All categories

3 years ago

What is the average range of the PH scale? THANKS!!

Chemistry

2 answers:

alex41 [277]3 years ago

8 0

<span>The pH scale goes from 0-14. 0-6.9 is acidic, 7 is neutral and 7.1-14 is basic</span>

g100num [7]3 years ago

3 0

The range of the PH scale is 0-14, so the average range is 7.

You might be interested in

Which of the following is the correct wedge and dash conformation for the following Newman projection?

kakasveta [241]

We have that the correct wedge and dash conformation for the following Newman projection is

From the Diagrams above

Two CH_3 groups points on opposite sides in plane

Two Br are on same side of plane

Two H also on same side of the plane so the plausible structure is IV

Therefore

The Correct option is IV

For more information on this visit

brainly.com/question/17756498?referrer=searchResults

8 0

3 years ago

A compound containing chromium, Cr; chlorine, Cl; and oxygen, O, is analyzed and found to be 33.6% chromium, 45.8% chlorine, and

Assoli18 [71]

Answer

The empirical formula is CrO₂Cl₂

Explanation:

Empirical formula is the simplest whole number ratio of an atom present in a compound.

The compound contain, Chromium=33.6%

Chlorine=45.8%

Oxygen=20.6%

And the molar mass of Chromium(Cr)=51.996 g mol.

Chlorine containing molar mass (Cl)= 35.45 g mol.

Oxygen containing molar mass (O)=15.999 g mol.

Step-1

Then,we will get,

Cr= $\frac{1}{51.996} \times33.6=0.64$ mol

Cl= $\frac{1}{35.45} \times45.8=1.29$ mol.

O= $\frac{1}{15.99} \times=1.28$ mol.

Step-2

Divide the mole value with the smallest number of mole, we will get,

Cr= $\frac{0.64}{0.64} =1$

Cl= $\frac{1.29}{0.64} =2$

O= $\frac{1.28}{0.64} =2$

Then, the empirical formula of the compound is CrO₂Cl₂ (Chromyl chloride)

4 0

3 years ago

Read 2 more answers

The acid dissociation constant Ka of boric acid (H3BO3) is 5.8 times 10^-10. Calculate the pH of a 4.4 M solution of boric acid.

madam [21]

Answer: The pH of a 4.4 M solution of boric acid is 4.3

Explanation:

$H_3BO_3\rightarrow H^+H_2BO_3^-$

at t=0 cM 0 0

at eqm $c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 4.4 M and $\alpha$ = ?

$K_a=5.8\times 10^{-10}$

Putting in the values we get:

$5.8\times 10^{-10}=\frac{(4.4\times \alpha)^2}{(4.4-4.4\times \alpha)}$

$(\alpha)=0.000011$

$[H^+]=c\times \alpha$

$[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M$

Also $pH=-log[H^+]$

$pH=-log[4.8\times 10^{-5}]=4.3$

Thus pH of a 4.4 M $H_3BO_3$ solution is 4.3

3 0

3 years ago

4HF(g)+SiO2(s)→SiF4(g)+2H2O(l)

Alecsey [184]

<span>Gallium-72 that is what i got when i did the math for it</span>

8 0

3 years ago

an unknown molecule is found to consist of 24.2% carbon by mass, 4.0% hydrogen by mass and the remaining mass is due to chlorine

Paha777 [63]

Answer:

C3 H6 Cl 3

Explanation:

C -24.2%

H - 4.0%

Cl - (100-24.2 - 4.0)=73.8 %

We can take 100g of the substance, then we have

C -24.2 g

H - 4.0 g

Cl - 73.8 g

Find the moles of these elements

C -24.2 g/12.0 g/mol =2.0 mol

H - 4.0 g/1.0 g/mol = 4. 0 mol

Cl - 73.8 g/ 35.5 g/mol = 2.1 mol

Ratio of these elements gives simplest formula of the substance

C : H : Cl = 2 : 4 : 2 = 1 : 2 : 1

CH2Cl

Molar mass (CH2Cl) = 1*12.0 +2*1.0 + 1*35.5 = 49.5 g/mol

Real molar mass = 150 g/mol

real molar mass/ Molar mass (CH2Cl) = 150 /49.5=3

So, Real formula should be C3 H6 Cl 3.

4 0

3 years ago

Read 2 more answers