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4 years ago

Pure water is neutral. if water is neither an acid nor base what is the ph of the water

Chemistry

2 answers:

SCORPION-xisa [38]4 years ago

7 0

Its pH is constant which is neither acid or base

Igoryamba4 years ago

7 0

The answer is 7.

Since pH values are made up of 14 numbers, and the smaller the number is, the more acidic a substance is, and the larger the pH value is, the more alkaline the substance is. So, the median number of the pH scale is 7.

We can easily test the pH value of simple substances (liquid the best) by using pH paper, which you can easily get in drugstores, or pH meter, where you should be able to find in most laboratories. You'll have a chart to match with and you should get the value of 7 if it's pure water that you're testing.

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If pea plants with the genotypes RR and rr cross, what genotype will their offspring all have?

Aliun [14]

I think it’ll be Rr

6 0

3 years ago

Read 2 more answers

An organic compound has a distribution coefficient of 1.5

jasenka [17]

Answer:

Three times with 5 ml will yield more

Explanation:

Let x represent the amount yield

Kd = (x/15) / ((50-x) / 15) where Kd = 1.5

1.5 = (x/15) / ((50-x) / 15)

x / (50 - x) = 1.5

x = 75 - 1.5x

x + 1.5x = 75

2.5x = 75

x = 75 / 2.5 = 30 mg

when extraction three times

1st extraction

(x1/5) / ((50 - x1) / 15) = 1.5

3x1 / 50 - x1 = 1.5

3x1 = 75 - 1.5x1

3x1 + 1.5x1 = 75

4.5x1 = 75

x1 = 75 / 4.5 = 16.67 gm

second extraction

(x2/ 5) / (33.33 - x2 ) / 15) = 1.5

3x2 / ( 33.33 - x2) = 1.5

3x2 = 1.5(33.33 - x2)

3x2 = 49.995 - 1.5x2

3x2 + 1.5x2 = 49.995

4.5x2 = 49.995

x2 = 49.995 / 4.5

x2 = 11.11 mg

Third extraction

(x3/5) / ((22.22 - x3) / 15) = 1.5

3x3 = 1.5 ( 22.22 - x3)

3x3 + 1.5x3 = 33.33

4.5x3 = 33.33

x3 = 33.33 / 4.5 = 7.41 mg

total extraction = x1 + x2 + x3 =16.67 + 11.11 + 7.41 = 35.19 mg

the three times extraction using 5ml yields 5.19 mg more

3 0

3 years ago

What did Josef Loschmidt and Amedeo Avogadro Contribute to our understanding of basic molecular numbers, sizes, and reaction rat

mario62 [17]

From Avogadro we obtained a physical constant of matter which is Avogadro's number, and from both scientists we understand that elementary gases such as hydrogen, nitrogen, and oxygen were composed of two atoms.

<h3>What is Avogadro's number?</h3>

Avogadro's number, or Avogadro's constant, is the number of particles found in one mole of a substance.

The Avogadro's number is given as 6.02 x 10²³.

Summary of Josef Loschmidt and Amedeo Avogadro Contribution to chemistry.

Equal volumes of gas contain equal numbers of molecules,
Elementary gases such as hydrogen, nitrogen, and oxygen were composed of two atoms.

Thus, from Avogadro we obtained a physical constant of matter which is Avogadro's number, and from both scientists we understand that elementary gases such as hydrogen, nitrogen, and oxygen were composed of two atoms.

Learn more about Avogadro's here: brainly.com/question/1581342

#SPJ1

4 0

2 years ago

The osmotic pressure of a solution containing 2.04 g of an unknown compound dissolved in 175.0 mLof solution at 25 ∘C is 2.13 at

kherson [118]

Answer: The molecular formula of the compound is $C_4H_{10}O_4$

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 2.13 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Given mass of compound = 2.04 g

Volume of solution = 175.0 mL

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$2.13atm=1\times \frac{2.04\times 1000}{\text{Molar mass of compound}\times 175.0}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\\text{Molar mass of compound}=\frac{1\times 2.04\times 1000\times 0.0821\times 298}{2.13\times 175.0}=133.9g/mol$

Calculating the molecular formula:

The chemical equation for the combustion of compound having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=36.26g$

Mass of $H_2O=14.85g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 36.26 g of carbon dioxide, $\frac{12}{44}\times 36.26=9.89g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 14.85 g of water, $\frac{2}{18}\times 14.85=1.65g$ of hydrogen will be contained.

Mass of oxygen in the compound = (22.08) - (9.89 + 1.65) = 10.54 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{9.89g}{12g/mole}=0.824moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.65g}{1g/mole}=1.65moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{10.54g}{16g/mole}=0.659moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.659 moles.

For Carbon = $\frac{0.824}{0.659}=1.25\approx 1$

For Hydrogen = $\frac{1.65}{0.659}=2.5$

For Oxygen = $\frac{0.659}{0.659}=1$

Converting the mole fraction into whole number by multiplying the mole fraction by '2'

Mole fraction of carbon = (1 × 2) = 2

Mole fraction of oxygen = (2.5 × 2) = 5

Mole fraction of hydrogen = (1 × 2) = 2

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 5 : 2

The empirical formula for the given compound is $C_2H_5O_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 133.9 g/mol

Mass of empirical formula = 61 g/mol

Putting values in above equation, we get:

$n=\frac{133.9g/mol}{61g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(2\times 2)}H_{(5\times 2)}O_{(2\times 2)}=C_4H_{10}O_4$

Hence, the molecular formula of the compound is $C_4H_{10}O_4$

4 0

3 years ago

How did the metric system came to existence? (explain how it started until it is being implemented)

ozzi

Ever heard of this thing called... RESEARCH!? You might want to try it buddy. Sayounara.

3 0

3 years ago