Answer:
Why ? Because 1 molecule of H2SO4 gives 2 H+ ions per molecule while only one H+ ion is required to neutralize 1 molecule of KOH. So, 1 molecule of H2SO4 can neutralize 2 molecules of KOH. Hence, we would require 525 ml of 0.03 M H2SO4 to neutralize 525 ml of 0.06 M KOH. How will we prepare 525 ml of 0.03 M H2SO4 ?
Explanation:
Now, we have 0.025 M H2SO4 and we do not know how much volume we have.
We will use the standard N1 X V1 = N2 X V2 for this calculation.
N1=0.025 M; V1=unknown; N2=0.03 M and V2=525 ml.
So V1= (0.03 X 525)/(0.025) = 630 ml.
Answer: Belongs to the group 2A
Explanation:
As you can see, the first two ionization energies are close and low, meaning that this element ionizates easily.
Not only loses easily the first electron, but the second too
To remove the third electron you requiered a huge amount of energy
Now, elements easily ionizable are the ones from group IA, group 2A and transition metals.
The last ones have mixed characteristics in matter of how many electrons you can remove from them, so they are not a family.
Now the question: group I or group II ?
The elements of group I have low ionization energies for the first electron but high energies for the second ones.
Being all that said, the unknown element belongs to the Group 2A
A colored line, as long as it is one single piece, not broken
<span>The answer to question 2 is C. A magnifying glass is an example of a plano-convex lens, where one side of the lens is flat and the other is a convex curve. The answer to question 3 is either B or C. A converging lens is curved on both sides and so the rays of light coming out of it converge at a point, which is known as the focal point. When the object is inside the focal point, the image is real and inverted. If it is inside the focal point, the image is virtual and upright. Therefore the image in this question will be upright. The focal length is the distance between the image that is being magnified and the centre of the magnifying lens. A real image can only be formed when the object is further away from the lens than the focal length. Therefore, in this question, the image is virtual, as the object is closer to the lens than the focal length. The answer to question 4 is D because the index of refraction cannot be less than 1. The answer to question 5 is D because only concave mirrors can produce real images; other types produce virtual images. For question 6, the answer is D. In the rainbow, each of the colours refracts at a slightly different angle; red has the smallest refractive index and violet the largest. Of the options, orange is closest to red. For question 12, A is the answer. A higher operating temperature is not a reason fluorescent lamps are better than incandescent lamps because they have a lower operating temperature. Question 15: all of these are characteristics of different electromagetic waves. For question 18, B is true - special care must be taken when low illuminance is required to reduce glare. The answer to question 19 is B - a compound microscope makes use of two lenses. For question 20, the answer is 5 meters away. The illuminance (E) is equal to light intensity (I) divided by the square distance from the light source (d). Therefore, 4 = 100/d squared. To switch this around, d squared is equal to 100/4 = 25. Then find the square root of 25, which is 5.</span>
Answer:
Think it's NC13
Explanation:
It's the only one missing in the molecule
