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3 years ago

A chemical formula shows the kinds and numbers of _____ in the smallest representative unit of a substance.

Chemistry

1 answer:

Firdavs [7]3 years ago

4 0

A chemical formula shows the kinds and numbers of atoms in the smallest representative unit of a substance.

Explanation:

In chemistry, a formula unit is the empirical formula of "ionic or covalent network solid compound" that is used as an independent entity for "stoichiometric calculations". This formula is a representation of a molecule that uses chemical symbols.

The unit is the lowest whole number ratio of ions represented in an ionic compound. It gives the numbers of atoms representing the "smallest representative" unit of a substance. The number of atoms also tells us about the chemical and physical properties of the compound formed.

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what mass of aluminium hydroxide is needed to decompose in order to produce 65.0 L of water at STP in stoichiometry?

pishuonlain [190]

Aluminium Hydroxide on decomposition produces Al₂O₃ and Water vapors.

2 Al(OH)₃ → Al₂O₃ + 3 H₂O

According to equation at STP,

67.2 L (3 moles) of H₂O is produced by = 78 g of Al(OH)₃
So,
65.0 L of H₂O will be produced by = X g of Al(OH)₃

Solving for X,
X = (65.0 L × 78 g) ÷ 67.2 L

X = 75.44 g of Al(OH)₂
Result:
75.44 g of Al(OH)₂ is needed to decompose in order to produce 65.0 L of water at STP in stoichiometry

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2 years ago

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How many electron pairs are shared by an oxygen molecule?

Lana71 [14]

Two electron pairs is the answer

6 0

2 years ago

What is the hypothesis for Is riding the bus or riding a bike a quicker way to get to school?

OleMash [197]

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5 0

2 years ago

Determine the empirical formulas for compounds with the following percent compositions:

Elis [28]

Answer:

For a: The empirical formula of the compound is $P_2O_5$

For b: The empirical formula of the compound is $KH_2PO_4$

Explanation:

For a:

We are given:

Percentage of P = 43.6 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of P = 43.6 g

Mass of O = 56.4 g

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

Step 1: Converting the given masses into moles.

Moles of Phosphorus = $\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{43.6g}{31g/mole}=1.406moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{56.4g}{16g/mole}=3.525moles$

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 1.406 moles.

For Phosphorus = $\frac{1.406}{1.406}=1$

For Oxygen = $\frac{3.525}{1.406}=2.5$

Converting the moles in whole number ratio by multiplying it by '2', we get:

For Phosphorus = $1\times 2=2$

For Oxygen = $2.5\times 2=5$

Step 3: Taking the mole ratio as their subscripts.

The ratio of P : O = 2 : 5

Hence, the empirical formula for the given compound is $P_2O_5$

For b:

We are given:

Percentage of K = 28.7 %

Percentage of H = 1.5 %

Percentage of P = 22.8 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of K = 28.7 g

Mass of H = 1.5 g

Mass of P = 43.6 g

Mass of O = 56.4 g

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

Step 1: Converting the given masses into moles.

Moles of Potassium = $\frac{\text{Given mass of Potassium}}{\text{Molar mass of Potassium}}=\frac{28.7g}{39g/mole}=0.736moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of hydrogen}}=\frac{1.5g}{1g/mole}=1.5moles$

Moles of Phosphorus = $\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{22.8g}{31g/mole}=0.735moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{47g}{16g/mole}=2.9375moles$

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 0.735 moles.

For Potassium = $\frac{0.736}{0.735}=1$

For Hydrogen = $\frac{1.5}{0.735}=2.04\approx 2$

For Phosphorus = $\frac{0.735}{0.735}=1$

For Oxygen = $\frac{2.9375}{0.735}=3.99\approx 4$

Step 3: Taking the mole ratio as their subscripts.

The ratio of K : H : P : O = 1 : 2 : 1 : 4

Hence, the empirical formula for the given compound is $KH_2PO_4$

3 0

3 years ago

Diana raises a 1000. N piano a distance of 5.00 m using a set of pulleys. She pulls

Alina [70]

Answer:

a) 250 N

b) 50 N

c) 5000 J

d) 4

e) 3.33

Explanation:

Given that weight of piano ( $F_r$ ) = 1000 N, distance moved by pulley ( $d_r$ ) = 5 m and the rope used ( $d_e$ ) =20 m

a) The effort applied ( $F_e$ ) if it was an ideal machine is:

$F_e=\frac{F_rd_r}{d_e} = \frac{1000*5}{20}=250N$

b) If the actual effort = 300 N

The force to overcome friction = actual effort - $F_e$ = 300 - 250 = 50 N

c) Work output = $F_rd_r=1000*5=5000J$

d) ideal mechanical advantage (IMA) = $\frac{d_e}{d_r} =\frac{20}{5} = 4$

e) Input force = 300 N, Therefore:

actual mechanical advantage = $F_r$ / input force = 1000 / 300 = 3.33

5 0

3 years ago