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djyliett [7]
3 years ago
8

Would you use baking powder to see whether or not it is an acid or a substance? yes/no?

Chemistry
2 answers:
alexandr1967 [171]3 years ago
8 0
Yes you would use it
Zanzabum3 years ago
4 0

Answer:

yes.

Explanation:

If you have a liquid and you want to know if it is acidic, an easy way to tell is to mix in a little baking soda. The baking soda reacts with acids to produce bubbles.

You may be familiar with building a homemade kitchen volcano. You mix vinegar (an acid) with baking soda. It foams up as the baking soda reacts with the acid. This is in essence what you can do to test if a solution is acidic or not. If there is no acid present, the solution will not bubble when you add the baking soda.

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Can someone please help me with the with this
solong [7]

Answer:

Chemical

Explanation:

4 0
3 years ago
How do you find solubility? I am given only the temperature and the substance is KNO3. I don’t need anyone to solve it for me, j
skad [1K]
There’s no formula that relates solubility to temperature, but you can look up the solubility constant Ksp of substance given and then take the square root of that to find solubility.
5 0
3 years ago
So in general what rule can you make about substances that dissolve in solvent?
spin [16.1K]
In general, SOLUBILITY<span> is an ability of a substance to dissolve. In the process of dissolving, the substance which is being dissolved is called a </span>solute<span> and the substance in which the solute is dissolved is called a </span>solvent.<span> A mixture of solute and solvent is called a </span><span>solution.</span>
3 0
3 years ago
Read 2 more answers
CO(g)+2H2(g)⇌CH3OH(g)CO(g)+2H2(g)⇌CH3OH(g) This reaction is carried out at a different temperature with initial concentrations o
Katarina [22]

Answer:

9.4

Explanation:

The equation for the reaction can be represented as:

CO_{(g)}    +      2H_2O_{(g)}   ⇄  CH_3OH_{(g)}

The ICE table can be represented as:

                                  CO_{(g)}    +      2H_2_{(g)}   ⇄  CH_3OH_{(g)}

Initial                          0.27             0.49              0.0

Change                      -x                  -2x                 x

Equilibrium               0.27 - x         0.49 -2x          x

We can now say that the concentration of  CH_3OH_{(g)} at equilibrium is x;

Let's not forget that at equilibrium  CH_3OH_{(g)} = 0.11 M

So:

x =  [CH_3OH_{(g)}] = 0.11 M

[CO_{(g)}] = 0.27 - x

[CO_{(g)}] = 0.27 - 0.11

[CO_{(g)}] = 0.16 M

[2H_2_{(g)}] = (0.49 - 2x)

[2H_2_{(g)}] = (0.49 - 2(0.11))

[2H_2_{(g)}] = 0.49 - 0.22

[2H_2_{(g)}] = 0.27 M

K_C = \frac{[CH_3OH]}{[CO][H_2]^2}

K_C = \frac{(0.11)}{(0.16)[(0.27)^2}

K_C = 9.4307

K_C = 9.4

∴ The equilibrium constant at that temperature = 9.4

8 0
3 years ago
In the compound h2s hydrogen has an oxidation number of 1+ and sulfur has an oxidation number of
elena-14-01-66 [18.8K]
Oxidation number is defined as the total number of electrons that are gained or lost by the atom to form a chemical bond. 
the oxidation number of the compound H₂S is 0.
the sum of the oxidation numbers of the individual elements should add up to the oxidation number of the compound. 

(oxidation number of H x 2 H ions) + oxidation number of S = 0
since we know the oxidation number of H, lets name the oxidation number of S = x
(+1 * 2 )+ (x) = 0
2 + x = 0
x = -2
oxidation number of S is -2
3 0
3 years ago
Read 2 more answers
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