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MW of H2SO4 is 98.08. 2M = 2 x 98.08 in 1 L of water (1 gram=1 ml). Take 500 ml water in a 1 L measuring cylinder. Add 196.16 ml slowly along the side into water in the measuring cylinder. Use 50 ml pipette with automated pipettor. If needed you may want to keep the cylinder in ice to take care of the heat generated. Then make up to volume to 1 L with water. Eq. wt for H2SO4 = 98.08/2 = 49.039. SO for 2N solution, 2 eq.wt in 1 L. 98.08 ml in 1 L water adopting the method cited above.
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How can you prepare 1N H2SO4 from H2SO4 95 percent?
To prepare 1N H2SO4 from 95% H2SO4, you would first need to dilute the 95% H2SO4 with water by adding the appropriate amount of water to achieve the desired concentration. To calculate the volume of 95% H2SO4 needed to make 1N solution, you need to use the formula: (Normality of stock solution) * (Volume of stock solution) = (Normality of diluted solution) * (Volume of diluted solution). Adjust the volumes accordingly to prepare the desired 1N solution.
How can you prepare H2SO4 0.25N?
To prepare a 0.25N solution of H2SO4, you would need to dilute concentrated sulfuric acid (typically around 95-98% purity) with water. For example, to prepare 1L of 0.25N H2SO4, you would mix 20.6mL of concentrated H2SO4 with approximately 980mL of water. Always add acid to water slowly and carefully while stirring to avoid splashing and heat generation.
How do you prepare 0.25N sulphuric acid from 2N sulphuric acid?
To prepare 0.25N sulphuric acid from 2N sulphuric acid, you can dilute the 2N solution by adding 7 parts of water to 1 part of the 2N solution. This will result in a final 0.25N sulphuric acid solution.
How do you prepare 2.5N h2so4?
how 2.5N H2SO4 prepared from concentrated H2SO4
What is 2N dilute sulphuric acid?
2N dilute sulfuric acid refers to a solution where the concentration of sulfuric acid (H2SO4) is equivalent to 2 moles per liter (2N). This solution is often used in chemical reactions or laboratory experiments that require a specific concentration of sulfuric acid.
2m - n - m - 2n?
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What is 2m plus 2n plus p?
It can't be simplified any further so it is:2m + 2n + p
If two odd numbers are added together what is the result?
Two odd numbers added together will always be an even number. I'll show how: Let M and N be any two integers. 2M is even, and 2N is even. So (2M +1) is an odd number, and (2N + 1) is an odd number. (2M +1) + (2N +1) = 2M + 1 + 2N +1 = 2M + 2N + 2 = 2(M + N + 1), which is even.
Is 2m-2n equals 0 linear?
Yes.
Is the sum of an even whole number and an odd whole number always odd?
Yes --------------------------------------------- Let n be an integer. Then 2n is an even number Let m be an integer. Then 2m is an even number and 2m + 1 is an odd number. Then: even + odd = (2n) + (2m + 1) = (2n + 2m) + 1 = 2(n + m) + 1 = 2k + 1 (where k = m + n) which is an odd number.
How do you solve L equals 2M plus 2N for N?
N=l-m
Why does an odd number times an odd number equal an odd number?
Let's do some algebra. Assume that "m" and "n" are any integers. An even number is divisible by 2, so 2m or 2n would be even. An odd number is one that is not divisible by 2, so 2m + 1, or 2n + 1, are odd numbers.Multiply those two odd numbers together: (2m+1)(2n+1) = 4mn + 2m + 2n + 1. Since the first three parts are even, the added 1 at the end makes the result odd.
Why do you get an odd number when you multiply two odd numbers?
Let one odd number be "2m + 1", the other odd number "2n + 1" (where "m" and "n" are integers). All odd numbers have this form. If you multiply this out, you get (2m+1)(2n+1) = 4mn + 2m + 2n + 1. Since each of the first three parts is even, the "+1" at the ends converts the result into an odd number.
Why odd number minus a odd number even?
Every odd number leaves a remainder of 1 when divided by 2. Therefore, every odd number is of the form 2k +1 where k is some integer.Suppose 2m + 1 and 2n + 1 are two odd numbers.Then (2m + 1) - (2n + 1) = 2m + 1 - 2n - 1 = 2m - 2n = 2*(m - n)By the closure property of integers under addition (and subtraction), m and n being integers implies than (m - n) is an integer. Therefore 2*(m - n) is an even integer.
How is the sum of three odd numbers always odd?
Algebraically, this can be seen: An odd number is one more than an even number, and all multiples of 2 are even numbers. Let three odd numbers be (2n + 1), (2m + 1) & (2k + 1) for some n, m & k. Adding them gives: (2n + 1) + (2m + 1) + (2k + 1) = 2n + 2m + 2k + 1 + 1 + 1 = 2n + 2m + 2k + 2 + 1 = 2(n + m + k + 1) + 1 which is one more than an even number - an odd number.
Why do you get an even number when adding two even numbers together?
The presentation of an even number is 2n (we can use any letter, such as 2m or 2r) Let the first even number be 2n and the second number be 2m. So we have, 2n + 2m = 2(n + m) let n + m = r, then by substituting r for n + m we have = 2r Since 2r represents an even number, then we can say that when are adding two even numbers we obtain another even number.
Why does two odd numbers equal an even number?
Suppose x and y are two odd numbers.Then x = 2m - 1 and y = 2n - 1 for some integers m and n. Then x + y = m2 - 1 + 2n - 1 = 2m + 2n - 2 = 2(m + n - 1) That is, x + y is a multiple of 2: in other words, it is even.
