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Anna71 [15]
3 years ago
12

WILL GIVE BRAINLIEST

Chemistry
1 answer:
dedylja [7]3 years ago
4 0

Answer:

In order to be able to solve this problem, you will need to know the value of water's specific heat, which is listed as

c=4.18Jg∘C

Now, let's assume that you don't know the equation that allows you to plug in your values and find how much heat would be needed to heat that much water by that many degrees Celsius.

Take a look at the specific heat of water. As you know, a substance's specific heat tells you how much heat is needed in order to increase the temperature of 1 g of that substance by 1∘C.

In water's case, you need to provide 4.18 J of heat per gram of water to increase its temperature by 1∘C.

What if you wanted to increase the temperature of 1 g of water by 2∘C ?

This will account for increasing the temperature of the first gram of the sample by n∘C, of the the second gramby n∘C, of the third gram by n∘C, and so on until you reach m grams of water.

And there you have it. The equation that describes all this will thus be

q=m⋅c⋅ΔT , where

q - heat absorbed

m - the mass of the sample

c - the specific heat of the substance

ΔT - the change in temperature, defined as final temperature minus initial temperature

In your case, you will have

q=100.0g⋅4.18Jg∘C⋅(50.0−25.0)∘C

q=10,450 J

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lakkis [162]

We have that you can you prepare 450 ml of 1 X buffer A by adding 45ml of 10X buffer A with 405ml(450-45) of water

From the question we are told

If you have 100 ml of 10X Buffer A, how could you prepare 450 ml of 1 X buffer A

Generally the equation for the  Concentration  Volume relationship is mathematically given as

C1V1=C2V2\\\\Where\\\\C1=10X\\\\C2=1X buffer A\\\\V2=450ml \\\\Therefore\\\\V1=\frac{450*1}{10}\\\\V1=45ml

Therefore

You can you prepare 450 ml of 1 X buffer A by adding 45ml of 10X buffer A with 405ml(450-45) of water

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