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andriy [413]
3 years ago
10

Understand how to represent chemical reactions using an equations Question If two compounds in a chemical reaction are listed wi

th the coefficients 2 and 5 respectively, which is NOT a possible numerical combination of these compounds that will result in their complete reaction?
Chemistry
1 answer:
konstantin123 [22]3 years ago
4 0

Answer: 8 and 10

Explanation:

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Is 2OH−+Ca2+−>2Ca(OH)2 balanced
wariber [46]

Answer:

No.

Explanation:

No.  There is 1 atom of Ca on the left and 2 Ca's on the right  and 2 OH's on the left and 4 on the right.

The balanced equation is:

4OH-  +  2Ca2+ ---->  2Ca(OH)2.

3 0
3 years ago
Read 2 more answers
If an object is accelerating the forces acting on the object are ?​
belka [17]

Answer:

If an object is accelerating the forces acting on the object are BALANCED.

Explanation

if an object is moving at a constant rate of acceleration, the the forces acting upon it are balanced .

7 0
3 years ago
NEED HELP ASAP ITS DUE TODAY THESE QUESTIONS
beks73 [17]

1.

Density can be defined as the mass of the substance in unit volume.

Density = mass / volume

Hence, g/mL and kg/L can be used as the units of density.

Those units are interchangeable because when converting one unit into other one nothing will happen to the value.

That is because when converting the g into kg, you have to multiply the value by 1 x 10⁻³. When converting mL into L, you should again multiply the volume by 1 x 10⁻³. Then those 1 x 10⁻³ will cancel off and the original value will remain as same.

2.

Answer is 2.70 g/mL.

<em>Explanation;</em>

Mass of the block = 146 g

Volume of the block =  length x width x height

                                  = 6.0 cm x 3.0 cm x 3.0 cm

                                  = 54 cm³ = 54 mL

Density = mass / volume

            = 146 g / 54 mL

            = 2.70 g/mL

3.

Answer is  3.39 g/mL.

<em>Explanation;</em>

When immersing an object in a solution, the increased volume indicates the volume of that object.

Hence,

  Volume of the object = increased volume of water level

                                      = final volume - initial volume

                                      = 27.8 mL - 21.2 mL

                                      = 6.6 mL

Mass of the object = 22.4 g

Density = Mass / Volume

            = 22.4 g / 6.6 mL

            = 3.39 g/mL

4.

Accepted value is the value that scientists and community accept as true. This is a theoretical value.

But the measured value is the value that you obtain from doing experiments. This is the actual value.

If your measured value is more close to the accepted value, then your measured value is more precise. But, if your measured value is far away from accepted value means that your value is not precise and there may have some errors.

5.

Answers : Percent error is 9.11 %

                 The element which has 7.13 g/cm³ as density is Zinc (Zn).

<em>Explanation;</em>

Percent error can be calculated by using following formula.

% error = ( (measured value - accepted value) / accepted value ) x 100%

            = ( (7.78 g/cm³ - 7.13 g/cm³) / 7.13 g/cm³ ) x 100%

            = 9.11 %

6 0
4 years ago
What is the pH of a solution with a 7.8 x 10-13 M hydronium ion concentration?
Ksivusya [100]

Answer:

hydronium concentration camps and the Base is it going there now and the Base

6 0
3 years ago
What is the ratio of [a–]/[ha] at ph 2.75? the pka of formic acid (methanoic acid, h–cooh) is 3.75?
Varvara68 [4.7K]

The dissociation of formic acid is:

HCOOH \rightleftharpoons HCOO^{-} + H^{+}

The acid dissociation constant of formic acid, k_a is:

k_a = \frac{[HCOO^{-}]  [H^{+}]}{HCOOH}

Rearranging the equation:

\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}

pH = 2.75

pH = -log[H^{+}]

[H^{+}]= 10^{-2.75} = 1.78 \times 10^{-3}

pk_a = 3.75

k_a = 10^{-3.75} = 1.78\times 10^{-4}

Substituting the values in the equation:

\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}

\frac{[HCOO^{-}]}{[HCOOH]} = \frac{1.78\times 10^{-4}}{1.78\times 10^{-3}}

Hence, the ratio is \frac{1}{10}.

4 0
3 years ago
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