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2 years ago

Help what are the answers FAST

Chemistry

2 answers:

Crank2 years ago

8 0

Since the other person didn’t answer question 1:

1- This is 0.28 x Avagadro’s Constant
= 0.28 x (6.02214086x10^23)
= 1.68619944x10^23

____ [38]2 years ago

3 0

Answer:

2: 0×yaren1.55×10

0×y

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2 years ago

1. Ibuprofen (C13H18O2) is the active ingredient in many nonprescription pain relievers. Each tablet contains 200 mg of ibuprofe

Dmitry_Shevchenko [17]

Answer :

The molar mass of ibuprofen is, 206.29 g/mole.

The number of moles of ibuprofen in a single tablet is, 0.000969 moles

The number of moles of ibuprofen in four doses is, 0.007752 moles

Solution : Given,

Molar mass of carbon = 12.01 g/mole

Molar mass of hydrogen = 1.01 g/mole

Molar mass of oxygen = 15.99 g/mole

1) Now we have to calculate the molar mass of ibuprofen.

Molar mass of ibuprofen, $C_{13}H_{18}O_2$ = $(13\times 12.01)+(18\times 1.01)+(2\times 15.99)=206.29g/mole$

The molar mass of ibuprofen = 206.29 g/mole

2) Now we have to calculate the moles of ibuprofen.

Formula used : $Moles=\frac{Mass}{\text{ Molar mass}}$

Given : Mass of ibuprofen = 200 mg = 0.2 g (1 mg = 1000 g)

$\text{ Moles of ibuprofen}=\frac{\text{ Mass of ibuprofen}}{\text{ Molar mass of ibuprofen}}=\frac{0.2g}{206.29g/mole}=0.000969moles$

The moles of ibuprofen = 0.000969 moles

3) Now we have to calculate the number of moles of ibuprofen for four doses.

Number of tablets in one dose = 2

Total number of tablets in 4 doses = 4 × 2 = 8

Number of moles of ibuprofen in 8 tablets =

$\text{ Number of moles of ibuprofen in 1 tablet}\times \text{ Total number of tablets}=0.000969\times 8=0.007752moles$

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2 years ago

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How many moles of carbon dioxide can be formed by the decomposition of 5.0 moles of aluminium carbonate

Ksenya-84 [330]

Answer:

30 moles of CO₂

Explanation:

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2 years ago

When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to

Reika [66]

Answer:

$0.4167\ \text{M}^{-1}\text{min}^{-1}$

Explanation:

Half-life

${t_{1/2}}A=2\ \text{min}$

${t_{1/2}}B=4\ \text{min}$

Concentration

${[A]_0}_A=1.2\ \text{M}$

${[A]_0}_B=0.6\ \text{M}$

We have the relation

$t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}$

$\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}$

Comparing the exponents we get

$1=n-1\\\Rightarrow n=2$

The order of the reaction is 2.

$t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}$

The rate constant is $0.4167\ \text{M}^{-1}\text{min}^{-1}$

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2 years ago

What element is in group 13, period 4

Ludmilka [50]

Answer:

Gallium

Explanation:

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3 years ago

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