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kaheart [24]
3 years ago
5

Depending upon the tip of a dropper pipet there are approximately 20 drops per milliliter of water. the experimental procedure p

art a and b indicates the addition of 5 -10 dropps of each solutions to the test tubes calculate the volume range in milliliters for the solution
Chemistry
1 answer:
Murrr4er [49]3 years ago
8 0

Answer : The volume range in milliliters for the solution is 0.25 - 0.5 mL

Explanation :

We have been given that there are approximately 20 drops per milliliter of water

This information can be used as a conversion factor as \frac{1mL}{20 drops}

We are using 5-10 drops of the solution.

Let us calculate milliliters in 5 drops.

5 drops \times\frac{1 mL}{20 drops} = 0.25 mL

Similarly, 10 drops would contain

10 drops \times\frac{1 mL}{20 drops} = 0.5 mL solution.

Therefore the volume range in milliliters for the solution is 0.25 - 0.5 mL

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Is MgCO3 organic or inorganic
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Its inorganic as MgCO3 is contains no carbon more hydrogen which is a crutial component of all organic compounds 
4 0
3 years ago
At 7 degrees Celsius the volume of gas is 49 liters. At the same pressure its volume is 74 mL at what temperature
Romashka-Z-Leto [24]
Charle's law gives the relationship between volume and temperature of gas.
It states that at constant pressure, volume of gas is directly proportional to temperature of gas.
V/T = k
where V - volume , T - temperature and k- constant 
\frac{V1}{T1} =  \frac{V2}{T2}
parameters for the first instance are on the left side of the equation and parameters for the second instance are on the right side of the equation 
T1 - 7 °C + 273 = 280 K
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\frac{49 mL}{280K} =  \frac{74 mL}{T}
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4 0
3 years ago
Given the following information: benzoic acid = C6H5COOH hydrocyanic acid = HCN C6H5COOH is a stronger acid than HCN (1) Write t
Maurinko [17]

Answer:

The net ionic equation is

C6H5COOH+ CN-= C6H5COO- + HCN

Explanation:

From the ionic equation

C6H5COOH + Na+ + CN- = C6H5COO- + Na+ + HCN

Only sodium is the spectator ion, so it cancels out, since C6H5COOH and HCN do not ionize completely they are left undissociated

5 0
3 years ago
(a) Given that Ka for acetic acid is 1.8 X 10^-5 and that for hypochlorous acid is 3.0 X 10^-8, which is the stronger acid? (b)
Gala2k [10]

Answer:

HOAc is stronger acid than HClO

ClO⁻ is stronger conjugate base than OAc⁻

Kb(OAc⁻) = 5.5 x 10⁻¹⁰

Kb(ClO⁻) = 3.3 x 10⁻⁷

Explanation:

Assume 0.10M HOAc => H⁺ + OAc⁻  with Ka = 1.8 x 10⁻⁵

=> [H⁺] = √Ka·[Acid] =√(1.8 x 10⁻⁵)(0.10) M = 1.3 x 10⁻³M H⁺

Assume 0.10M HClO => H⁺ + ClO⁻ with Ka = 3 x 10⁻⁸

=> [H⁺] = √(3 x 10⁻⁸)(0.10)M = 5.47 x 10⁻⁵M H⁺

HOAc delivers more H⁺ than HClO and is more acidic.

Kb = Kw/Ka, Kw = 1 x 10⁻¹⁴

Kb(OAc⁻) = 5.5 x 10⁻¹⁰

Kb(ClO⁻) = 3.3 x 10⁻⁷

4 0
3 years ago
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