Answer:
Explanation:
1. Given that,
Initial speed of a bicycle, u = 4 m/s
The final speed of a bicycle, v = 6 m/s
Time, t = 6 s
We need to find the acceleration of the bicycle. We know that, acceleration is equal to the change in speed divided by time taken. So,
So, the acceleration of the bicycle is equal to 
.
The other students in the lab should be notified next in this type of scenario.
<h3>What is an acid?</h3>
This is a substance which donates protons and is very corrosive. It also turns blue litmus paper red.
When it was spilled and baking soda was used to neutralize it on the floor , it is best to inform the other students too so as to prevent them from being exposed by mistake thereby reducing risk of injury.
Read more about Acid here brainly.com/question/25148363
#SPJ1
Answer:
159 mg caffeine is being extracted in 60 mL dichloromethane
Explanation:
Given that:
mass of caffeine in 100 mL of water = 600 mg
Volume of the water = 100 mL
Partition co-efficient (K) = 4.6
mass of caffeine extracted = ??? (unknown)
The portion of the DCM = 60 mL
Partial co-efficient (K) = 
where; 
solubility of compound in the organic solvent and 
= solubility in aqueous water.
So; we can represent our data as:
÷ 
Since one part of the portion is A and the other part is B
A+B = 60 mL
A+B = 0.60
A= 0.60 - B
4.6=
÷ 
4.6 = 
4.6 × =
4.6 B 
= 0.6 - B
2.76 B = 0.6 - B
2.76 + B = 0.6
3.76 B = 0.6
B = 
B = 0.159 g
B = 159 mg
∴ 159 mg caffeine is being extracted from the 100 mL of water containing 600 mg of caffeine with one portion of in 60 mL dichloromethane.
One mole of hydrogen peroxide contains 6.02 x 10^23 molecules of hydrogen peroxide. And each molecule contains 4 atoms, so the answer is 4 x 6.02 x 10^23.
Compound 1: Sodium borohydride
In sodium borohydride (NaBH4), B is a central metal.
Electronic configuration of B is 1s2 2s2 2p1.
B undergoes sp3 hybridization in NaBH4, to generate 4 hybrid orbitals. These hybrid orbitals, forms sigma bond with 4 'H' atoms. Due to this, the structure of sodium borohydride in tetrahedral.
........................................................................................................................
Compound 2: B<span>oron trifluoride
</span>In boron trifluoride (BF3), B is a central metal.
Electronic configuration of B is 1s2 2s2 2p1.
B undergoes sp2 hybridization in NaBH4, to generate 3 hybrid orbitals. These hybrid orbitals, forms sigma bond with 3 'H' atoms. Due to this, the structure of <span>boron trifluoride</span> is <span>triangular planner</span>.
