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3 years ago

Describe how Electronegativity is related to metallic character of an element

1 answer:

8 0

Metallic character decreases as you move across the periodic table from left to right. This occurs as atoms more readily accept electrons to fill a valence shell than lose them to remove the unfilled shell. Metallic character increases as you move down the periodic table.

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Based on the information given below, about how many trips around the Milky Way has the Sun made since the emergence of life on

miss Akunina [59]

It has oribited the milky way more then 20 times

3 0

3 years ago

A gas has a density of 1.57 g/L at 40.0 Â°C and 2.00 atm of pressure. What is the identity of the gas?

Naddika [18.5K]

Answer:

Neon

Explanation:

Step 1: Given and required data

Density of the gas (ρ): 1.57 g/L

Temperature (T): 40.0°C

Pressure (P): 2.00 atm

Ideal gas constant (R): 0.08206 atm.L/mol.K

Step 2: Convert T to Kelvin

We will use the following expression.

K = °C + 273.15 = 40.0 + 273.15 = 313.2 K

Step 3: Calculate the molar mass of the gas (M)

For an ideal gas, we will use the following expression.

ρ = P × M/R × T

M = ρ × R × T/P

M = 1.57 g/L × 0.08206 atm.L/mol.K × 313.2 K/2.00 atm

M = 20.17 g/mol

The gas with a molar mass of 20.17 g/mol is Neon.

6 0

2 years ago

For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi

natka813 [3]

Answer: The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

Explanation:

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

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

For Iron:

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:

$\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol$

For the given chemical reaction:

$2Fe(s)+O_2(g)\rightarrow 2FeO(s)$

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = $\frac{2}{2}\times 0.0771=0.0771mol$ of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:

$0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g$

To calculate the percentage yield of iron (ii) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

$\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%$

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0

2 years ago

Name the following: Simple chemical molecules made of single unit

Stolb23 [73]

Answer:

atom is the answer I think

7 0

3 years ago

15g divided by 15ML plllssss what is it!! Need to know

natulia [17]

Answer:

The answer should be 1000 kg / m3

8 0

3 years ago