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3 years ago

Why did mendeleev switch iodine and tellurium?

1 answer:

6 0

An example would be iodine has a lower relative atomic mass than tellurium, so it should come before tellurium in Mendeleev's table. In order to get iodine in the same group as the other elements with similar properties such as fluorine, chlorine, and bromine, he had to put it after tellurium, which broke his own rules.

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What is the temperature of 0.645 mole of neon in a 2.00 L vessel at 4.68 atm?

storchak [24]

Answer:

—96.03°C

Explanation:

We'll begin by writing out the information provided by the question. This includes:

Number of mole (n) = 0.645 mole

Volume (V) = 2.00 L

Pressure (P) = 4.68 atm

Temperature (T) =?

Recall: that the gas constant = 0.082atm.L/Kmol

With the ideal gas equation PV = nRT, the temperature of the gas can be obtained as follow:

PV = nRT

4.68 x 2 = 0.645 x 0.082 x T

Divide both side 0.645 x 0.082

T = (4.68 x 2) /(0.645 x 0.082)

T = 176.97 K

Now, We can also express the temperature obtained in celsius as shown below:

Temperature (celsius) = temperature (Kelvin) - 273

Temperature (celsius) = 176.97 - 273

Temperature (celsius) = —96.03°C

The temperature of the Neon gas is

—96.03°C

7 0

3 years ago

What is a control group?

Arte-miy333 [17]

Answer:

A control group is someone who takes control over something so a tyrant or a dictator.

Explanation:

If this isnt what your looking for let me know, there wasnt alot of context.

8 0

3 years ago

Read 2 more answers

Hydrazine (N₂H₄), a rocket fuel, reacts with oxygen to form nitrogen gas and water vapor. The reaction is represented with the e

Vinvika [58]

At STP one mol weighs 22.4L

Moles of O_2

4.2/22.4
2.1/11.2
0.19mol

1 mol.O_2 can create 2mol water

moles of water

2(0.19)
0.38mol

Volume of water

0.38(22.4)
8.512L

8 0

1 year ago

Read 2 more answers

N2H4 + N2O4 --> N2 + H2O

ahrayia [7]

Answer:

The limiting reagent is N2O4
14,09g

Explanation:

First, we adjust the reaction.

$2N_{2} H_{4}$ + $N_{2} O_{4}$ ⇄ $6N_{2} + 4H_{2}O$

Second, we assume that the participating moles are equal to the stoichiometric ratios because we do not know the amounts of the reagents.

We can determinate what is the limiting reagent comparing of product amounts which can be formed from each reactant.

Using $N_{2} H_{4}$ to form $H_{2}O$

$molH_{2} O = 1mol N_{2} H_{4} } . \frac{4 mol H_{2} O}{2mol N_{2} H_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}H_{4} }{32,04\frac{g}{mol} N_{2} H_{4} }$

$molH_{2} O = 1, 125 mol$

Using $N_{2} O_{4}$ to form $H_{2} O$

$molH_{2} O = 1mol N_{2} O_{4} } . \frac{4 mol H_{2} O}{1mol N_{2} O_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}O_{4} }{92\frac{g}{mol} N_{2} O_{4} }$

$molH_{2} O = 0,783 mol$

The limiting reagent is N2O4, because can produce only 0, 783 mol of H2O.

This is the minimum measure can be formed of each product.

∴ $MassOfH_{2}O = 0,783mol . 18\frac{g}{mol}$

$MassOfH_{2}O = 14,09g$

5 0

3 years ago

I’ll mark as brainliest

zysi [14]

First, let's count mole of 10 g Calcium Carbonate

mole = Mass / Molecular Mass

Calcium Carbonate = CaCO₃

Molecular Mass = Ar Ca + Ar C + (3 x Ar O)

Molecular Mass = 40 + 12 + (3 x 16)

Molecular Mass = 100

Mole of CaCO₃ = 10 gram / 100

Mole of CaCO₃ = 0,1 mol

then equal the reaction equation first

CaCO₃ + 2 HCl ==> CaCl₂ + CO₂ + H₂O (Equal)

To count the mass of carbon dioxide that produced we must know the mole of CO₂ first

we can count by coefficient comparison

mole CO₂ = $\frac{coefficient \ of \ CO_2}{coefficient \ of \ CaCO_3}$ x mole CaCO₃

mole CO₂ = (1/1) x 0,1 mole

mole CO₂ = 0,1 mole

Mass of CO₂ = mole CO₂ x Molecular Mass of CO₂

Mass of CO₂ = 0,1 mole x (12 + (2 x 16))

Mass of CO₂ = 0,1 mole x 44

Mass of CO₂ = 4,4 g

so, mass of carbon dioxide that's produced by 10 g of calcium carbonate on reaction with chloride acid is 4,4 g.

6 0

3 years ago