Answer:
(2R,3S)-2-chloro-3,5-dimethylhexane
Explanation:
As first step we have the <u>attack of the OH group</u> to the P atom in the PCl3 and one of the Cl atoms would leave. Then we will have a <u>rearrangement</u> to produce a <u>double bond </u>with the oyxgen on the OH. Finally the Cl produced will a<u>ttack the carbon</u> in a <u>Sn2 substitution reaction</u> to produce the halide with an <u>opposite configuration</u>.
Answer:
Arsenic.
Explanation:
Hello there!
In this case, since insecticides are substances that act as poisons to get rid of insects in order to prevent their presence and/or reproduction in houses, companies, crops and others, a substance that has been widely used is the metalloid arsenic due to its direct affection of the insect's body (movement, performance, cellular functions).
In addition, high levels of arsenic in food could cause arsenic poisoning in humans as well, that is why such practice must be properly performed and by using the correct security protocol.
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a) Group 2 elements have 2 electrons on their outer shell, so they form a 2+ charge.
b) they lose 2 electrons as they are transferred to the non metal.
c)They obtain this charge as when they are made into an ionic compound the 2 electrons on the outer shell are transferred to the non metal, meaning there are 2 more protons that electrons, giving it a positive charge.
hope this helps! :)
Answer:The volume of the remaining gas that is ammonia is 23.85 L.
Explanation:

Moles of 
Moles of HCl of gas = 
According to reaction 1 mole of HCl reacts with 1 mol of
then 2.06 moles of HCl will react with = 2.06 moles of
Moles left of ammonia left = 4.43 - 2.06 = 2.36 moles
Volume of the gas will be given by Ideal gas equation: PV=nRT
Pressure = 752 mmHg = 752 × 0.0031 atm = 2.33 atm
R = 0.08026 L atm/K mol
V = ? , n = number of moles of ammonia
Temperature = 14 °C = 14 + 273 K = 287 K(0°C = 273K)

The volume of the remaining gas that is ammonia is 23.85 L.
Answer:
8.934 g
Step-by-step explanation:
We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 192.12 44.01
H₃C₆H₅O₇ + 3NaHCO₃ ⟶ Na₃C₆H₅O₇ + 3H₂O + 3CO₂
m/g: 13.00
For ease of writing, let's write H₃C₆H₅O₇ as H₃Cit.
(a) Calculate the <em>moles of H₃Cit
</em>
n = 13.00 g × (1 mol H₃Cit /192.12 g H₃Cit)
n = 0.067 67 mol H₃Cit
(b) Calculate the <em>moles of CO₂
</em>
The molar ratio is (3 mol CO₂/1 mol H₃Cit)
n = 0.067 67 mol H₃Cit × (3 mol CO₂/1 mol H₃Cit)
n = 0.2030 mol CO₂
(c) Calculate the <em>mass of CO₂
</em>
m = 0.2030 mol CO₂ × (44.01 g CO₂/1 mol CO₂)
m = 8.934 g CO₂