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1 year ago

What is the oxidation state of co in [co(oh2)6]2?

Chemistry

1 answer:

Ahat [919]1 year ago

7 0

Correct compound is [Co(H2O)6]^+3

The oxidation state of Co is +3

[Co(H2O)6]^+3

Let

Co be x

H2O = 0

x + (0)6 = 3

x = 3

What is oxidation state?

The oxidation state, commonly known as the oxidation number, is the potential charge that an atom would have if all of its bonds to other atoms were fully ionic. It describes the degree of atom oxidation within a chemical compound.The oxidation state may be positive, negative, or zero in theory.

The oxidation number, sometimes referred to as the oxidation state, is the total number of electrons that an atom gains or loses while uniting with another atom to form a chemical bond.

To learn more about oxidation state click the given link

brainly.com/question/8990767

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Consider 80.0-g samples of two different compounds consisting of only carbon and oxygen. One of the compounds consists of 21.8 g

goldenfox [79]

Answer:

Ratio of the mass carbon that combines with 1.00 g of oxygen in compound 2 to the mass of carbon that combines with 1.00 g of oxygen in compound 1 = 2

Explanation:

First, detemine the mass of oxygen in the two samples by difference:

mass of oxygen = mass of sample - mass of carbon

Item Compound 1 Compound 2

Sample 80.0 g 80.0 g

Carbon 21.8 g 34.3 g

Oxygen: 80.0 g - 21.8g = 58.2 g 80.0 g - 34.3 g = 45.7 g

Second, determine the ratios of the masses of carbon that combine with 1.00 g of oxygen:

For each sample, divide the mass of carbon by the mass of oxygen determined above:

Sample Mass of carbon that combines with 1.00 g of oxygen

Compound 1 21.8 g / 58.2 g = 0.375

Compound 2 34.3 g / 45.7 g = 0.751

Third, determine the ratio of the masses of carbon between the two compounds.

Divide the greater number by the smaller number:

Ratio = 0.751 / 0.375 = 2.00 which in whole numbers is 2

6 0

4 years ago

Balance the equation N2O5 + H2 -> NH3 + H2O

Lubov Fominskaja [6]

Answer:

I hope this is it. I'm not really sure.

4 0

3 years ago

Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia. N 2 ( g ) + 3 H 2 ( g )

Kisachek [45]

Answer:

After complete reaction, 0.280 moles of ammonia are produced

Explanation:

Step 1: Data given

Number of moles N2 = 0.140 moles

Number of moles H2 = 0.434 moles

Step 2: The balanced equation

N2(g) + 3H2 (g) ⟶ 2NH3 (g)

Step 3: Calculate the limiting reactant

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

N2 is the limiting reactant. It will completely be consumed (0.140 moles).

H2 is in excess. There will react 3*0.140 = 0.420 moles

There will remain 0.434 - 0.420 = 0.014 moles

Step 4: Calculate moles NH3

For 0.140 moles N2 we'll have 2*0.140 = 0.280 moles NH3

After complete reaction, 0.280 moles of ammonia are produced

5 0

3 years ago

If a gas is initially at a pressure of 9 atm, a volume of 21 liters, and a temperature of 253 K, and then the pressure is raised

alex41 [277]

Answer:

15.04 mL

Explanation:

Using Ideal gas equation for same mole of gas as

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

Given ,

V₁ = 21 L

V₂ = ?

P₁ = 9 atm

P₂ = 15 atm

T₁ = 253 K

T₂ = 302 K

Using above equation as:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac {{9}\times {21}}{253}=\frac {{15}\times {V_2}}{302}$

Solving for V₂ , we get:

V₂ = 15.04 mL

3 0

3 years ago

Burka [1]

Answer:

$\large \boxed{\text{E) 721 K; B) 86.7 g}}$

Explanation:

Question 7.

We can use the Combined Gas Laws to solve this question.

a) Data

p₁ = 1.88 atm; p₂ = 2.50 atm

V₁ = 285 mL; V₂ = 435 mL

T₁ = 355 K; T₂ = ?

b) Calculation

$\begin{array}{rcl}\dfrac{p_{1}V_{1}}{T_{1}}& =&\dfrac{p_{2}V_{2}}{T_{2}}\\\\\dfrac{1.88\times285}{355} &= &\dfrac{2.50\times 435}{T_{2}}\\\\1.509& = &\dfrac{1088}{T_{2}}\\\\1.509T_{2} & = & 1088\\T_{2} & = & \dfrac{1088}{1.509}\\\\ & = & \textbf{721K}\\\end{array}\\\text{The gas must be heated to $\large \boxed{\textbf{721 K}}$}$

Question 8. I

We can use the Ideal Gas Law to solve this question.

pV = nRT

n = m/M

pV = (m/M)RT = mRT/M

a) Data:

p = 4.58 atm

V = 13.0 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 385 K

M = 46.01 g/mol

(b) Calculation

$\begin{array}{rcl}pV & = & \dfrac{mRT}{M}\\\\4.58 \times 13.0 & = & \dfrac{m\times 0.08206\times 385}{46.01}\\\\59.54 & = & 0.6867m\\m & = & \dfrac{59.54}{0.6867 }\\\\ & = & \textbf{86.7 g}\\\end{array}\\\text{The mass of NO$_{2}$ is $\large \boxed{\textbf{86.7 g}}$}$

7 0

3 years ago