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2 years ago

6. An average Mastiff puppy weighs 2.72 kilograms. How many pounds is an average Mastiff puppy? (1lb = 453.6 g)

Chemistry

1 answer:

Sloan [31]2 years ago

4 0

The number of pounds present in 2.72 kg of Mastiff puppy is 5.996 pounds

<h3>What is weight?</h3>

Weight can be defined as the gravitation pull on a body. or it can also be defined as the product of the mass of a body and its acceleration due to gravity.

<h3>Units of weight</h3>

weight has several units and they include

Newton
pounds
ounces
Tons.

These units can be converted.

From the question,

1 lb = 453.6 g

But,

1 kg = 1000 g

Then,

2.72 kg = 2720 g

Therefore,

2.72 kg of Mastiff = (2720/453.6) pounds of average Mastiff = 5.996 pounds

Hence, The number of pounds present in 2.72 kg of Mastiff puppy is 5.996 pounds.

Learn more about weight here: brainly.com/question/2337612

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A student takes a sample of KOH solution and dilutes it with 100.00 mL of water. The student determines that the diluted solutio

Firlakuza [10]

Answer:

2.2mL

Explanation:

First, let us analyse what was given from the question:

C1 = 2.09M

V1 =?

C2 = 0.046M

V2 =100mL

Using dilution formula (C1V1 = C2V2), we can calculate the volume of the original solution as follows:

C1V1 = C2V2

2.09 x V1 = 0.046 x 100

Divide both side by the coefficient of V1 ie 2.09, we have:

V1 = (0.046 x 100) / 2.09

V1 = 2.2mL

8 0

4 years ago

Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79

BARSIC [14]

Answer : The mole fraction and partial pressure of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of $CH_4$ = 1.79 mole

Moles of $CO_2$ = 1.20 mole

Moles of $He$ = 3.71 mole

Now we have to calculate the mole fraction of $CH_4,CO_2$ and $He$ gases.

$\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267$

and,

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179$

and,

$\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554$

Thus, the mole fraction of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of $CH_4,CO_2$ and $He$ gases.

According to the Raoult's law,

$p_i=X_i\times p_T$

where,

$p_i$ = partial pressure of gas

$p_T$ = total pressure of gas = 5.78 atm

$X_i$ = mole fraction of gas

$p_{CH_4}=X_{CH_4}\times p_T$

$p_{CH_4}=0.267\times 5.78atm=1.54atm$

and,

$p_{CO_2}=X_{CO_2}\times p_T$

$p_{CO_2}=0.179\times 5.78atm=1.03atm$

and,

$p_{He}=X_{He}\times p_T$

$p_{He}=0.554\times 5.78atm=3.20atm$

Thus, the partial pressure of $CH_4,CO_2$ and $He$ gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0

3 years ago

Answer questions a-c about the Bronsted acid-base reaction below using the identifying letters A-D below each structure. The pKa

GuDViN [60]

Answer:

a H2CO3 b HCO3- and c H+ and HCO3-

Explanation:

As the pKa value of phenol is more than that of carbonic acid(H2CO3), the carbonic acid will have high Ka value than that of phenol.

The acid that contain high Ka value act as stong acid.From that point of view H2CO3 is a strong acid than phenol as the Ka value of carbonic acid is greater than that of phenol.

The conjugate base of H2CO3 is bicarbonate ion(HCO3-)

c The species that predorminates at equilibrium are H+ and HCO3-

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3 years ago

The process of blank is when two atoms combine together form a larger atom releasing energy

Bogdan [553]

NUCLEAR FUSION :):):):)

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3 years ago

What volume of concentrated (10.2 M) HCl would be required to prepare 1.11 x 104 mL of 1.5 M HC1? Enter your answer in scientifi

Tomtit [17]

Answer:

The required volume is 1.6 x 10³mL.

Explanation:

When we want to prepare a dilute solution from a concentrated one, we can use the dilution rule to find out the required volume to dilute. This rule states:

C₁ . V₁ = C₂ . V₂

where,

C₁ and V₁ are the concentration and volume of the concentrated solution

C₂ and V₂ are the concentration and volume of the dilute solution

In this case, we want to find out V₁:

C₁ . V₁ = C₂ . V₂

$V_{1} = \frac{C_{2}.V_{2}}{C_{1}} = \frac{1.5M \times1.11.10^{4}mL }{10.2M} =1.6\times10^{3} mL$

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4 years ago