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3 years ago

How are a mole and a dozen similar

1 answer:

6 0

A mole is the same idea as a dozen, but the number involved is a lot bigger than 12. In fact, one mole is equal to 6.02 x 1023 items. This number is also called Avogadro's Number, and is sometimes abbreviated NA. Memorize this conversion factor!

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What is the mass of a sample of metal that is heated from 58.8°C to 88.9°C with a

Vadim26 [7]

Answer:

$\boxed {\boxed {\sf 333 \ grams}}$

Explanation:

We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

$Q= mc \Delta T$

The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.

ΔT= final temperature - initial temperature

The sample was heated from 58.8 degrees Celsius to 88.9 degrees Celsius.

ΔT= 88.9 °C - 58.8 °C = 30.1 °C

Now we know three variables:

Q= 4500.0 J
c= 0.4494 J/g°C
ΔT = 30.1 °C

Substitute these values into the formula.

$4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)$

Multiply on the right side of the equation. The units of degrees Celsius cancel.

$4500.0 \ J = m (13.52694 J/g)$

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

$\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}$

The units of Joules cancel.

$\frac {4500.0 \ J }{13.52694 J/g}= m$

$332.6694729 \ g =m$

The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

$333 \ g \approx m$

The mass of the sample of metal is approximately 333 grams.

8 0

2 years ago

When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular

Luda [366]

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_4O_2$

Explanation:

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

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=1.03g$

Mass of $H_2O=0.14g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 1.03 g of carbon dioxide, $\frac{12}{44}\times 1.03=0.28g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.14 g of water, $\frac{2}{18}\times 0.14=0.016g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles$

Step 2: 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 0.00775 moles.

For Carbon = $\frac{0.023}{0.00775}=2.96\approx 3$

For Hydrogen = $\frac{0.016}{0.00775}=2.06\approx 2$

For Oxygen = $\frac{0.00775}{0.00775}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is $C_3H_{2}O_1=C_3H_2O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

$n=\frac{108.10g/mol}{54g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_2O$ and $C_6H_4O_2$

6 0

3 years ago

the student records the concentration of the stock solution of H2O2 to be 0.950 M. The student proceeds to prepare the reaction

lana [24]

Answer:

0.0400M of KI

Explanation:

Molarity is an unit of concentration defined as the ratio between moles of solute and liters of solution.

When you add 10.0 mL of 0.10M KI and 15.0mL, total volume is:

25.0mL = 0.025L of solution

And moles of KI are:

0.0100L × 0.10M = 0.00100 moles of KI

Thus, molarity is:

0.00100 moles / 0.025L = 0.0400M of KI

6 0

3 years ago

A county creates an park to protect a forest.

Hatshy [7]

Great for the ecosystem

3 0

3 years ago

Where do the OH- ions go as additional acid is added?

qwelly [4]

The result when the OH- ions go as additional acid is added is that they remain in solution but change concentration. This is because when you add more acid to hydroxide, it will change its concentration due to the acidity added.

8 0

3 years ago

Read 2 more answers

How are a mole and a dozen similar​

How are a mole and a dozen similar