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2 years ago

Compounds containing potassium and oxygen

1 answer:

4 0

Answer: The law of proportion states that elements combine in whole number ratios. The gram readings for K are multiples of each other, both in grams and moles.

Let us compare the ratios:

2.44 grams/1.22 grams = 2

4.89 grams/2.44 grams = 2

Therefore, Potassium always combines with Oxygen in a ratio of 2 is to 1.

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Having accurate Specific date can give scientist what they need for more reliabliltiy.

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3 years ago

Read 2 more answers

For the following formula, C4H9Cl, calculate the IHD and select all the types of unsaturation that might be present in the molec

Airida [17]

Answer:

The IHD = 0

There is no unsaturation in the compound C4H9Cl

Explanation:

IHD = Index for Hydrogen Deficiency .It determines total number of unsaturation or Pi- bond present in the molecule.It is calculated using the formula:

For Molecule:

$C_{c}H_{h}O_{o}N_{n}X_{x}$

Here . C = carbon

H = Hydrogen

O = Oxygen

N = Nitrogen

X = Halogen

The IHD is calculated as :

$\frac{2c-h-x+n+2}{2}$ ...............(1)

For example :

C4H8O2

Here c = 4 , h = 8 , o = 2 .

n = x=0

Put the value in equation (1) and find IHD

$\frac{2(4)-8-0+0+2}{2}$

$\frac{8-8+2}{2}$

$\frac{2}{2}$

IHD = 1

For C4H9Cl

c = 4 , h = 9 and x = 1

IHD for this molecule can be calculated as :

$\frac{2c-h-x+n+2}{2}$

$\frac{2(4)-9-1+2}{2}$

$\frac{8-9-1+2}{2}$

= 0

This means there is no unsaturation in the molecule.

Each bond is sigma bond and properly saturated.

5 0

3 years ago

For the following reaction, 9.30 grams of glucose (C6H12O6) are allowed to react with 13.8 grams of oxygen gas. glucose (C6H12O6

amid [387]

Answer:

13.7 g of CO₂

Limiting reactant: C₆H₁₂O₆

3.81 g of O₂

Explanation:

We convert the mass of the reactants to moles, in order to find out the limiting reactant and the excess reagent

9.30 g / 180 g/mol = 0.052 moles of glucose

13.8 g / 32 g/mol = 0.431 moles of oxygen

The equation is: C₆H₁₂O₆(s) + 6O₂ (g) → 6CO₂ (g) + 6H₂O (l)

Ratio is 1:6. Let's consider this rule of three:

1 mol of glucose reacts with 6 moles of oxygen

Then, 0.052 moles of glucose must react with (0.052 . 6) /1 = 0.312 moles

We have 0.431 moles of oxygen and we only need 0.312 moles. This means that an amount of oxygen still remains after the reaction is complete:

0.431 - 0.312 = 0.119 moles. We convert the moles to mass:

0.119 mol . 32 g / 1mol = 3.81 g

In conclussion, the limiting reactant is the glucose.

6 moles of oxygen react with 1 mol of glucose

0.431 moles of O₂ will react with (0.431 . 1) /6 = 0.072 moles of glucose

We only have 0.052 moles, so it is ok to say, that glucose is the limiting cause we do not have enough glucose.

Let's verify, the maximum amount of carbon dioxide that can be formed:

1 mol of glucose can produce 6 moles of CO₂

Therefore 0.052 moles of glucose will produce (0.052 . 6) /1 = 0.312 moles

We convert the moles to mass → 0.312 mol . 44 g /1 mol = 13.7 g

6 0

3 years ago

If 1.2 moles of ideal gas occupy a volume of 18.2 L at a pressure of 1.8 ATM what is the temperature of gas in degrees Celsius.

Novosadov [1.4K]

Answer: A 59.5 degree celcius

The equation that we will use to solve this problem is :

PV = nRT where:

P is the pressure of gas = 1.8 atm

V is the volume of gas = 18.2 liters

n is the number of moles of gas = 1.2 moles

R is the gas constant = 0.0821

T is the temperature required (calculated in kelvin)

Using these values to substitute in the equation, we find that:

(1.8)(18.2) = (1.2)(0.0821)(T)

T = 332.5 degree kelvin

The last step is to convert the degree kelvin into degree celcius:

T = 332.5 - 273 = 59.5 degree celcius

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3 years ago

Lithium is a metal,chlorine is a non metal,so LiCl would be a ______bond

ivann1987 [24]

Ionic is the answer. This is because lithium has a positive charge, while chlorine has a negative charge, meaning the compound doesn’t necessarily have an overall charge.

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3 years ago