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3 years ago

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How does one calculate the pH of a solution? O A. The pH is the product of [H+] and [OH-]. O B. The pH is the negative log of [O

H-]. O c. The pH is the log of [H+). D. The pH is the negative log of [ht].

1 answer:

blagie [28]3 years ago

8 0

Answer:

I believe it is A

Explanation:

pH = -log[H+] and pOH = -log[OH−].

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Hello Friends! think you guys can help a sister out lol > Based on the following image, what is the chemical formula of the c

kiruha [24]

Answer:

$CO_2$

Explanation:

Looking at the atom, you see that there is 1 carbon atom and 2 oxygen atoms.

The chemical formula of Carbon is just C (from the periodic table), and the chemical formula of Oxygen is just O (from the periodic table).

When you create the chemical formula of a compound, you want to indicate how many atoms of each element there are. Here, since there are 2 oxygen atoms, you want to have some form of $O_2$ , and since there's only 1 carbon atom, you want to have some form of C.

Now, we just put it together: $CO_2$ (this is btw carbon dioxide)

Hope this helps!

3 0

4 years ago

A bar of gold is 5.0mm thick, 10.0cm long and 2.0cm wide. It has a mass of exactly 193.0g. What is the desity of gold?

Tanzania [10]

<h3>Answer:</h3>

19.3 g/cm³

<h3>Explanation:</h3>

Density of a substance refers to the mass of the substance per unit volume.

Therefore, Density = Mass ÷ Volume

In this case, we are given;

Mass of the gold bar = 193.0 g

Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm

We are required to get the density of the gold bar

Step 1: Volume of the gold bar

Volume is given by, Length × width × height

Volume = 0.50 cm × 10.0 cm × 2.0 cm

= 10 cm³

Step 2: Density of the gold bar

Density = Mass ÷ volume

Density of the gold bar = 193.0 g ÷ 10 cm³

= 19.3 g/cm³

Thus, the density of the gold bar is 19.3 g/cm³

3 0

3 years ago

If enought heat was removed from water. What will it change into and why

Savatey [412]

If enought heat was removed by water, its will probably evaporate or turn into steam.

7 0

4 years ago

Read 2 more answers

10.0 g of gaseous ammonia and 6.50 g of oxygen gas are introduced into a previously evacuated 5.50 L vessel. If the ammonia and

Shalnov [3]

Answer:

The density is 3g/L

Explanation:

The reaction that occurs in the vessel is:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

10,0g of NH₃ are:

$10,0g * \frac{1mol}{17,031g} = 0,587 moles$

6,50 g of O₂ are:

$6,50g * \frac{1mol}{32g} = 0,203 moles$

For a complete reaction of O₂ there are necessaries:

$0,203 mol * \frac{4molNH_{3}}{5molO_{2}}= 0,163 moles of NH_{3}$

O₂ is limiting reactant. The excess moles of NH₃ are:

0,587 - 0,163 = <em>0,424 moles of NH₃</em>

These moles are:

$0,424mol * \frac{17,031g}{1mol} =$ <em>7,22g of NH₃</em>

Knowing O₂ is limiting reactant, mass of NO and H₂O are:

$0,203molO_{2}*\frac{4molNO}{5molO_{2}}*\frac{30,01g}{1molNO} =$ <em>4,87g of NO</em>

$0,203molO_{2}*\frac{6molH_{2}O}{5molO_{2}}*\frac{18,02g}{1molH_{2}O} =$ <em>4,39g of H₂O</em>

The total mass is: 7,22g + 4,87g + 4,39g = 16,48g ≡ <em>16,5g </em>

<em>-</em><em>The same mass add in the first. By matter conservation law-</em>

As vessel volume is 5,50L, density is:

16,5g/5,50L = <em>3g/L</em>

I hope it helps!

7 0

4 years ago

This is my cookie >:( you cannot have it

Fofino [41]

Hmm, welll I think it’s heterogeneous

6 0

3 years ago