Answer:
2.25×10¯³ mm.
Explanation:
From the question given above, we obtained the following information:
Diameter in micrometer = 2.25 μm
Diameter in millimetre (mm) =?
Next we shall convert 2.25 μm to metre (m). This can be obtained as follow:
1 μm = 1×10¯⁶ m
Therefore,
2.25 μm = 2.25 μm / 1 μm × 1×10¯⁶ m
2.25 μm = 2.25×10¯⁶ m
Finally, we shall convert 2.25×10¯⁶ m to millimetre (mm) as follow:
1 m = 1000 mm
Therefore,
2.25×10¯⁶ m = 2.25×10¯⁶ m /1 m × 1000 mm
2.25×10¯⁶ m = 2.25×10¯³ mm
Therefore, 2.25 μm is equivalent to 2.25×10¯³ mm.
Answer:

Explanation:
C = Allowable concentration = 1.1 mg/L
= Flow rate of river = 
= Discharge from plant = 
= Background concentration = 0.69 mg/L
= Maximum concentration that of the pollutant
The concentration of the mixture will be

The maximum concentration that of the pollutant (in mg/L) that can be safely discharged from the wastewater treatment plant is
.
Find the number of moles
C = n / V
C(Concentration) = 0.30 moles / L
V ( Volume) = 2 L
n = ??
n = C * V
n = 0.30 mol / L * 2 L
n = 0.60 mol
Find the molar mass
2Na = 23 * 2 = 46 grams
1S = 32 * 1 = 32 grams
O4 = 16 * 4 = 64 grams
Total = 142 grams / mol
Find the mass
n = given mass / molar mass
n = 0.06 mol
molar Mass = 142 grams / mol
given mass = ???
given mass = molar mass * mols
given mass = 142 * 0.6
given mass = 85.2 grams.
85.2 are in a 2 L solution that has a concentration of 0.6 mol/L
Answer:
So 1st it is B then D then E then a then C
Answer:
See explanation below
Explanation:
The first two pictures show the reagents used in these reactions a) and b). As it was stated, An E2 reaction proceeds with an antiperiplanar stereochemistry, so in the case of reaction a) it fill form a product with the groups in opposite directions. In other words, a Trans product.
In the case of reaction b) we have the same reaction, with the difference that we have changed the CH3 and phenyl group of positions. This will cause that the reaction will proceed the same but the stereochemistry of the final product will be changed too. In this case, and according to the picture 3 attached, we can see that the product formed is a cis product. So we can conclude that the relation of product a) and b) is that they are isomers, the trans and cis isomers respectively. See picture below for mechanism and products