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3 years ago

What is the valence charge of Neon

Chemistry

2 answers:

8_murik_8 [283]3 years ago

5 0

Are electrons located at the outermost shell of an atom

Alexeev081 [22]3 years ago

3 0

Neon has 0 electrons in its valence shell. Its called a rare gas. This means that it doesnt accept any electrons. I guess youcould say its valence charge is 0.

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A gas occupies 18.5L at stp. what volume will it occupy at 735 torr and 57°c?

olga nikolaevna [1]

The volume that will be occupied at 735 torr and 57 c is 23.12 L

calculation

At STP temperature=273 k and pressure=760 torr

by use of combined gas formula

that is P1V1/T1= P2V2/T2

where; P1 =760 torr

T1= 273 K

V1= 18.5 L

P2= 735 torr

T2= 57+273= 330 K

V2=?

by making V2 the formula of subject

V2= T2P1V1/P2T1

V2= [(18.5L x 330 k x 760 torr)/(735 torr x 273 k)]= 23.12 L

5 0

3 years ago

A prairie dog eats grass. A coyote eats the prairie dog. This is an example

Oduvanchick [21]

Answer:

This is an example of a food chain

Explanation:

Think of it as a chain reaction. The grass feeds and nourishes the prairie dog. Upon eating the prairie dog, the coyote gets the nutrients from both the grass the prairie dog ate and from the prairie dog itself.

5 0

3 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

3 years ago

Enter your answer in the provided box.

Lina20 [59]

Answer:

5.06atm

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (Litres)

V2 = final volume (Litres)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question;

P1 = 1.34 atm

P2 = ?

V1 = 5.48 L

V2 = 1.32 L

T1 = 61 °C = 61 + 273 = 334K

T2 = 31 °C = 31 + 273 = 304K

Using P1V1/T1 = P2V2/T2

1.34 × 5.48/334 = P2 × 1.32/304

7.34/334 = 1.32P2/304

Cross multiply

334 × 1.32P2 = 304 × 7.34

440.88P2 = 2231.36

P2 = 2231.36/440.88

P2 = 5.06

The final pressure is 5.06atm

6 0

3 years ago

8. __H2 + __O2-> __H2O

Rus_ich [418]

Anser:

Explanation:

hope this helps

8 0

2 years ago