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nalin [4]
4 years ago
8

How do scientists do background research on topics they have question about

Chemistry
1 answer:
Aliun [14]4 years ago
3 0
They use information from other things that relate to that topic and then use that to do their research
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It is a covalant bond. Because any compound made up of nonmetals will be covalant. Compounds made up of a non metal and metal willl form a ionic compund. :)
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3 years ago
A 25.0g piece of aluminum sits in a room and cools. It loses 4300.0 J of heat. If the initial temperature of aluminum is 125.3°C
Vesna [10]

Answer:

-65.8C

Explanation:

8 0
3 years ago
Acetylene is hyrodgenated to form ethane. The feed to the reactor contains 1.60 mol H2/mol C2H2. The reaction proceeds to comple
Black_prince [1.1K]

Answer:

Explanation:

C₂H₂     +   2H₂   =     C₂H₆

1 mole     2 mole        1 mole

Feed of reactant is 1.6 mole H₂ / mole C₂H₂

or 1.6 mole of H₂ for 1 mole of C₂H₂

required ratio as per chemical reaction written above

2 mole of  H₂ for 1 mole of C₂H₂

So H₂ is in short supply . Hence it is limiting reagent .

1.6 mole of H₂ will react with half of 1.6 mole or .8 mole of C₂H₂ to form .8 mole of C₂H₆

a )Calculate the stoichiometric reactant ratio =  mole H₂ reacted/mole C₂H₂ reacted

= 1.6 / .8 = 2 .

b )

yield ratio = mole C₂H₆ formed / mole H₂ reacted ) = 0.8 / 1.6 = 1/2 = 0.5 .

4 0
3 years ago
Interesting facts about coal
babunello [35]
Coal will eventually run out as it is an non-renewable energy. these are when they cannot be reproduced to make more,unlike would as u can replant them to create trees
8 0
3 years ago
Read 2 more answers
Titanium has an HCP crystal structure, a c/a ratio of 1.669, an atomic weight of 47.87 g/mol, and a density of 4.51 g/cm3. Compu
IrinaK [193]

Answer : The atomic radius for Ti is, 1.45\times 10^{-8}cm

Explanation :

Atomic weight = 47.87 g/mole

Avogadro's number (N_{A})=6.022\times 10^{23} mol^{-1}

First we have to calculate the volume of HCP crystal structure.

Formula used :  

\rho=\frac{Z\times M}{N_{A}\times V} .............(1)

where,

\rho = density  = 4.51g/cm^3

Z = number of atom in unit cell (for HCP = 6)

M = atomic mass  = 47.87 g/mole

(N_{A}) = Avogadro's number  

V = volume of HCP crystal structure = ?

Now put all the values in above formula (1), we get

4.51g/cm^3=\frac{6\times (47.87g/mol)}{(6.022\times 10^{23}mol^{-1}) \times V}

V=1.06\times 10^{-22}cm^3

Now we have to calculate the atomic radius for Ti.

Formula used :

V=6R^2c\sqrt{3}

Given:

c/a ratio = 1.669 that means,  c = 1.669 a

Now put (c = 1.669 a) and (a = 2R) in this formula, we get:

V=6R^2\times (1.669a)\sqrt{3}

V=6R^2\times (1.669\times 2R)\sqrt{3}

V=(1.669)\times (12\sqrt{3})R^3

Now put all the given values in this formula, we get:

1.06\times 10^{-22}cm^3=(1.669)\times (12\sqrt{3})R^3

R=1.45\times 10^{-8}cm

Therefore, the atomic radius for Ti is, 1.45\times 10^{-8}cm

3 0
3 years ago
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