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alexandr1967 [171]
2 years ago
8

A Rectangular Prism has a height of 4cm, a width of 5cm, a length of 1.2 cm, and a mass of 85,000 g. What is it's density?

Chemistry
1 answer:
wolverine [178]2 years ago
4 0

Answer:

The answer is D.

Explanation:

Volume (V) = l x w x h

V = 4 x 5 x 1.2

V = 20 x 1.2

V = 24

Density = Mass/Volume  or D = M/V  

D = 85,000/24

D = 3541.66666666

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Number 1 is 4.875 Number 2 is 3.47 repeating

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How many joules it take to melt 35g of ice at 0
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We can't even see half of the question
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Calculate the volume of the gas when the pressure of the gas is 1.60 atm at a temperature of 298 K. There are 160. mol of gas in
LekaFEV [45]

Answer:

2445 L

Explanation:

Given:  

Pressure = 1.60 atm

Temperature = 298 K

Volume = ?

n =  160 mol

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 08206 L.atm/K.mol

Applying the equation as:

1.60 atm × V = 160 mol × 0.08206 L.atm/K.mol × 298 K  

<u>⇒V = 2445.39 L</u>

Answer to four significant digits, Volume = 2445 L

6 0
3 years ago
What is the percent composition of hydrogen in NH4HCO3?
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To find the Percent Composition of an atom, you use this formula:
Mass of element in the compound you're studying on ( in this case it's 5 since there is 5 Hydrogens) over the mass of the compound (which is here 79), Multiplied by 100 since you want a percent. 
So we get: 
\frac{5}{79} * 100
So you get about: 
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6.3

So, the percent composition of Hydrogen in NH4HCO3 is 6.3%

Hope this Helps! :D


7 0
3 years ago
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has a standard free‑energy change of − 3.59 kJ / mol at 25 °C. What are the concentrations of A , B , and C at equilibrium if, a
Mamont248 [21]

Answer: The concentrations of A , B , and C at equilibrium are 0.1583 M, 0.2583 M, and 0.1417 M.

Explanation:

The reaction equation is as follows.

               A + B \rightarrow C

Initial :     0.3   0.4          0

Change:  -x       -x           x

Equilbm: (0.3 - x)  (0.4 - x)  x  

We know that, relation between standard free energy and equilibrium constant is as follows.

      \Delta G = -RT ln K

Putting the given values into the above formula as follows.

      \Delta G = -RT ln K

      -3.59 kJ/mol = -8.314 \times 10^{-3} kJ/mol K ln (\frac{x}{(0.3 - x)(0.4 - x)})

                x = 0.1417

Hence, at equilibrium

  •  [A] = 0.3 - 0.1417

       = 0.1583 M

  •  [B] = 0.4 - 0.1417

       = 0.2583 M

  •  [C] = 0.1417 M
5 0
2 years ago
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