How many N2O4 molecules are in 76.3 g N2O4?

A metal X from two oxide A and B .3.oogm of A and B contain 0.72 and 1.16g of oxygen respectively.calculate the maases of metal

devlian [24]

Answer:

Explanation:

Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.

Metal X can form 2 oxides (A and B).

A + B = 3g

The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.

The mass of metal X in the two oxides will be the same because it's the same metal.

Thus, we represent the mass of the metal in the two oxides as 2X.

2X + 0.72 + 1.16 = 3

2X + 1.88 = 3

2X = 3 - 1.88

2X = 1.12

X = 0.56

<u>Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.</u>

Thus, mass of metal (X) in 1g of oxygen in A is

0.56g ⇒ 0.72g

X ⇒ 1

X = 1 × 0.56/0.72

X = 0.78 g

Hence, 0.78g of the metal will combine with 1g of oxygen for A

Also, mass of metal (X) in 1g of oxygen in B is

0.56g ⇒ 1.16g

X ⇒ 1g

X = 1×0.56/1.16

X = 0.48 g

Thus, 0.48g of the metal will combine with 1g of oxygen for B

6 0

2 years ago

What would be the bond order for He 2 2- molecule

LuckyWell [14K]

Answer:

In He2 molecule,

Atomic orbitals available for making Molecular Orbitals are 1s from each Helium. And total number of electrons available are 4.

Molecular Orbitals thus formed are:€1s2€*1s2

It means 2 electrons are in bonding molecular orbitals and 2 are in antibonding molecular orbitals .

Bond Order =Electrons in bonding molecular orbitals - electrons in antibonding molecular orbitals /2

Bond Order =Nb-Na/2

Bond Order =2-2/2=0

Since the bond order is zero so that He2 molecule does not exist.

Explanation:

8 0

2 years ago

Mrs. Stark weighs 70kg. How much Potential Energy does she have if she stands on the roof of an 80m tall building?

8_murik_8 [283]

E = mgh

E = 70*9.81*80

E = 54936 J

5 0

2 years ago

Study the Lewis dot diagram for nitrogen. Explain what is correct and what is incorrect about the diagram. Use details to suppor

Verdich [7]

Answer:

The Lewis dot diagram is supposed to have dots on each side. What's incorrect is that there isn't a dot on the bottom, only the left and right side and the top. What's correct about this is that there are 5 outer valence electrons, and they correctly put 5 dots, even though they're in the wrong place.

Explanation:

4 0

2 years ago

SO

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

2 years ago