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grin007 [14]
3 years ago
14

A 4.000 g sample of an unknown metal, M, was completely burned in excess O2 to yield 0.02225 mol of the metal oxide, M2O3. What

is the metal
Chemistry
1 answer:
Solnce55 [7]3 years ago
7 0
4X + 3O₂ = 2X₂O₃

n(X₂O₃)=0.02225 mol
m(X)=4.000 g
x - the molar mass of metal

m(X)/4x=n(X₂O₃)/2

x=m(X)/{2n(X₂O₃)}

x=4.000/{2*0.02225)=89.89 g/mol

X=Y (yttrium)



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Which statement describes vascular plants?
alexandr1967 [171]

Answer:

A vascular plant is any one of a number of plants with specialized vascular tissue. The two types of vascular tissue, xylem and phloem, are responsible for moving water, minerals, and the products of photosynthesis throughout the plant.

Explanation:

7 0
3 years ago
A fish company delivers 22 kg of salmon, 5.5 kg of crab and 3.48 kg of oysters to your seafood restaurant. What is the mass, in
Rainbow [258]

Answer:

Mass of sea food = 30.98 Kg

Mass of sea food in pound = 68.31 lbs

Explanation:

Salmon, crab and oysters all are sea food.

Mass of sea food = Mass of salmon + Mass of crab + mass of oyster

Mass of salmon = 22 kg

Mass of crab = 5.5 kg

Mass of oysters = 3.48 kg

Mass of sea food = Mass of salmon + Mass of crab + mass of oyster

                             = 22 + 5.5 + 3.48

                             = 30.98 Kg

1 Kg = 2.205 lbs

Therefore, 30.98 kg = 30.98 × 2.205

                                 = 68.31 lbs

8 0
3 years ago
The kinetic energy within molecules of an object _ when heat is added? Increases decreases or remains the same
kotegsom [21]

The kinetic energy of an object increases when heat is added.

6 0
3 years ago
What are three reasons why a population might change?
Tom [10]
If people have children, If people die, and if there is a fire or volcanic eruption.

4 0
3 years ago
What is the density of CHCL3 vapor at 1.00atm and 298K?
Advocard [28]

Answer:

4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.

Explanation:

By ideal gas equation:

PV=nRT

Number of moles (n)

can be written as: n=\frac{m}{M}

where, m = given mass

M = molar mass

PV=\frac{m}{M}RT\\\\PM=\frac{m}{V}RT

where,

\frac{m}{V}=d which is known as density of the gas

The relation becomes:

PM=dRT    .....(1)

We are given:

M = molar mass of chloroform= 119.5 g/mol

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the gas = 298K

P = pressure of the gas = 1.00 atm

Putting values in equation 1, we get:

1.00atm\times 119.5g/mol=d\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\\\d=4.88g/L

4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.

7 0
3 years ago
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