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3 years ago

Antoinc Laurent Lavoisicr's contribution to chemistry was. a.the law of conservation of mass.

Chemistry

1 answer:

Ulleksa [173]3 years ago

8 0

The answer for the following question is explained below.

The option for the following answer is "d".

Explanation:

The phlogiston theory of combustion:(Greek word phlogiston means "BURN")

Phlogiston theory states that phlogisticated substances are the substances that contain phlogiston and dephlogisticate when burned.

Dephlogisticating is the process of releasing stored phlogiston,which is absorbed by the air.

Growing plants then absorb this phlogiston,which is why air does not spontaneously combust and also plant matte burns as well as it does.

The prevailing theory was that flammable materials contained a substance called phlogiston that was released during the combustion.

For example:

phlogiston was transferring into the surrounding air.

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What is the net force on this object?

blondinia [14]

The answer is 3N up.

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3 years ago

Calculate the concentration of an aqueous solution of naoh that has a ph of 11.50

Reika [66]

NaOH is a strong base and complete dissociation into Na⁺ and OH⁻ ions.
Therefore [NaOH] = [OH⁻]
To calculate the [OH⁻], we can first find the pOH as NaOH is a basic solution.
pH + pOH = 14
Since pH = 11.50
pOH = 14 - 11.50
pOH = 2.50
We can calculate [OH⁻] by knowing pOH
pOH = -log[OH⁻]
[OH⁻] = antilog(-pOH)
[OH⁻] = 3.2 x 10⁻³ M
therefore [NaOH] = 3.2 x 10⁻³ M

7 0

3 years ago

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Name 5 metals used to make the penny over the years.

Irina18 [472]

These 5 metals are: zinc,copper,tin,bronze,nickel

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3 years ago

The solubility of magnesium phosphate at a given temperature is 0.173 g/L. Calculate the Ksp at this temperature. After you calc

jek_recluse [69]

Answer: $K_{sp}=1.25\times 10^{-14}$

$pK_{sp}=13.90$

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as $K_{sp}$

The equation for the ionization of magnesium phosphate is given as:

$Mg_3(PO_4)_2\rightarrow 3Mg^{2+}+2PO_4^{3-}$

When the solubility of $Mg_3(PO_4)_2$ is S moles/liter, then the solubility of $Mg^{2+}$ will be 3S moles\liter and solubility of $PO_4^{3-}$ will be 2S moles/liter.

Thus S = 0.173 g/L or $\frac{0.173g/L}{262.8g/mol}=0.00065mol/L$

$K_{sp}=(3S)^3\times (2S)^2$

$K_{sp}=108S^5$

$K_{sp}=108\times (0.00065)^5=1.25\times 10^{-14}$

$pK_{sp}=-log(K_{sp})=\log (1.25\times 10^{-14})=13.90$

5 0

3 years ago

How many moles of NaCl are equivalent to 15.6g NaCl

aev [14]

Hey there!:

Molar mass NaCl = 58.44 g/mol

Number of moles:

n = mass of solute / molar mass

n = 15.6 / 58.44

n = 0.2669 moles of NaCl

hope this helps!

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3 years ago

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