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lianna [129]
3 years ago
14

What substance is added to cordial to make it more dilute?

Chemistry
1 answer:
shusha [124]3 years ago
7 0

The answer is water.

That is water is added to cordial to make it more dilute.

Cordial is basically used to describe sweeter distilled spirits . Or it can be used to describe non-alcoholic drink. These drinks can be diluted using water.

So the answer is water, that is water is added to cordial to make it more dilute.

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If 10.0 g water at 0.0°C is mixed with 20.0 g of water at 30.0°C what is the final temperature of the mixture?
Bond [772]

Answer:

1.67 gradius Celcius

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

7 0
3 years ago
Which hydrocarbon belongs to the Alkyne homologous series?
Ira Lisetskai [31]

Answer:

(A) CHCH

Explanation:

Hello,

Alkynes are formed when triple bonds are formed between carbons, it means that due to the triple bonds, three bonds are already exhausted and carbon atoms need four bonds to attain the octet, for that reason, an extra bond is allowed. Now, such extra bond, when talking about hydrocarbons is filled with hydrogens, for that reason the alkyne should be (A) CHCH as it also accomplishes the following formula which is not accomplished by the rest of the options:

C_nH_{2n-2}

Since n=2, we have:

C_2H_2

Which matches with CHCH.

Best regards.

3 0
3 years ago
If the only thing we are interested in studying is how the number of particles affects pressure, what variables do we have to be
andre [41]
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6 0
4 years ago
An environmental scientist developed a new analytical method for the determination of cadmium (cd^2+) in mussels. To validate th
Anettt [7]

Answer:

The method is accurate  in the calculation of the Cu^+2

Explanation:

As a first step we have to calculate the <u>average concentration </u>of Cu^+2 find it by the method.

\frac{0.782+0.762+0.825+0.838+0.761 }{5} =0.79 ppm

Then we have to find the<u> standard deviation:</u>

s=\sqrt{\frac{1}{N-1}\sum_{i=1}^N(x_i-\bar{x})^2}=0.0359

For the confidence interval we have to use the formula:

μ=Average±\frac{t*s}{\sqrt{n} }

Where:

t=t student constant with 95 % of confidence and 5 data=2.78

μ= 0.79  ±  \frac{2.78*0.0359}{\sqrt{5} }

upper limit:  0.84

lower limit: 0.75

If we compare the limits of the value obtanied by the method (Figure 1 Red line) with the reference material (Figure 1 blue line) we can see that the values obtained by the method are within the values suggested by the reference material. So, it's method is accurate.

7 0
3 years ago
What type of matter is in this
malfutka [58]
Pretty sure it’s Plasma
5 0
3 years ago
Read 2 more answers
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