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3 years ago

Find the mass ratios and atomic ratios of the following compounds.

Chemistry

1 answer:

geniusboy [140]3 years ago

3 0

I am typing this to get more answers for one of my tests good luck dawg

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Which part of the atom is directly involved in chemical changes? A. the electron B. the energy levels C. the nucleus D. the char

7nadin3 [17]

The part of the atom that is involved in chemical changes is A. electron. The electrons that are in the most outer shells are called valence electrons which are easily removed or shared to form bonds. Valence electrons are related to the number of valence electrons

8 0

3 years ago

Determine how many atoms of pure silver will be created when 19.83 x 1023 atoms of copper are used in the following reaction:

Lubov Fominskaja [6]

Answer:

$\boxed{3.966 \times 10^{24}\text{ atoms of Ag}}$

Explanation:

(a) Balanced equation

Cu + 2AgNO₃ ⟶ Cu(NO₃)₂+ 2Ag

(b) Calculation

You want to convert atoms of Cu to atoms of Ag.

The atomic ratio is ratio is 2 atoms Ag:1 atom Cu

$\text{Atoms of Ag} = 19.83 \times 10^{23}\text{atoms Cu} \times \dfrac{\text{2 atoms Ag}}{\text{1 atom Cu}}\\\\= 3.966 \times 10^{24}\text{ atoms of Ag}\\\\\text{The reaction will produce }\boxed{\mathbf{3.966 \times 10^{24}}\textbf{ atoms of Ag}}$

4 0

3 years ago

Read 2 more answers

For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell potential at 25 ∘C if the concentrations are [

sveta [45]

Explanation:

The given data is as follows.

$E^{o}$ = 0.483, $[Co^{3+}]$ = 0.173 M,

$[Co^{2+}]$ = 0.433 M, $[Cl^{-}]$ = 0.306 M,

$P_{Cl_{2}}$ = 9.0 atm

According to the ideal gas equation, PV = nRT

or, P = $\frac{n}{V}RT$

Also, we know that

Density = $\frac{mass}{volume}$

So, P = MRT

and, M = $\frac{P}{RT}$

= $\frac{9.0 atm}{0.0820 L atm/mol K \times 298 K}$

= $\frac{9.0}{24.436}$

= 0.368 mol/L

Now, we will calculate the cell potential as follows.

E = $E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}$

= $0.483 - \frac{0.0591}{2} log \frac{(0.433)^{2}(0.368)}{(0.173)(0.306)^{2}}$

= $0.483 - 0.02955 log \frac{0.0689}{0.0162}$

= $0.483 - 0.02955 \times 0.628$

= 0.483 - 0.0185

= 0.4645 V

Thus, we can conclude that the cell potential of given cell at $25^{o}C$ is 0.4645 V.

4 0

3 years ago

If 100.0g of nitrogen is reacted with 100.0g of hydrogen, what is the excess reactant? What is the limiting reactant? Show your

Drupady [299]

N₂ : limiting reactant

H₂ : excess reactant

<h3>Further explanation</h3>

Given

mass of N₂ = 100 g

mass of H₂ = 100 g

Required

Limiting reactant

Excess reactant

Solution

Reaction

mol N₂(MW=28 g/mol) :

$\tt mol=\dfrac{mass}{MW}=\dfrac{100}{28}=3.571$

mol H₂(MW= 2 g/mol) :

$\tt mol=\dfrac{100}{2}=50$

A method that can be used to find limiting reactants is to divide the number of moles of known substances by their respective coefficients, and small or exhausted reactans become a limiting reactants

From the equation, mol ratio N₂ : H₂ = 1 : 3, so :

$\tt \dfrac{3.571}{1}\div \dfrac{50}{3}=3.571\div 16.6$

N₂ becomes a limiting reactant (smaller ratio) and H₂ is the excess reactant

5 0

2 years ago

a gas has a volume of 750ml at pressure of 2.15atm what will the pressure be if the volume is 1.25ml

Dmitry [639]

Answer:

1290 atm

Explanation:

P1V1=P2V2

P2= (P1V1)/V2

P2=(2.15 atm * 750 mL)/(1.25 mL)

P2=1290 atm

6 0

3 years ago