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3 years ago

CO2 is an element. True False

Chemistry

2 answers:

Reil [10]3 years ago

7 0

Answer:False

Explanation:co2 or carbon dioxide is a compound not an element, you can even check this on the periodic table

V125BC [204]3 years ago

3 0

Answer:

false

Explanation:

Carbon dioxide is a compound made up of the elements carbon and oxygen. Its chemical formula is CO2, which specifies that one molecule of CO2 contains one carbon atom and two oxygen atoms.

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Vanadium crystallizes in a body-centered cubic lattice, and the length of the edge of a unit cell is 305 pm. What is the density

Anestetic [448]

Answer:

5.96 g/cm^3

Explanation:

Corner atom = 1/8

Atoms in center = 1

Atoms in face of the cube= 1/2

Molar mass of V = 50.94 g/mol (from period table)

1 mole = 6.02x10^23

In BCC unit cell:

(8 x 1/8)+ 1=2 per 1 unit cell

Mass: 2(50.94g)/6.02x10^23 = 1.69x10^-22 g/unit cell

305pm=(305x10^-12m÷10^-2m) x (1mL÷1cm^3)

= 2.837 x 10^-23 mL

1pm=10^-12m

1cm=10^-2m

1mL=1cm^3

density=mass/volume

density of V = 1.69x10^-22g÷2.837x10^-23mL

=5.957g/mL

=5.96g/cm^3

5 0

2 years ago

My professor gave me two questions to solve using the Van Der Waals Equation. She told us to solve for P and the second one we h

Fed [463]

Answer:

P=atm

$b=\frac{L}{mol}$

Explanation:

The problem give you the Van Der Waals equation:

$(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT$

First we are going to solve for P:

$(P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}$

$P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}$

Then you should know all the units of each term of the equation, that is:

$P=atm$

$n=mol$

$R=\frac{L.atm}{mol.K}$

$a=atm\frac{L^{2}}{mol^{2}}$

$b=\frac{L}{mol}$

$T=K$

$V=L$

where atm=atmosphere, L=litters, K=kelvin

Now, you should replace the units in the equation for each value:

$P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}$

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

$P=\frac{L.atm}{L-L}-atm$

Then operate the fraction subtraction:

P= $P=\frac{L.atm-L.atm}{L}$

$P=\frac{L.atm}{L}$

And finally you can find the answer:

P=atm

Now solving for b:

$(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT$

$(V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}$

$nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}$

$b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}$

Replacing units:

$b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}$

Multiplying and dividing units,(please see the second photo below), we have:

$b=\frac{L-\frac{L.atm}{atm}}{mol}$

$b=\frac{L-L}{mol}$

$b=\frac{L}{mol}$

7 0

3 years ago

Pb(NO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaNO3 (aq)

Blizzard [7]

Pb(NO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaNO3 (aq)
Starting with with 200.0 grams of Pb(NO3)2 and 120.0 grams of NaI:
A. What is the limiting reagent?
B. How many grams of PbI2 is theoretically formed?
C. How many grams of the excess reactant remains?
D. If 48 grams of NaNO3 actually formed in the reaction, what is the percent yield of this reaction?

8 0

3 years ago

A chemical reaction is shown below:

BabaBlast [244]

Answer:

Mass = 8.46 g

Explanation:

Given data:

Mass of water produced = ?

Mass of glucose = 20 g

Mass of oxygen = 15 g

Solution:

Chemical equation:

C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂

Number of moles of glucose:

Number of moles = mass/molar mass

Number of moles = 20 g/ 180.16 g/mol

Number of moles = 0.11 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 15 g/ 32 g/mol

Number of moles = 0.47 mol

now we will compare the moles of water with oxygen and glucose.

C₆H₁₂O₆ : H₂O

1 : 6

0.11 : 6/1×0.11 = 0.66

O₂ : H₂O

6 : 6

0.47 : 0.47

Less number of moles of water are produced by oxygen thus it will limit the yield of water and act as limiting reactant.

Mass of water produced:

Mass = number of moles × molar mass

Mass = 0.47 mol ×18 g/mol

Mass = 8.46 g

8 0

3 years ago

A cube of plastic 1.2x10^-5 km on a side has a mass of 1.1 g. what is its density in g/cm^3

Vesna [10]

Length of 1 side 1.2*10^-5km =1.2*10^-5*10^5 =1.2cm

volume of the cube (1.2)^3=1.728 cm^3 

density= mass/volume= 1.1/1.728=0.636 g/cm^3

7 0

3 years ago

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