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Wittaler [7]
3 years ago
13

Look at the CHEMICAL EQUATION below...

Chemistry
1 answer:
galina1969 [7]3 years ago
8 0

Answer:

H=12

O=6+12=18

C=6

and the equation is balanced

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How many moles of ethane gas, C2H6(g), will occupy 15.0 L at 30 degrees Celsius and 5.0 atm?
Alenkinab [10]

Answer:

3 mol

Explanation:

Given data:

Volume of ethane = 15.0 L

Temperature = 30°C

Pressure = 5.0 atm

Number of moles of ethane = ?

Solution:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will convert the temperature.

30+273 = 303 K

5.0 atm × 15.0 L = n×0.0821 atm.L/ mol.K   × 303 K

75 atm.L =  n× 24.87 atm.L/ mol

n = 75 atm.L / 24.87 atm.L/ mol

n = 3 mol

8 0
2 years ago
What is stp? provide an answer using at least two different units of pressure and complete sentences?
kogti [31]
STP stands for the Standard Temperature and Pressure. The standard temperature is 273 Kelvin or 0 degrees Celcius or 32 Fahrenheit and the standard pressure is 1 atm pressure or 760 mmHg, or 10,325 Pascals.  STP is most commonly used when performing calculations on gases. 
3 0
3 years ago
Which of the following best describes careers that use chemistry?
telo118 [61]

Answer:

D

Explanation:

7 0
2 years ago
Try the same filtering method with the salt water. What happened? Taste the
enyata [817]

Yeah, what? I don't think Brainly knows that. Maybe you should taste the filtered salt water and find out the answer.

3 0
2 years ago
Read 2 more answers
A student dissolves 3.9g of aniline (C6H5NH2) in 200.mL of a solvent with a density of 1.05 g/mL . The student notices that the
Tresset [83]

Answer:

a. Molarity= M =2.1x10^{-1}M

b. Molality= m=2.0x10^{-1}m

Explanation:

Hello,

In this case, given the information about the aniline, whose molar mass is 93g/mol, one could assume the volume of the solution is just 200 mL (0.200 L) as no volume change is observed when mixing, therefore, the molarity results:

M=\frac{n_{solute}}{V_{solution}} =\frac{3.9g*\frac{1mol}{93g} }{0.2L} =2.1x10^{-1}M

Moreover, the molality:

m=\frac{n_{solute}}{m_{solvent}} =\frac{3.9g*\frac{1mol}{93g} }{0.2L*\frac{1.05kg}{1L} } =2.0x10^{-1}m

Best regards.

7 0
3 years ago
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