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3 years ago

A milligram is a millionth of a gram true or false

Chemistry

1 answer:

Musya8 [376]3 years ago

3 0

Answer:

false

Explanation:

One gram is equal to 1000 milligrams. One microgram is equal to a millionth of a gram.

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A stock solution is 28.2 percent ammonia (NH3) by mass, and the solution has a density of 0.8990 grams per milliliter. What volu

Savatey [412]

Answer:

$\boxed{\text{20.2 mL}}$

Explanation:

Assume that the volume of the stock solution is 1 L.

1. Mass of stock solution

$\text{Mass} = \text{1000 mL} \times \dfrac{\text{0.8990 g}}{\text{1 mL}} = \text{899.0 g}$

2.Mass of NH₃

$\text{Mass of NH}_{3} = \text{899.0 g stock} \times \dfrac{\text{28.2 g NH}_{3}}{\text{100 g stock}} = \text{253.5 g NH}_{3}$

3. Moles of NH₃

$\text{Moles of NH}_{3} = \text{253.5 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.89 mol NH}_{3}$

4. Molar concentration of stock solution

$c = \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{14.89 mol}}{\text{1 L}} = \text{14.89 mol/L}$

5. Volume of stock needed for dilution

Now that you know the concentration of the stock solution, you can use the dilution formula .

$c_{1}V_{1} = c_{2}V_{2}$

to calculate the volume of stock solution.

Data:

c₁ = 14.89 mol·L⁻¹; V₁ = ?

c₂ = 0.500 mol·L⁻¹; V₂ = 600 mL

Calculations:

$\begin{array}{rcl}14.89V_{1} & = & 0.500 \times 600\\14.89V_{1} & = & 300\\V_{1} & = & \text{20.2 mL}\\\end{array}\\\text{You will need $\boxed{\textbf{20.2 mL}}$ of the stock solution.}$

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Calculate the oxidation number of “C” in CH3COCH3 please

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Answer:

The oxidation number of C (carbon) is +4

Explanation:

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Valence electrons are important when drawing molecular structures because

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Answer:

Explanation:

Valence electrons are in the outermost shell. They form chemical bonds with other elements, not any other electrons. Hope this helps!

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3 years ago

Some one help me please I'm stuck Show using two conversion factors how you would convert from 0.020 kg into mg.

julia-pushkina [17]

Answer:

$=2.0x10^4mg$

Explanation:

Hello there!

In this case, when performing units conversions involving two proportional factors we need to make sure we first convert to the base unit and then to the target one; thus, since 1 kg = 1000 g and 1 g = 1000 mg, we set up the following expression:

$=0.020kg*\frac{1000g}{1kg} *\frac{1000mg}{1g} \\\\=2.0x10^4mg$

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