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Airida [17]
2 years ago
6

Calculate the molarity of a solution made by dissolving 5.00 g of glucose (C6H12O6) in sufficient water to form exactly 100 mL o

f solution.
Chemistry
2 answers:
matrenka [14]2 years ago
6 0
I think it’s 2.7 moldm-3
avanturin [10]2 years ago
4 0

Answer:

0.28 mol {dm}^{ - 3}

Explanation:

given; 5.00g equivalent to 100mL

5.00g = 100mL

convert 100mL to dm³

100mL is 100milliLitre

milli = 10^-3

100mL = 100 × (10^-3)

= 10^2 × 10^-3

=

{10}^{2 - 3}

= 10^-1 L

1l  = 1 {dm}^{3}

therefore, 10^-1 L = 10^-1 dm³

Relative Atomic Mass of;

C=12, H=1, O=16

Molar mass of glucose (C6H12O6)

= 12×6 + 1×12 + 16×6

= 180g/mol

Now convert the mass given to mole using;

mole \:  =  \frac{mass}{molar \: mass}

mole \:  =  \frac{5.00}{180}

mole = 0.028 mol

therefore, 0.028 mol is equivalent to 10^-1 dm³

10^-1 dm³ = 0.028mol

divide both sides by 10^-1 to get 1dm³

1 {dm}^{3}  =   \frac{0.028 \: mol}{ {10}^{ - 1} }

= 0.28 mol

molarity \:  = 0.28mol {dm}^{ - 3}

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What mass of CaSO3 must have been present initially to produce 14.5 L of SO2 gas at a temperature of 12.5°C and a pressure of 1.
german
When the reaction equation is:

CaSO3(s) → CaO(s) + SO2(g)

we can see that the molar ratio between CaSO3 & SO2 is 1:1 so, we need to find first the moles SO2.

to get the moles of SO2 we are going to use the ideal gas equation:

PV = nRT

when P is the pressure =  1.1 atm

and V is the volume = 14.5 L 

n is the moles' number (which we need to calculate)

R ideal gas constant = 0.0821

and T is the temperature in Kelvin = 12.5 + 273 = 285.5 K

so, by substitution:

1.1 * 14.5 L = n * 0.0821 * 285.5

∴ n = 1.1 * 14.5 / (0.0821*285.5)

       = 0.68 moles SO2

∴ moles CaSO3 = 0.68 moles

so we can easily get the mass of CaSO3:

when mass = moles * molar mass

and we know that the molar mass of CaSO3= 40 + 32 + 16 * 3 = 120 g/mol


∴ mass = 0.68 moles* 120 g/mol = 81.6 g
7 0
3 years ago
How many moles of O₂ are needed to react completely with 35.0 mol of FeCl₃? *
densk [106]

Answer:

26.3 moles of O₂ are needed to react completely with 35.0 mol of FeCl₃

Explanation:

To determine the number of moles of O₂ that are needed to react completely with 35.0 mol of FeCl₃, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction),  and rule of three as follows: if 4 moles of FeCl₃ react with 3 moles of O₂, 35 moles of FeCl₃ with how many moles of O₂ will it react?

molesofO_{2} =\frac{35 moles of FeCl_{3}*3 moles of O_{2}  }{4 moles of FeCl_{3}}

moles of O₂= 26.25 ≅ 26.3

<u><em>26.3 moles of O₂ are needed to react completely with 35.0 mol of FeCl₃</em></u>

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Vocabulary crossword puzzle properties of minerals
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Answer: what were

Explanation:

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According to markovnikov's rule of the electrophilic addition to an alkene, the electrophile, usually a proton, is more likely t
stealth61 [152]

According to markovnikov's rule of the electrophilic addition to an alkene, the electrophile, usually a proton, is more likely to add to the less-substituted carbon in a double bond.

With additional substituents present in this configuration, the intermediate carbocation is stabilised by being located on the more-substituted carbon.

The nucleophile will then end up in a double bond on the more-substituted carbon in a reaction that follows Markovnikov's rule.The outcome of some addition reactions is described by Markovnikov's rule or Markownikoff's rule in organic chemistry. Vladimir Markovnikov, a Russian scientist, created the rule in 1870.

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8 0
2 years ago
identify the reagents you would use to convert each of the following compounds into pentanoic acid: (a) 1-pentene (b) 1-bromobut
Morgarella [4.7K]

a)BH3.THF is used to convert 1-pentane to pentanoic acid and b)NaCN is used to convert Bromobutane to pentanoic acid.

a) The conversion of 1-pentane to pentanoic acid using BH3, also known as hydroboration-oxidation, is a two-step reaction involving the reaction of 1-pentane with borane (BH3), followed by oxidation of the resulting 1-pentylborane with hydrogen peroxide or other oxidizing agents.

In the first step, 1-pentane reacts with borane (BH3) to form 1-pentylborane, through a process known as hydroboration. This reaction is catalyzed by a Lewis acid, such as aluminum chloride, and proceeds via a hydride transfer from the borane to the 1-pentane.

In the second step, the 1-pentylborane is oxidized to pentanoic acid using hydrogen peroxide (H₂O₂) or other suitable oxidizing agents. The oxidation is catalyzed by an acid, such as hydrochloric acid (HCl), and proceeds via a proton transfer from the 1-pentylborane to the hydrogen peroxide. The end result is the conversion of 1-pentane to pentanoic acid.

The overall chemical reaction for the conversion of 1-pentane to pentanoic acid using borane (BH₃) and hydrogen peroxide (H₂O₂) is as follows:

1-pentane + BH₃ + H₂O₂ → pentanoic acid + H₂O + BH₂

b)The conversion of 1-Bromo butane to pentanoic acid using sodium cyanide (NaCN) proceeds via a nucleophilic substitution reaction. The reaction mechanism involves the following steps:

1. Attack of the nucleophile, NaCN, on the carbon atom of 1-Bromo butane to form a tetrahedral intermediate.

2. Loss of a proton from the tetrahedral intermediate to form a carbanion.

3. Protonation of the carbanion by water (or another proton source) to form pentanoic acid.

The overall reaction can be represented as follows:

1-Bromo butane + NaCN → Pentanoic Acid + NaBr

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3 0
1 year ago
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