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3 years ago

Write the empirical formula of at least four binary iconic compounds that could be formed from the following ions:

Chemistry

1 answer:

Nat2105 [25]3 years ago

7 0

Explanation:

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A gas mixture with a total pressure of 745 mmHg contains each of the following gases at the indicated partial pressures: CO2, 24

Setler79 [48]

<u>Answer:</u>

<u>For Part A:</u> The partial pressure of Helium is 218 mmHg.

<u>For Part B:</u> The mass of helium gas is 0.504 g.

<u>Explanation:</u>

<u>For Part A:</u>

We are given:

$p_{CO_2}=245mmHg\\p_Ar}=119mmHg\\p_{O_2}=163mmHg\\P=745mmHg$

To calculate the partial pressure of helium, we use the formula:

$P=p_{CO_2}+p_{Ar}+p_{O_2}+p_{He}$

Putting values in above equation, we get:

$745=245+119+163+p_{He}\\p_{He}=218mmHg$

Hence, the partial pressure of Helium is 218 mmHg.

<u>For Part B:</u>

To calculate the mass of helium gas, we use the equation given by ideal gas:

PV = nRT

or,

$PV=\frac{m}{M}RT$

where,

P = Pressure of helium gas = 218 mmHg

V = Volume of the helium gas = 10.2 L

m = Mass of helium gas = ? g

M = Molar mass of helium gas = 4 g/mol

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

T = Temperature of helium gas = 283 K

Putting values in above equation, we get:

$218mmHg\times 10.2L=\frac{m}{4g/mol}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 283K\\\\m=0.504g$

Hence, the mass of helium gas is 0.504 g.

6 0

3 years ago

The periodic table

astra-53 [7]

it's Lithium or Li

..............................

5 0

3 years ago

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The student's lab manual says to mix some of his Na2CO3 solution with an aqueous solution of copper(II) sulfate (CuSO4)

lord [1]

Explanation:

When the student mixed the solution sodium carbonate with solution of copper(II) sulfate ; Copper Hydroxocarbonate , sodium sulfate and carbon dioxide gas was obtained as a products.

The balanced chemical reaction

$2Na_2CO_3+2CuSO_4\rightarrow Cu_2(OH)_2CO_3+2Na_2SO_4+CO_2$

Where:

$Cu_2(OH)_2CO_3$ = Copper(II) Hydroxocarbonate

$Na_2CO_3$ = Sodium carbonate

$CuSO_4$ = Copper(II) sulfate

$Na_2SO_4$ = Sodium sulfate

$CO_2$ = Carbon-dioxide

3 0

3 years ago

Do free electrons appear anywhere in the balanced equation for a redox reaction?

Andru [333]

Yes, free electrons appear in balanced redox reaction equations. However, this is only true for half-reactions. This is because redox reactions primarily involve the transfer of electrons, which are better visualized if explicitly shown in the balanced reactions. In reduction reactions, electrons are placed on the left side of the equation. Oxidation reactions show electrons on the right side of the equation.

Explanation:

A half reaction is either the chemical reaction or reduction reaction part of an oxidoreduction reaction. A half reaction is obtained by considering the amendment in chemical reaction states of individual substances concerned within the oxidoreduction reaction. Half-reactions are usually used as a way of leveling oxidoreduction reactions.The half-reaction on the anode, wherever chemical reaction happens, is Zn(s) = Zn2+ (aq) + (2e-).

The metal loses 2 electrons to create Zn2+. The half-reaction on the cathode wherever reduction happens is Cu2+ (aq) + 2e- = Cu(s).

Here, the copper ions gain electrons and become solid copper.

4 0

3 years ago

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Antique paintings will deteriorate more rapidly in a humid environment. Art museums will combat this by placing a desiccant into

stira [4]

233.1g

Explanation:

Given:

Formula units of FeCl₃ = 8.65 x 10²³ formula units

Unknown:

Mass of the FeCl₃ = ?

Solution:

The formula unit is the same as the amount of subatomic particles in an atom.

1 mole of a substance = 6.02 x 10²³ formula units

For 8.65 x 10²³ formula units; the number of moles;

Number of moles = $\frac{8.65 x 10 ^{23} }{6.02 x 10^{23} }$ = 1.44moles

Now to find the mass of FeCl₃;

Mass of FeCl₃ = 1.44 x 162.2 = 233.1g

learn more:

Number of moles brainly.com/question/1841136

#learnwithBrainly

3 0

2 years ago