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Aneli [31]
3 years ago
6

Can someone help me please

Chemistry
1 answer:
strojnjashka [21]3 years ago
7 0
Yes what do u need help with
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The use of evaporation and condensation to separate the partsof a mixture is called _____
Keith_Richards [23]

Answer: Evaporation Condensation is used to seperate the parts of a mixture

Explanation: This is because it is the first order of the distilation process

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2 years ago
Stock naming please help
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Answer:

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3 years ago
Calculate the ph of a solution with a [H3O+]=5.6x10-9M
Kisachek [45]

Answer:

               pH  =  8.25

Explanation:

                   The acidity or basicity of a solution is measured by its pH. The pH scale ranges from 0 to 14. Solutions having pH from 0-6.9 are considered acidic, at 7 neutral and basic when ranging from 7.1-14.

pH is calculated as,

                                 pH  =  - log [H⁺]   ---- (1)

Where;

            [H⁺]  =  concentration of Acid

Also, for bases pH i calculated using following formula,

                                 pH  =  14 - pOH

Therefore, Putting value of H⁺ in equation 1,

                                 pH  =  - log [5.6 × 10⁻⁹]

                                pH =  8.25

The solution provided is basic in nature.

6 0
3 years ago
What is 5.2034 x 10^8 in standard format
777dan777 [17]
Five hundred twenty million, three hundred and forty thousand. hope it helps!

8 0
3 years ago
Read 2 more answers
When a 3.00 grams sample of a compound containing only c, h, and o was completely burned, 1.17 grams of h2o and 2.87 grams of co
SVEN [57.7K]

Answer:- CH_2O_2

Solution:- From given masses of carbon dioxide and water we could calculate the moles that helps to calculate the moles of C and H.

Molar mass of carbon dioxide = 44 gram per mol

molar mass of water = 18.02 gram per mol

From given info, combustion of compound gives 1.17 grams of water and 2.87 grams of carbon dioxide. Let's calculate the moles of these:

1.17gH_2O(\frac{1mol}{18.02g})

= 0.0649molH_2O

Similarly, 2.87gCO_2(\frac{1mol}{44g})

= 0.0652molCO_2

One mol of water has two moles of H. So, the moles of H would be two times the moles of water as calculated above.

So, moles of H = 2* 0.0649 = 0.1298 mol

One mol of carbon dioxide contains one mol of C. So, the moles of C would be equal to the moles of carbon dioxide calculated above.

moles of C = 0.0652 mol

Let's convert the moles of H and C to grams so that we could calculate the amount of oxygen present in the sample as:

grams of H in sample = 1.008 x 0.1298 = 0.1308 g

grams of C in sample = 12*0.0652 = 0.7824 g

If we subtract the sum of the masses of C and H from sample mass then it would give as the mass of oxygen since the sample has only C, H and O.

mass of O in sample = 3.00g - (0.1308 g + 0.7824 g)

= 3.00 g - 0.9132 g

= 2.0868 g

Let's convert these grams of oxygen to moles on dividing by it's atomic mass as:

2.0868gO(\frac{1mol}{15.999g})

= 0.130 mol O

Now, we have the moles of all the three atoms and we know that an empirical formula is the simplest whole number ratio of the moles of atoms. So, let's calculate the ratio. For this, we divide the moles of each by the least one of them.Looking at the moles, the least value is for carbon. So, let's divide the moles of each by the moles of C as:

C = \frac{0.0652}{0.0652}  = 1

H = \frac{0.1298}{0.0652}  = 2

O = \frac{0.130}{0.0652}  = 2

The ratio of C, H and O is 1:2:2. So, the simplest formula of the compound is CH_2O_2 .



3 0
3 years ago
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