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Art [367]
3 years ago
8

I NEED HELP PLEASE! ITS DUE IN 30 MINUTES

Chemistry
1 answer:
notka56 [123]3 years ago
7 0

Answer:

The  correct answer is A

Explanation:

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A pure compound is found to be 40.0% carbon by mass, 6.73% hydrogen by mass, and 53.3% oxygen by mass. determine the empirical f
ratelena [41]
Since there is no weight, I would assume that this is a 100g of pure compound.
Okay so I would be changing the percentage to gram to solve for the mole.
So
40.0g C (1 mol C/12.01 g C) = 3.33 mol C
6.73g H (1 mol H/1.01 g H ) = 6.66 mol H
53.3g O (1 mol O/16.00 g O) = 3.33 mol O

With that, two of our moles is 3.33, so we consider that are our 1, as it is also the lowest. Therefore the empirical formula is CH2O
3 0
3 years ago
Solve the ideal gas law equation for pressure.
posledela

Answer:

p=\frac{nRT}{V}

Explanation:

The ideal gas law equation is an equation that relates some of the quantities that describe a gas: pressure, volume and temperature.

The equation is:

pV=nRT

where

p is the pressure of the gas

V is the volume of the gas

n is the number of moles of the gas

R is the gas constant

T is the absolute temperature of the gas (must be expressed in Kelvin)

Here we want to solve the equation isolating p, the pressure of the gas.

We can do that simply by dividing both terms by the volume, V. We find:

p=\frac{nRT}{V}

So, we see that:

- The pressure is directly proportional to the temperature of the gas

- The pressure is inversely proportional to the volume of the gas

6 0
3 years ago
Where is energy stored in living sytems
EastWind [94]

Answer:

energy is stored in the chemical bonds in molecules

Explanation:

6 0
3 years ago
If the pH of a 1.00-in. rainfall over 1800 miles2 is 3.70, how many kilograms of sulfuric acid, H2SO4, are present, assuming tha
NARA [144]
There are 2.32 x 10^6 kg sulfuric acid in the rainfall. 

Solution: 
We can find the volume of the solution by the product of 1.00 in and 1800 miles2: 
     1800 miles2 * 2.59e+6 sq m / 1 sq mi = 4.662 x 10^9 sq m 
     1.00 in * 1 m / 39.3701 in = 0.0254 m  
     Volume = 4.662 x 10^9 m^2 * 0.0254 m
                  = 1.184 x 10^8 m^3 * 1000 L / 1 m3
                  = 1.184 x 10^11 Liters 

We get the molarity of H2SO4 from the concentration of [H+] given by pH = 3.70: 
     [H+] = 10^-pH = 10^-3.7 = 0.000200 M 
     [H2SO4] = 0.000100 M  
 
By multiplying the molarity of sulfuric acid by the volume of the solution, we can get the number of moles of sulfuric acid: 
     1.184 x 10^11 L * 0.000100 mol/L H2SO4 = 2.36 x 10^7 moles H2SO4 

We can now calculate for the mass of sulfuric acid in the rainfall: 
     mass of H2SO4 = 2.36 x 10^7 moles * 98.079 g/mol
                               = 2.32 x 10^9 g * 1 kg / 1000 g
                               = 2.32 x 10^6 kg H2SO4
3 0
3 years ago
Read 2 more answers
For the reaction o(g) + o2(g) → o3(g) δh o = −107.2 kj/mol given that the bond enthalpy in o2(g) is 498.7 kj/mol, calculate the
Aneli [31]
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
8 0
3 years ago
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