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lesya [120]
3 years ago
9

How do I balance chemical equations ?​

Chemistry
1 answer:
Maru [420]3 years ago
8 0

Answer & explanation:

To balance a chemical equation, we must make sure that there are equal amounts of each element in the equation on either side.

To balance an equation step-by-step we first start by counting the amount of each element on each side. If they aren't equal, we must make them amounts equal on either side.

After you make sure all the elements are balanced, the net charge on both sides must also be equal.

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Calculate the density of an object with a mass of 220 and a volume of 145mL.
LiRa [457]

Answer:

1.52 g/mL

Explanation:

The formula for finding the density of an object is d = m/v, or density = mass divided by volume.

Therefore, we can input the two values and solve for d.

d = 220/145

d = 44/29

d ≈ 1.52 g/mL

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2 years ago
A transverse and longitudinal wave combine to form __________. a sound wave a radio wave a surface wave an electromagentic wave
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A transverse and longitudinal wave combine to form a Water Wave.
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Convert 2.0 micrometer to centimeters and to inches
Dima020 [189]

Answer:

2x10⁻⁴cm and 7.9x10⁻⁵ inches

Explanation:

Micrometer, μm, is an unit of length. Could be related with another units of lenght as, for example:

1.0μm = 1x10⁻⁴cm

1.0μm = 3.93701x10⁻⁵ inches

That means, 2.0μm are:

2.0μm * (1x10⁻⁴cm / 1μm) = 2x10⁻⁴cm

And:

2.0μm * (3.93701x10⁻⁵ inches / 1μm) = 7.9x10⁻⁵ inches

7 0
2 years ago
0.030 moles of a weak acid, HA, was dissolved in 2.0 L of water to form a solution. At equilibrium, the concentration of HA was
nikitadnepr [17]

Answer: The value of k_{a} for the weak acid is 3.07 \times 10^{-4}.

Explanation:

First, we will calculate the molarity of HA as follows.

     [HA] = \frac{\text{no. of moles}}{\text{volume}}

             = \frac{0.03 mol}{2 L}

             = 0.015

At equilibrium,

              HA + H_{2}O \rightleftharpoons H_{3}O^{+} + A^{-}

Initial:   0.015                  0          0

Change:  -x                     +x         +x

Equilibm: 0.015 - x          x           x

It is given that, at equilibrium

          [HA] = 0.015 - x = 0.013

             - x = 0.013 - 0.015

                  = 0.002

Now, expression for k_{a} of this reaction is as follows.

          k_{a} = \frac{[A^{-}][H_{3}O^{+}]}{[HA]}

                      = \frac{x^{2}}{[HA]}

                      = \frac{(0.002)^{2}}{0.013}

                      = 3.07 \times 10^{-4}

Thus, we can conclude that the value of k_{a} for the weak acid is 3.07 \times 10^{-4}.

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