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3 years ago

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An industrial chemist is studying a sample of an unknown metal. Describe two ways he could change the metal physically and two w

ays he could change the metal chemically to try to identify it.

1 answer:

Illusion [34]3 years ago

5 0

Answer:idk what they cpuld do but we can say this is the right answer :))))

Explanation:

I need PoInTs

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An excess of oxygen reacts with 451.4 g of lead, forming 374.7 g of lead(II) oxide. Calculate the percent yield of the reaction.

Stels [109]

Answer: The percent yield of the reaction is 77.0 %

Explanation:

$2Pb+O_2\rightarrow 2PbO$

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

$\text{Number of moles of lead}=\frac{451.4g}{207.2g/mol}=2.18moles$

$\text{Number of moles of lead oxide}=\frac{374.7g}{223.2g/mol}=1.68moles$

According to stoichiometry:

2 moles of $Pb$ produces = 2 moles of $PbO_2$

2.18 moles of $Pb$ is produced by= $\frac{2}{2}\times 2.18=2.18moles$ of $PbO_2$

Mass of $PbO_2$ = $moles\times {\text {Molar mass}}=2.18\times 223.2g/mol=486.6$

percent yield = $\frac{374.7g}{486.6g}\times 100=77.0\%$

3 0

3 years ago

Any two difference between short sightedness and long sightednes

SSSSS [86.1K]

Explanation:

tala it is also called myopia lekhnu hai

Äni Arko MA it is also called hypermetropia .

3 0

3 years ago

some infrared radiation reaches earth and sum does not with part does reach earth longer wave length or short wave length

Ede4ka [16]

The answer is long wave length because long wave lengths contians less energy, and would not harm living things such as plants and animals. the more engey a wave length has, the less harmful it is.

short wave length: lots of energy, extremely hot. (examples: gamma rays, and UV (ultraviolet) rays.

long wave lengths: not much energy, safe for humans and other life on Earth.

hopefully this helps.

4 0

3 years ago

Swer questions 12 B D 12 Classify each model A-D as either an ele- ment, a compound, or a mixture. Explain your reasoning for ea

MrRissso [65]

Answer:

13-b

Explanation:

because 13-b is close to a and 12

3 0

3 years ago

A 25.0 mL NaOH solution of unknown concentration was titrated with a 0.189 M HCl solution. 19.6 mL HCl was required to reach the

butalik [34]

Answer:

0.172 M

Explanation:

The reaction for the first titration is:

HCl + NaOH → NaCl + H₂O

First we <u>calculate how many HCl moles reacted</u>, using the <em>given concentration and volume</em>:

19.6 mL * 0.189 M = 3.704 mmol HCl

As one HCl mol reacts with one NaOH mol, <em>there are 3.704 NaOH mmoles in 25.0 mL of solution</em>. With that in mind we <u>determine the NaOH solution concentration</u>:

3.704 mmol / 25.0 mL = 0.148 M

As for the second titration:

H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O

We <u>determine how many NaOH moles reacted</u>:

34.9 mL * 0.148 M = 5.165 mmol NaOH

Then we <u>convert NaOH moles into H₃PO₄ moles</u>, using the <em>stoichiometric coefficients</em>:

5.165 mmol NaOH * $\frac{1mmolH_3PO_4}{3mmolNaOH}$ = 1.722 mmol H₃PO₄

Finally we <u>determine the H₃PO₄ solution concentration</u>:

1.722 mmol / 10.0 mL = 0.172 M

5 0

3 years ago