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xxTIMURxx [149]
3 years ago
9

1. How many cm are equal to 1.45 m?

Chemistry
1 answer:
geniusboy [140]3 years ago
8 0

Explanation:

(1) 1 m = 100 cm

1.45 m = 100 × 1.45 cm = 145 cm

(2) 1 kg = 1000 g

325 g = 0.325 kg

(3) 1 L = 1000 mL

0.0024 L= 2.4 mL

(4) 1 km = 1000 m

1.55\times 10^4\ m=1.55\times 10^4\times 0.001=15.5\ km

(5) 1 mm = 0.001 m

4.75\times 10^{-2}\ m=4.75\times 10^{-2}\times 1000\ mm\\\\=47.5\ mm

(6) 1 kg = 100000 cg

0.459 kg = 45900 cg

(7) 1 dm = 10⁻⁴ dm

5,995 dm = 5,995 × 10⁻⁴ km

(8) 1 g = 1000 mg

450 g = 450000 mg

(9) 1 dm = 100 m

0.003 dm = 0.3 mm

(10) 1 mL = 0.001 l

4.567 × 10⁴ mL = 45.67 L

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If 252 grams of iron are reacted with 321 grams of chlorine gas, what is the mass of the excess reactant leftover after the reac
mario62 [17]

Answer:

Iron is in excess.

1) The mass of the iron remaining = 83.38 grams

2) Ethane is in excess. There will remain 90.06 grams ethane

Explanation:

Step 1: Data given

Mass of iron = 252 grams

Mass of Cl2 = 321 grams

Molar mass of Fe = 55.845

Molar mass of Cl2 = 70.9 g/mol

Step 2: The balanced equation

2Fe(s)+3Cl2(g)⟶2FeCl3(s)

Step 3: Calculate moles

Moles = mass / molar mass

Moles Fe = 252.0 grams / 55.845 g/mol = 4.512 moles

Moles Cl2 = 321.0 grams / 70.90 g/mol = 4.528 moles

Step 4: Calculate the limiting reactant

For 2 moles Fe we need 3 moles Cl2 to produce 2 moles Fecl3

Cl2 is the limiting reactant. It will completely be consumed (4.528 moles).

Fe is in excess. There will 4.528 * 2/3 = 3.019 moles be consumed

There will remain 4.512 - 3.019 = 1.493 moles of Fe

The mass of the iron remaining = 1.493 * 55.845 g/mol =83.38 grams

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If 152 grams of ethane (C2H6) are reacted with 231 grams of oxygen gas, what is the excess reactant?

Step 1: Data given

Mass of ethane = 152.0 grams

mass of O2 =231.0 grams

Molar mass of ethane = 30.07 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

2C2H6(g) + 7O2(g) ⟶ 4CO2(g) +  6H2O(g)

Step 3: Calculate moles

Moles = mass / molar mass

Moles ethane = 152.0 grams / 30.07 g/mol = 5.055 moles

Moles O2 = 231.0 grams / 32.0 g/mol = 7.22 moles

Step 4: Calculate limiting reactant

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed (7.22 moles).

Ethane is in excess. There will react 7.22 * 2/7 = 2.06 moles

There will remain 5.055 - 2.06 = 2.995 moles ethane

2.995 moles ethane = 2.995 * 30.07 g/mol = 90.06 grams ethane

8 0
4 years ago
Read 2 more answers
Show your work. Write a balanced net ionic equation for the following reaction. H3PO4(aq) + Ca(OH)2(aq) Ca3(PO4)2(aq) + H2O(l)
viktelen [127]
To do a balancing to result to a balanced equation, we do elemental balance from each side of the equation. Hence we do separate balances for H, P, O and Ca. In this case, the final balanced equation after balancing elementally is <span>3 Ca(OH)2(aq) + 2 H3Po4(aq) = 6 H2O(l) + Ca3(Po4)2(s)</span>
6 0
4 years ago
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What is the percentage yield of O2 if 12.3 g of KClO3 (molar mass 123 g) is decomposed to produce 3.2 g of O2 (molar mass 32 g)
My name is Ann [436]

Answer:

The percentage yield of O2 is 66.7%

Explanation:

Reaction for decomposition of potassium chlorate is:

2KClO₃ →  2KCl  +  3O₂

The products are potassium chloride and oxygen.

Let's find out the moles of chlorate.

Mass / Molar mass = Moles

12.3 g / 123 g/mol = 0.1 mol

So ratio is 2:3, 2 moles of chlorate produce 3 mol of oxygen.

Then, 0.1 mol of chlorate may produce (0.1  .3)/ 2 = 0.15 moles

Let's convert the moles of produced oxygen, as to find out the theoretical yield.

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To calculate the percentage yield, the formula is

(Produced Yield / Theoretical yield) . 100 =

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Answer:

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3 years ago
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The answer would be: C) Suspension

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3 years ago
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