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3 years ago

It refers to the length of the entire path the object travelled

Chemistry

1 answer:

kiruha [24]3 years ago

3 0

Answer:

Path length is the overall distance traveled following the path of where the object travel. ... Displacement is the distance from the starting point of the object to its final point irregardless where it travels.

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Match the name of the greenhouse gas with one of the human activities that produces it.

ss7ja [257]

Answer:

Hope this will help you

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4 years ago

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A student is given a 2.002 g sample of unknown acid and is told that it might be butanoic acid, a monoprotic acid (HC4H7O2, equa

Elina [12.6K]

Answer: The identity of the unknown acid is butanoic acid or ascorbic acid.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$

Molarity of NaOH solution = 0.570 M

Volume of solution = 39.55 mL

Putting values in above equation, we get:

$0.570M=\frac{\text{Moles of NaOH}\times 1000}{39.55}\\\\\text{Moles of NaOH}=\frac{0.570\times 39.55}{1000}=0.0225mol$

The chemical equation for the reaction of NaOH and monoprotic acid follows:

$NaOH+HX\rightarrow NaX+H_2O$

By Stoichiometry of the reaction:

1 mole of NaOH reacts with 1 mole of HX

So, moles of monoprotic acid = 0.0225 moles

The chemical equation for the reaction of NaOH and diprotic acid follows:

$2NaOH+H_2X\rightarrow 2NaX+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of diprotic acid

So, moles of diprotic acid = $\frac{0.0225}{2}=0.01125moles$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For butanoic acid:

Mass of butanoic acid = 2.002 g

Molar mass of butanoic acid = 88 g/mol

Putting values in above equation, we get:

$\text{Moles of butanoic acid}=\frac{2.002g}{88g/mol}=0.02275mol$

For L-tartaric acid:

Mass of L-tartaric acid = 2.002 g

Molar mass of L-tartaric acid = 150 g/mol

Putting values in above equation, we get:

$\text{Moles of L-tartaric acid}=\frac{2.002g}{150g/mol}=0.0133mol$

For ascorbic acid:

Mass of ascorbic acid = 2.002 g

Molar mass of ascorbic acid = 176 g/mol

Putting values in above equation, we get:

$\text{Moles of ascorbic acid}=\frac{2.002g}{176g/mol}=0.01137mol$

As, the number of moles of butanoic acid and ascorbic acid is equal to the number of moles of acid getting neutralized.

Hence, the identity of the unknown acid is butanoic acid or ascorbic acid.

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3 years ago

rank the four gases (air, exhaled air, gas produced from the decomposition of H2O2, gas from decomposition of NaHCO3, in order o

SVEN [57.7K]

Answer: H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)

Initial important note:

Although NaHCO₃ contents oxygen atoms, and you can calculate its compositoin, the resulting gas does not containg pure oxygen gas (O₂). For the comparisson it is not useful to calculate the content of oxygent atoms, but the concentration of O₂ gas. As such, the gas from NaHCO₃ contains 0% of pure O₂, that is why it is ranked last.

1) Air:

Source: internet

Approximate 23%. It is variable, because air is not a pure substance but a mixture of gases, whose compositon is not unique.

2) Exhaled air:

Source: internet.

Approximate 13%. The compositon of the air changes in our lungs, due to the respiration process: we inhale fresh air with around 23% of oxygen, part of this oxygen pass to the cells (lungs - blood - heart - cells) and then it is exhaled with a lower content of air and a greater content of CO₂

3) Air from the decomposition of H₂O₂.

In this case we can do a chemical calculation, since we can state the chemical equation of the reaction:

i) Chemical Equation:

H₂O₂ (g) → H₂ (g) + O₂ (g)

ii) mole ratio of the products 1 mol H₂ : 1 mol O₂

iii) convert moles into mass (grams)

1 mol H₂ × 2 × 1.008 g/mol = 2.016 g

1 mol O₂ × 2 × 15.999 g/mol = 31.998 g

Composition, % = [31.998 g / (2.016 g + 31.998 g) ] × 100 ≈ 94%

4) Air from the decomposition of NaHCO₃:

i) chemical equation:

2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)

ii) mole ratio: take into account only the gases in the products:

1 mol CO₂ (g) : 1 mol H₂O

iii) mass in grams

CO₂: molar mass ia approximately 44.01 g/mol

H₂O: molar mass is approximately 18.02 g/mol

iii) Those gases although have oxygen atoms, do not hae free oxygen gas, which is what we are compariing. That means, that from the decomposition of NaHCO₃ you get 0% oxygen gas.

5) The result is:

H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)

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3 years ago

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How many blizzards are in 13m?

muminat

What do you mean by that? Can you explain further?

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3 years ago

Magnesium (Mg): 1s22s22p63s2 longhand notation noble-gas notation

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