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2 years ago

Which compound is an exception to the octet rule?

Chemistry

2 answers:

slavikrds [6]2 years ago

8 0

Answer:

the answer is ClF3 on edge

Explanation:

viva [34]2 years ago

5 0

Answer:

CIF3

Explanation:

I just took the test

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How many grams of copper (II) nitrate would be produced from 0.80 g of copper metal reacting with excess nitric acid?

zaharov [31]

Answer:

$m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2$

Explanation:

Hello!

In this case, since the chemical reaction between copper and nitric acid is:

$2HNO_3+Cu\rightarrow Cu(NO_3)_2+H_2$

By starting with 0.80 g of copper metal (molar mass = 63.54 g/mol) and considering the 1:1 mole ratio between copper and copper (II) nitrate (molar mass = 187.56 g/mol) we can compute that mass via stoichiometry as shown below:

$m_{Cu(NO_3)_2}=0.80gCu*\frac{1molCu}{63.54gCu} *\frac{1molCu(NO_3)_2}{1molCu} *\frac{187.56gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2$

However, the real reaction between copper and nitric acid releases nitrogen oxide, yet it does not modify the calculations since the 1:1 mole ratio is still there:

$4HNO_3+Cu\rightarrow Cu(NO_3)_2+2H_2O+2NO_2$

Best regards!

7 0

2 years ago

If you have ever peeled the label off a glass jar, you may have noticed that the glue does not easily wash off with water. Howev

shepuryov [24]

Answer:

It can be removed by acidic chemicals

Explanation:

8 0

2 years ago

You have 125 g of a certain seasoning and are told that it contains 62.0 g of salt. what is the percentage of salt by mass in th

r-ruslan [8.4K]

62.0g/125g= 0.496 x 100 = 49.6%

4 0

2 years ago

A 1.0 x 102- gram sample is found to be pure alanine, an amino acid found in proteins. How many moles of alanine are in the samp

Romashka-Z-Leto [24]

Answer:

1.123x10⁻⁴ moles of alanine

Explanation:

In order to convert grams of alanine into moles, we need to know its molecular weight:

The formula for alanine is C₃H₇NO₂, meaning its molecular weight would be:

12*3 + 7*1 + 14 + 16*2 = 89 g/mol

Then we divide the sample mass by the molecular weight, to do the conversion:

1.0x10⁻² g ÷ 89 g/mol = 1.123x10⁻⁴ moles

4 0

1 year ago

Acetone has a boiling point of 56.5 celcius. How many grams of the acetone vapor would occupy the 250 mL Erlenmeyer flask at 57

vodomira [7]

Answer:

0.515 g

Explanation:

Acetone (C₃H₆O) has a boiling point of 56.5 °C. How many grams of the acetone vapor would occupy the 250 mL Erlenmeyer flask at 57 °C and 730 mmHg?

Step 1: Given data

Temperature (T): 57°C

Pressure (P): 730 mmHg

Volume (V): 250 mL

Step 2: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15 = 57°C + 273.15 = 330 K

Step 3: Convert "P" to atm

We will use the conversion factor 1 atm = 760 mmHg.

730 mmHg × (1 atm/760 mmHg) = 0.961 atm

Step 4: Convert "V" to L

We will use the conversion factor 1 L = 1,000 mL.

250 mL × (1 L/1,000 mL) = 0.250 L

Step 5: Calculate the moles (n) of acetone

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 0.961 atm × 0.250 L/(0.0821 atm.L/mol.K) × 330 K

n = 8.87 × 10⁻³ mol

Step 6: Calculate the mass corresponding to 8.87 × 10⁻³ moles of acetone

The molar mass of acetone is 58.08 g/mol.

8.87 × 10⁻³ mol × 58.08 g/mol = 0.515 g

8 0

1 year ago