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ololo11 [35]
2 years ago
8

Of the concentration units below, only ________ uses kg of solvent in its calculation.

Chemistry
1 answer:
oksano4ka [1.4K]2 years ago
6 0

Answer:

E) molality

Explanation:

Molality -

Molarity of a substance , is the number of moles present in a Kg of solvent .

Hence , the formula for molality is given as follow -

m = n / s

m = molality

s = mass of solvent in Kg ,

n = moles of solute ,

Hence , from the given information of the question,

The concentration unit which have Kg of solvent , is molality.

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The name "penicillin" is used for several closely-related antibiotics. A 1.2177 g sample of one of these compounds is burned, pr
masya89 [10]

<u>Answer:</u> The empirical formula for the given organic compound is C_7H_{11}O_2SN

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

S_vN_wC_xH_yO_z+O_2\rightarrow CO_2+H_2O+SO_2+NO

where, 'v', 'w' 'x', 'y' and 'z' are the subscripts of sulfur, nitrogen, carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=0.2829g

Mass of H_2O=0.1159g

Mass of SO_2=0.4503g

Mass of NO = 0.2109 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

Molar mass of sulfur dioxide = 64 g/mol

Molar mass of nitrogen monoxide = 30 g/mol

  • <u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 2.1654 g of carbon dioxide, \frac{12}{44}\times 2.1654=0.590g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.6965 g of water, \frac{2}{18}\times 0.6965=0.077g of hydrogen will be contained.

  • <u>For calculating the mass of sulfur:</u>

In 64 g of sulfur dioxide, 32 g of sulfur is contained.

So, in 0.4503 g of sulfur dioxide, \frac{32}{64}\times 0.4503=0.225g of sulfur will be contained.

  • <u>For calculating the mass of nitrogen:</u>

In 30 g of nitrogen monoxide, 14 g of nitrogen is contained.

So, in 0.2109 g of nitrogen monoxide, \frac{14}{30}\times 0.2109=0.098g of nitrogen will be contained.

  • Mass of oxygen in the compound = (1.2177) - (0.590 + 0.077 + 0.225 + 0.098) = 0.2277 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.590g}{12g/mole}=0.049moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.077g}{1g/mole}=0.077moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.2277g}{16g/mole}=0.0142moles

Moles of Sulfur = \frac{\text{Given mass of Sulfur}}{\text{Molar mass of sulfur}}=\frac{0.225g}{32g/mole}=0.007moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{0.098g}{14g/mole}=0.007moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.007 moles.

For Carbon = \frac{0.049}{0.007}=7

For Hydrogen  = \frac{0.077}{0.007}=11

For Oxygen  = \frac{0.0142}{0.007}=2.03\approx 2

For Sulfur  = \frac{0.007}{0.007}=1

For Nitrogen  = \frac{0.007}{0.007}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O : S : N = 7 : 11 : 2 : 1 : 1

Hence, the empirical formula for the given compound is C_7H_{11}O_2S_1N_1=C_7H_{11}O_2SN

4 0
2 years ago
Consider the following hypothetical reaction: 2 P + Q → 2 R + S The following mechanism is proposed for this reaction: P + P Q
aleksandr82 [10.1K]

Answer: the answer is option (D). k[P]²[Q]

Explanation:

first of all, let us consider the reaction from the question;

2P + Q → 2R + S

and the reaction mechanism for the above reaction given thus,

P + P ⇄ T     (fast)

Q + T → R + U    (slow)

U → R + S    (fast)

we would be applying the Rate law  to determine the mechanism.

The mechanism above is a three step process where the slowest step seen is the rate determining step. From this, we can see that this slow step involves an intermediate T as reactant and is expressed in terms of a starting substance P.

It is important to understand that laws based on experiment do not allow for intermediate concentration.  

The mechanism steps for the reactions in the question  are given below when we add them by cancelling the intermediates on the opposite side of the equations then we get the overall reaction equation.

adding this steps gives a final overall reaction reaction.

2P + Q ------------˃ 2R + S

Thus the rate equation is given as

Rate (R) = K[P]²[Q]

cheers, i hope this helps

3 0
2 years ago
A 24.00 g sample contains 14.60 g Cl and 9.400 g B. What is the percent composition (by mass) of boron in this sample?
vaieri [72.5K]

Answer:

39.2 %

Explanation:

The following data were obtained from the question:

Mass of sample = 24 g

Mass of Cl = 14.6 g

Mass of B = 9.4 g

Percentage composition of boron =?

The percentage composition (by mass) of boron in the sample can be obtained as illustrated below:

Percentage composition of boron = mass of B /mass of sample × 100

Percentage composition of boron = 9.4/24 × 100

Percentage composition of boron = 39.2 %

Therefore, the percentage composition (by mass) of boron in the sample is 39.2 %

8 0
2 years ago
Write down the dissolution equation for rubidium chromate dissolving in water. (Chromate is a polyatomic ion with the formula Cr
ycow [4]

Answer:

Four moles of the cation

Explanation:

2Rb2CrO4(s)<--------> 4Rb^+(aq) + 2CrO4^2-(aq)

Now looking at the reaction equation, it can be seen that one mole of rubidium chromate contains two moles of rubidium ions and one mole of chromate ions.

The dissolution of two moles of rubidium chromate should then yield four moles of rubidium ions and two moles of chromate ions since the ratio of ions present is 2:1.

This explains the reaction equation written above for the dissolution of two moles of rubidium chromate as shown.

5 0
2 years ago
What is the pOH of a 1 x 10^-8 M solution of HNO3?
Sedaia [141]

Explanation:

to find POH u need PH first

so ph=-log(h3o)

so

ph =  -  log(1 \times  {10}^{ - 8} )  \\  \\ ph = 8 \\ ph + poh =14 \\ poh = 14 - 8 \\ poh = 6

8 0
1 year ago
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