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3 years ago

PLEASE HELP!! 20 POINTS!

Chemistry

1 answer:

Vinvika [58]3 years ago

3 0

Answer:

Energy transformation, also known as energy conversion, is the process of changing energy from one type of energy to another. In physics, energy is a quantity that provides the capacity to perform work (e.g. lifting an object) or provides heat. In addition to being convertible, according to the law of conservation of energy, energy is transferable to a different location or object, but it cannot be created or destroyed.

The energy in many of its forms may be used in natural processes, or to provide some service to society such as heating, refrigeration, lighting or performing mechanical work to operate machines. For example, to heat a home, the furnace burns fuel, whose chemical potential energy is converted into thermal energy, which is then transferred to the home's air to raise its temperature.

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A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio

Pavlova-9 [17]

Answer:

$C_7H_6O_2$

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

$n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC$

And the moles of hydrogen due to the produced 1.50 grams of water:

$n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.166molH$

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

$m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO$

Thus, the subscripts in the empirical formula are:

$C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2$

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

$n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol$

Which are equal to the moles of the acid:

$n_{acid}=0.0279mol$

And the molar mass:

$MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol$

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

$C_7H_6O_2$

Best regards!

8 0

3 years ago

216 grams of carbon 36 grams of hydrogen and 288 grams of oxygen whats the empirical formula

schepotkina [342]

Divide each wight by the relative atomic mass

C = 216 / 12 = 18
H = 36 / 1 = 36
O = 288/16 = 18

Ratio of C:h:O = 1:2:1

Empirical formula is CH2O Could be formaldehyde HCHO.

4 0

3 years ago

An atomega has a mass of 53 an an atomic number of 26. How many neutrons does it have

Ghella [55]

26, protons and nuetrons will always be the same

4 0

3 years ago

Calculate the percent error if the experimental value for the density of zinc is 9.95g/cm3, but the accepted value is 7.13g/cm3

Pani-rosa [81]

In order to calculate the experimental percent error, we follow these steps:
1- Subtract one value from the other (order does not matter as we take absolute)
2- Divide the obtained number by the accepted or true value.
3- Multiply the fraction you got from step 2 by 100 to get the percentage of error.

Now, we will apply these steps on our problem:
1- Subtract one value from the other:
9.95 - 7.13 = 2.82
2- Divide by accepted value:
2.82 / 7.13 = 0.3955
3- Multiply by 100 to get the error percentage:
error percentage = 0.3955 x 100 = 39.55%

7 0

3 years ago

What is the number of moles of kf in a 29-gram sample of the compound.

Irina18 [472]

Given

Weight of pottasium fluoride (KF) = 29 gms

To find

Number of moles of pottasium fluoride = ?

Solution

we know that molar mass of KF = 58.0967 g

Now,

$number \: of \: moles = \ \\ \frac{given \: weight \: of \: subtance}{molar \: mass}$

Number of moles = 29/58 = 0.5 moles

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6 0

2 years ago