Answer:
<em>a)</em> <em>1.392 x 10^6 g/cm^3</em>
<em>b) 8.69 x 10^7 lb/ft^3</em>
<em></em>
Explanation:
mass of the star m = 2.0 x 10^36 kg
radius of the star (assumed to be spherical) r = 7.0 x 10^5 km = 7.0 x 10^8 m
The density of substance ρ = mass/volume
The volume of the star = volume of a sphere = 
==> V = 
= 1.437 x 10^27 m^3
density of the star ρ = (2.0 x 10^36)/(1.437 x 10^27) = 1.392 x 10^9 kg/m^3
in g/cm^3 = (1.392 x 10^9)/1000 = <em>1.392 x 10^6 g/cm^3</em>
in lb/ft^3 = (1.392 x 10^9)/16.018 = <em>8.69 x 10^7 lb/ft^3</em>
Answer:
<h3>The answer is 30 cm³</h3>
Explanation:
The volume of a substance when given the density and mass can be found by using the formula
From the question
mass = 180 g
density = 6 g/cm³
We have
We have the final answer as
<h3>30 cm³</h3>
Hope this helps you
Answer:
I
Explanation:
The complete question can be seen in the image attached.
We need to understand what is actually going on here. In the first step that yields product A, the sodamide in liquid ammonia attacks the alkyne and abstracts the acidic hydrogen of the alkyne. The second step is a nucleophilic attack of the C6H5C≡C^- on the alkyl halide to yield product B (C6H5C≡C-CH3CH2).
Partial reduction of B using the Lindlar catalyst leads to syn addition of hydrogen to yield structure I as the product C.
Answer:
1. matches with elements.
2. matches with compounds.
3. matches with atoms
4. matches with weight
5. matches with gas
6. matches with carbon dioxide
7. matches with Mendeleev (there's an element named after him)
8. matches with IUPAC
Hope that helped :)
Answer:
hey there
Explanation:
CN- (aq) + H+ (aq) → HCN(I)
The reactants are aqueous solutions:
NaCN(aq) and HBr(aq)
When you mix these compounds you make pure HCN (I)
The molecular equation is:
NaCN(aq) + HBr(aq) → NaBr(aq) + HCN(I)
When you dissociate the reactants, you have: Nat(aq) +CN¯(aq) + H*(aq) + Br−(aq) → Nat(aq) + ->
Br (aq) + HCN(I)
Sodium bromide, it is a salt, that can also be
dissociated in the solution
To make, the net ionic equation you remove the repeated ions
CN- (aq) + H+ (aq) → HCN(I)
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