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adoni [48]
3 years ago
14

A sample of urea contains 1.121 g of N, 0.161 g of H, 0.48 g of C, and 0.64 g of O. What is the empirical formula of urea?

Chemistry
1 answer:
dybincka [34]3 years ago
5 0

Answer:

The empirical formula is N2 CH4 O

Explanation:

You might be interested in
Please help!!
Olegator [25]
Molarity =  Moles/Liter

Use the molecular atomic mass of NaCl to convert from grams to moles.
Molecular mass of NaCl is the sum of its atomic masses. Look at the periodic table to find these. Na is 23 g/mol and Cl is 35.5 g/mol ,
so NaCl = 23 + 35.5 = 58.5 g/mol

multiply to cancel out grams
76 g NaCl * (1mol / 58.5 g NaCl) = 1.3 mol NaCl

over 1 Liter is just 1.3 M NaCl
Hope this helps!
8 0
3 years ago
Which molecule below has hydrogen bonding? Which molecule below has hydrogen bonding? H2 CH3CH2OH HCl CH4 all of the above
QveST [7]

Answer:

CH_3CH_2OH

Explanation:

Hydrogen bonding:-

Hydrogen bonding is a special type of the dipole-dipole interaction and it occurs between hydrogen atom that is bonded to highly electronegative atom which is either fluorine, oxygen or nitrogen atom.

Partially positive end of the hydrogen atom is attracted to partially negative end of these atoms which is present in another molecule. It is strong force of attraction between the molecules.

Thus, hydrogen must be linked to electronegative atom which is oxygen, fluorine and nitrogen which is in CH_3CH_2OH and thus, it will shown hydrogen bonding.

3 0
3 years ago
How many millimeters (mm) is the length of a standard table if it is
matrenka [14]

Answer:

1,500 mm

Explanation:

if 1 meter = 1000 mm, 0.5 meters is 500 mm, so 1.50 meters is 1,500 mm

5 0
3 years ago
The orbital radii of four planets in our solar system is shown in the following table. Orbital Radii Planet Orbital Radii (AU) W
nalin [4]

Answer:

w and x

Explanation:

6 0
3 years ago
What is the volume of 33.25g of butane gas at 293 C and 10.934 kPa?
PSYCHO15rus [73]

Answer:

V = 240.79 L

Explanation:

Given data:

Volume of butane = ?

Temperature = 293°C

Pressure = 10.934 Kpa

Mass of butane = 33.25 g

Solution:

Number of moles of butane:

Number of moles = mass/ molar mass

Number of moles = 33.25 g/ 58.12 g/mol

Number of mole s= 0.57 mol

Now we will convert the temperature and pressure units.

293 +273 = 566 K

Pressure = 10.934/101 = 0.11 atm

Volume of butane:

PV = nRT

P= Pressure

V = volume

n = number of moles

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

T = temperature in kelvin

V = nRT/P

V = 0.57 mol × 0.0821 atm.L/ mol.K  ×566 K  / 0.11 atm

V = 26.49 L/0.11

V = 240.79 L

6 0
3 years ago
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