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2 years ago

I’m confused how to do this

Chemistry

1 answer:

Stels [109]2 years ago

7 0

Answer:

B. P (Phosphorus)

Explanation:

The element is be phosphorus.

This is true because phosphorus has a larger atomic radius than carbon. If you move than the group, elements gain additional electron shell. That additional electron shell keeps electrons far from the nucleus of the atom. This actually increases the atomic radius. Also, phosphorus is more electronegative than aluminum because phosphorus itself is a non-metal and non-metals are generally electronegative while aluminum is a metal and electropositive.

Phosphorus has a lower ionization energy than argon because the ionization energy across the period (i.e from left to right) increases and across period 3, you will find phosphorus first before argon in the periodic table.

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Calculate how many grams of the first reactant are necessary to completely react with 17.3 g of the second reactant. the reactio

soldier1979 [14.2K]

Taking into account the reaction stoichiometry, 16.611 grams of Na₂CO₃ are necessary to completely react with 17.3 g of CuCl₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Na₂CO₃ + CuCl₂ → CuCO₃ + 2 NaCl

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Na₂CO₃: 1 mole
CuCl₂: 1 mole
CuCO₃: 1 mole
NaCl: 2 moles

The molar mass of the compounds is:

Na₂CO₃: 129 g/mole
CuCl₂: 134.45 g/mole
CuCO₃: 123.55 g/mole
NaCl: 58.45 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Na₂CO₃: 1 mole ×129 g/mole= 129 grams
CuCl₂: 1 mole ×134.45 g/mole= 134.45 grams
CuCO₃: 1 mole ×123.55 g/mole= 123.55 grams
NaCl: 2 mole ×58.45 g/mole=116.9 grams

<h3>Mass of CuCl₂ required</h3>

The following rule of three can be applied: If by reaction stoichiometry 134.35 grams of CuCl₂ react with 129 grams of Na₂CO₃, 17.3 grams of CuCl₂ react with how much mass of Na₂CO₃?

mass of Na₂CO₃= (17.3 grams of CuCl₂× 129 grams of Na₂CO₃)÷ 134.35 grams of CuCl₂

<u><em>mass of Na₂CO₃= 16.611 grams</em></u>

Finally, 16.611 grams of Na₂CO₃ is required.

Learn more about the reaction stoichiometry:

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6 0

8 months ago

Identify the neutral element represented by this excited-state electron configuration, then write the ground-state electron conf

sineoko [7]

Answer:

1s² 2s² 2p³

Nitrogen

Explanation:

Excited state configuration: 1s² 2s² 2p² 3s¹

Unknown:

Ground state configuration = ?

Element symbol = ?

Solution:

Let us start by understanding what a ground state configuration entails:

A ground state configuration shows the lowest allowed energy levels of an atom. The excited state denotes when electrons have moved to higher energy levels away from their ground state.

The superscript in the configuration depicts the number of electrons in each of the sublevels.

We can use this number to identify the atom we are dealing with:

Total number of electrons = 2 + 2 + 2 + 1 = 7 electrons

The element with 7 electrons on the periodic table is Nitrogen.

Now, the ground state configuration:

In writing the electronic configuration of an atom, certain rules must be complied with.

Aufbau's principle states that the sublevels with lower energies are filled up before those with higher energies. The order of filling is:

1s 2s 2p 3s 3p 4s etc

To fill the s-orbital = 2 electrons

p-orbital = 6 electrons

The outermost shell electrons are usually the ones excited.

Now, the 3s¹ shell is an excited one taking 1 electron from the second energy level;

The ground state configuration is 1s² 2s² 2p³

5 0

2 years ago

Elements in which the d-sublevel is being filled have the properties of

Nadya [2.5K]

Answer:

Elements in which the d-sublevel is being filled have the properties of metals

4 0

2 years ago

Read 2 more answers

A. 1720 kJ<br> B. 125.6 kJ<br> C. 3440 kJ<br> D. 4730 kJ

Feliz [49]

Answer:

Q = 3440Kj

Explanation:

Given data:

Mass of gold = 2kg

Latent heat of vaporization = 1720 Kj/Kg

Energy required to vaporize 2kg gold = ?

Solution:

Equation

Q= mLvap

It is given that heat required to vaporize the one kilogram gold is 1720 Kj thus, for 2 kg

by putting values,

Q= 2kg × 1720 Kj/Kg

Q = 3440Kj

7 0

2 years ago

Using the following equation for the combustion of octane calculate the heat associated with the formation of 100.0 g of carbon

natali 33 [55]

Answer:

The right solution is "-602.69 KJ heat".

Explanation:

According to the question,

The 100.0 g of carbon dioxide:

= $\frac{100.0 \ g}{114.33\ g/mole}$

= $0.8747 \ moles$

We know that 16 moles of $CO_2$ formation associates with -11018 kJ of heat, then

0.8747 moles $CO_2$ formation associates with,

= $-\frac{0.8747}{16}\times 11018 \ KJ \ of \ heat$

= $-0.0547\times 11018$

= $-602.69 \ KJ \ heat$

8 0

2 years ago