M
Putting values in above equation, we get:
Hence, the volume of solution required is 0.0275 L.
Answer:
See below.
Step-by-step explanation:
Ethers react with HI at high temperature to produce an alky halide and an alcohol.
R-OR' + HI ⟶ R-I + H-OR'
<em>Benzylic ethers</em> react by an Sₙ1 mechanism by forming the stable benzyl cation.
- PhCH₂-OR + HI ⟶ PhCH₂-O⁺(H)R + I⁻ Protonation of the ether
- PhCH₂-O⁺(H)R ⟶ PhCH₂⁺ + HOR Sₙ1 ionization of oxonium ion
- PhCH₂⁺ + I⁻ ⟶ PhCH₂-I Nucleophilic attack by I⁻
If there is excess HI, the alcohol formed in Step 2 is also converted to an alkyl iodide:
ROH +HI ⟶ R-I + H-OH
Thus, benzyl ethyl ether reacts to form benzyl iodide (a) and ethanol (b).
The ethanol reacts with excess HI in an Sₙ2 reaction to form ethyl iodide (c).
