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2 years ago

Why is it useful to group elements based on the way they behave in chemical reactions?

1 answer:

8 0

<em>The elements in a group often look and behave similarly, because they have the same number of electrons in their outermost shell — the face they show to the world. Group 18 elements, on the far right side of the table, for example, have completely full outer shells and rarely participate in chemical reactions.</em>

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You might be interested in

If a gas is 0.8L at 900 mm Hg and 400K, what is the new pressure if the volume decreases to 0.5L and 300K?

mezya [45]

Answer:

The new pressure of the gas is 1080 mmHg

Explanation:

The general equation provides a way to calculate the relationship between changesnp in volume, temperature and pressure of a given mass of gas.

The general gas equation is given as P1V1/T1 = P2V2/T2

Where P1 is initial pressure of gas

V1 is initial volume of gas

T1 is initial temperature of gas

P2 is final pressure of gas

V2 is final volume of gas

T2 is final temperature of gas

From the given question, P1 = 900 mmHg, V1 = 0.8 L, T1 = 400 K, P2 = ?, V2 = 0.5 L, T2 = 300 K

Making P2 subject of the general gas equations:

P2 = P1V1T2/V2T1

P2 = (900 × 0.8 × 300) / (0.5 × 400)

P2 = 1080 mmHg

Therefore, the new pressure of the gas is 1080 mmHg.

5 0

3 years ago

Calcium oxide (CaO), an important ingredient in cement, is produced by decomposing calcium carbonate (CaCO3) at high temperature

alisha [4.7K]

Answer:

0.0018 mol/min is the average rate of carbon dioxide gas production in moles per minute during the 10 minutes.

Explanation:

Pressure of carbon dioxide gas = P = 0.25 atm

Volume of carbon dioxide gas = V = 5.0 L

Moles of carbon dioxide gas = n

Temperature of the carbon dioxide gas,T = 550°C = 550+273 K = 823 K

Using an ideal gas equation :

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.25 atm\times 5.0 L}{0.0821 atm L/mol K\times 823 K}$

n = 0.018 mol

In 10 minutes pressure of carbon dioxide gas reached to 0.25 atm

Average rate of production of moles of carbon dioxide gas per minute in duration of 10 minutes:

$=\frac{0.018 mol}{10 min}=0.0018 mol/min$

0.0018 mol/min is the average rate of carbon dioxide gas production in moles per minute during the 10 minutes.

8 0

3 years ago

How many moles of carbon are in 3.7 moles of c8h11no2

sleet_krkn [62]

Answer:- 29.6 moles of carbon.

Solution:- We have been given with 3.7 moles of $C_8H_11NO_2$ and asked to calculate the moles of C.

Looking at the formula of the compound, there are 8 carbons in it means 1 mol of he compound has 8 moles of C. So, if we multiply the given moles of the compound by 8 then we get the moles of C.

$3.7molC_8H_11NO_2(\frac{8molC}{1molC_8H_11NO_2})$

= 29.6 mol C

Hence. there are 29.6 moles of C in 3.7 moles of $C_8H_11NO_2$ .

8 0

3 years ago

IF anyone didn't do this E D G E N U I T Y thing here you guys go pass it around to people that need help this is what it is cal

SashulF [63]

Answer:

<h2>My My name: CORN CORNELIUS CORNWALL</h2><h2>My Age: 209374329 years old</h2><h2>Fav song : Baby by justin bieber</h2><h2>most legendary thing that i got : club penguin membership</h2>

5 0

2 years ago

The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.

enot [183]

Answer:

$\large \boxed{\text{21.6 L}}$

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 180.16

C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g: 24.5

(a) Moles of C₆H₁₂O₆

$\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}$

(b) Moles of CO₂

$\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}$

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37 °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

$\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}$

7 0

2 years ago