Answer:
A) 0.50 M
Explanation:
The computation of the number of molarity of the NaOH is given below:
As we know that
Molarity = Moles ÷ liters
where
Moles is 20
And, the liters is 40
So, the molarity is
= 20 ÷ 40
= 0.50 M
hence, the molarity of the NaOH solution is 0.50 M
Therefore the A option is correct
A lead ball added to a graduated cylinder containing 30.8 ml of water, causing the level of the water to increase to 89.9 ml, has a volume of 59.1 ml.
Volume is defined as the amount of space that matter occupies. There are three different ways to find the volume:
1) Volume by Space - by measuring its physical dimensions
2) Volume by Density and Mass - by dividing its weight by its density
3) Volume by Displacement - by measuring the volume displaced when immersed in a liquid or gas
To solve for the volume of the lead ball, we can use the displacement method by subtracting the initial volume from the final volume.
If the level of the water increased from 30.8 ml to 89.9 ml upon adding the lead ball, the displaced volume is:
volume = final volume - initial volume
volume = 89.9 - 30.8
volume = 59.1 ml
Hence, the volume of the lead ball, using displacement method, is 59.1 ml.
To learn more about volume by displacement: brainly.com/question/1945909
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Answer: meteorological phenomenon
Explanation: Hello, there! A rainbow is a meteorological phenomenon that is caused by reflection, refraction and dispersion of light in water droplets resulting in a spectrum of light appearing in the sky. It takes the form of a multicoloured circular arc. Rainbows caused by sunlight always appear in the section of sky directly opposite the sun.
I hoped i help and im here for any more questions you might have! thankyou!
Answer: a) The maximum amount of calcium sulfate that can be formed is 112.9 grams
b) The formula for the limiting reagent is 
c) 0.033 moles of excess reagent 
are left unreacted.
Explanation:
To calculate the moles, we use the equation:
a) moles of 
b) moles of
According to stoichiometry :
1 moles of 
require 1 mole of
Thus 0.377 moles of ![Ca(OH)_2/tex] will require=[tex]\frac{1}{1}\times 0.377=0.377moles](https://tex.z-dn.net/?f=Ca%28OH%29_2%2Ftex%5D%20will%20require%3D%5Btex%5D%5Cfrac%7B1%7D%7B1%7D%5Ctimes%200.377%3D0.377moles)
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent as (0.410-0.377)= 0.033 moles of are left unreacted.
As 1 mole of give = 1 mole of 
Thus 0.83 moles of give =
of
Mass of 
Thus 112.9 g of will be produced from the given masses of both reactants.
