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nikklg [1K]
3 years ago
13

In which process are simple materials chemically combined to form more complex materials?

Chemistry
1 answer:
allsm [11]3 years ago
3 0

Answer:

A) synthesis

I hope this helps!

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The cylinder of aluminum in class was approximately 13mm in circumference and 42mm in length. What would be it's approximate mas
Roman55 [17]

Answer:

Mass, m = 1.51 grams

Explanation:

It is given that,

The circumference of Aluminium cylinder, C = 13 mm = 1.3 cm

Length of the cylinder, h = 4.2 cm

We know that the density of the Aluminium is 2.7 g/cm³

Circumference, C = 2πr

r=\dfrac{C}{2\pi}\\\\r=\dfrac{1.3}{2\pi}\\\\r=0.206\ cm

Density is equal to mass per unit volume.

d=\dfrac{m}{V}

m is mass of the cylinder

V is the volume of the cylinder

V=\pi r^2h\\\\V=\dfrac{22}{7}\times0.206^2\times 4.2\\\\V=0.5601\ cm^3

So,

m=d\times V\\\\m=2.7\times 0.5601\\\\m=1.51\ g

So, the mass of the cylinder is 1.51 grams.

5 0
4 years ago
in example 5.11 of the text the molar volume of n2 at STP is given as 22.42 L/mol how is this number calculatd how does the mola
Valentin [98]

Answer:

V = 22.42 L/mol

N₂ and H₂ Same molar Volume at STP

Explanation:

Data Given:

molar volume of N₂ at STP = 22.42 L/mol

Calculation of molar volume of N₂ at STP  = ?

Comparison of molar volume of H₂ and N₂ = ?

Solution:

Molar Volume of Gas:

The volume occupied by 1 mole of any gas at standard temperature and pressure and it is always equal to 22.42 L/ mol

Molar volume can be calculated by using ideal gas formula  

                               PV = nRT

Rearrange the equation for Volume

                            V = nRT / P . . . . . . . . . (1)

where

P = pressure

V = Volume

T= Temperature

n = Number of moles

R = ideal gas constant

Standard values

P = 1 atm

T = 273 K

n = 1 mole

R = 0.08206 L.atm / mol. K

Now put the value in formula (1) to calculate volume for 1 mole of N₂

                   V = 1 x 273 K x 0.08206 L.atm / mol. K / 1 atm

                   V = 22.42 L/mol

Now if we look for the above calculation it will be the same for H₂ or any gas. so if we compare the molar volume of 1 mole N₂ and H₂ it will be the same at STP.

6 0
3 years ago
If 3.5 grams of NaN3 decomposed, how many grams of N2 would be produced?
Wewaii [24]

Answer:

5.25 moles.

Explanation:

The decomposition reaction of NaN₃ is as follows :

2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)

We need to find how many grams of N₂ produced in the process.

From the above balanced chemical reaction, we conclude that the ratio of moles of sodium azide and nitrogen gas are 2 : 3.

2 moles of sodium azide decomposes to give 3 moles of nitrogen gas. So,

3.5 moles of sodium azide decomposes to give \dfrac{3}{2}\times 3.5=5.25 moles of nitrogen gas.

Hence, the number of moles produced is 5.25 moles.

6 0
3 years ago
Hydrogen gas (H2) reacts with oxygen gas (O2) and produces water. If one mol of hydrogen reacts with one mole of oxygen, what is
Amiraneli [1.4K]
HI 

So, the formula for water is H2O

When you have the same amount of the reactants , hydrogen will be the limiting reactant. 
Limiting reactant is the thing that runs out first.
3 0
3 years ago
Please help me would really appreciate it thank you :)
tatyana61 [14]

Answer:

e. 8.04*10^4

Explanation:

80.4 g converted to mg is 80,400. 8.04*10^4= 80,400

4 0
3 years ago
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