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3 years ago

A cylinder of oxygen gas has a pressure of 6.00 atm at 25.0

Chemistry

1 answer:

Readme [11.4K]3 years ago

4 0

P1/T1=P2/T2 Gal Lussac's Law
25 C= 298K (just add 273)
0 C= 273 k

6.00atm/298=P2/273

P2=5.50 atm

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The abiotic components of an

DaniilM [7]

Answer:

C. nonliving.

Explanation:

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2 years ago

Determine if the bond between each pairs of atoms would be pure covalent, polar covalent, or ionic. n and se

Sveta_85 [38]

Types of Bonds can be predicted by calculating the difference in electronegativity.

If, Electronegativity difference is,

Less than 0.4 then it is Non Polar Covalent Bond

Between 0.4 and 1.7 then it is Polar Covalent Bond

Greater than 1.7 then it is Ionic

For N and Se,

E.N of Nitrogen = 3.04

E.N of Selenium = 2.55

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E.N Difference 0.49 (Weakly Polar Covalent)

As you have not provided remaining pairs, so if you have any of them, follow the method as mentioned above.

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3 years ago

27.4 g of Aluminum nitrite and 169.9 g of ammonium chloride react to form aluminum chloride, nitrogen, and water. How many grams

GarryVolchara [31]

Answer: The mass of excess reagent (ammonium chloride) remained after the reaction is 62.7 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For aluminium nitrite:

Given mass of aluminium nitrite = 27.4 g

Molar mass of aluminium nitrite = 41 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium nitrite}=\frac{27.4g}{41g/mol}=0.668mol$

For ammonium chloride:

Given mass of ammonium chloride = 169.9 g

Molar mass of ammonium chloride = 53.5 g/mol

Putting values in equation 1, we get:

$\text{Moles of ammonium chloride}=\frac{169.9g}{53.5g/mol}=3.176mol$

The chemical equation for the reaction of aluminium nitrite and ammonium chloride follows:

$Al(NO_2)_3+3NH_4Cl\rightarrow AlCl_3+3N_2+6H_2O$

By Stoichiometry of the reaction:

1 mole of aluminium nitrite reacts with 3 moles of ammonium chloride

So, 0.668 moles of aluminium nitrite will react with = $\frac{3}{1}\times 0.668=2.004mol$ of ammonium chloride.

As, given amount of ammonium chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium nitrite is considered as a limiting reagent because it limits the formation of product.

Excess moles of ammonium chloride = (3.176 - 2.004) mol = 1.172 moles

Calculating the mass of ammonium chloride by using equation 1, we get:

Excess moles of ammonium chloride = 1.172 moles

Molar mass of ammonium chloride = 53.5 g/mol

Putting values in equation 1, we get:

$1.172mol=\frac{\text{Mass of ammonium chloride}}{53.5g/mol}\\\\\text{Mass of ammonium chloride}=(1.172mol\times 53.5g/mol)=62.7g$

Hence, the mass of excess reagent (ammonium chloride) remained after the reaction is 62.7 grams

3 0

3 years ago

The electron configurations of two unknown elements X and Y are shown . X: 1s^ 2 2s^ 2 2p^ 6 3s^ 1 Y: 1s^ 2 2s^ 1 Which statemen

PIT_PIT [208]

Answer:

See below.

Explanation:

They both have 1 valency electron so will be metallic and in the same Group (Group 1) of the Periodic Table, so will have similar properties.

8 0

3 years ago

Read 2 more answers

What is the empirical formula of a compound that contains 32.39 percent sodium, 22.53 percent sulfur, and 45.07 percent oxygen?

djverab [1.8K]

To get the empirical formula of this compound, we take a basis of 100 grams which means each percentage is equivalent to 1 gram. Hence there is 32.39 grams sodium, 22. 53 grams sulfur and 45.07 grams oxygen. We convert each mass to their moles by dividing by their respective molar mass. Na: 1.408, S:0.704 and O:2.82. divide each with the lowest: Na: 2: S: 1 and O:4. Hence the formula is Na2SO4.

5 0

3 years ago