The mass in grams of butane at standard room temperature is 53.21 grams.
<h3>How can we determine the mass of an organic substance at room temperature?</h3>
The gram of an organic substance at room temperature can be determined by using the ideal gas equation which can be expressed as:
PV = nRT
- Pressure = 1.00 atm
- Volume = 22.4 L
- Rate = 0.0821 atm*L/mol*K
- Temperature = 25° C = 298 k
1 × 22.4 L = n × (0.0821 atm*L/mol*K× 298 K)
n = 22.4/24.4658 moles
n = 0.91556 moles
Recall that:
- number of moles = mass(in grams)/molar mass
mass of butane = 0.91556 moles × 58.12 g/mole
mass of butane = 53.21 grams
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Answer: atoms do have 3 subatomic particles but the nucleus is positive and the nucleus consists of protons and neutrons. Its positive because neutrons have no charge and protons have a positive charge. There are only electrons on the shells so no neutrons or protons on the shells
Explanation:
Answer:
Bromine mollecules are held together by van der waals forces while a water molecule constitutes both van der waals forces and hydrogen bomnding
Explanation:
This makes the water molecule recquire more heat energy to break the bond thus a higher boiling point while bromine structure requires just litttle heat energy
The molar Concentration of KMnO₄ is 0.000219 M
Concentration is the abundance of a constituent divided by means of the overall extent of an aggregate. numerous styles of mathematical description may be outstanding: mass awareness, molar awareness, variety concentration, and quantity awareness.
y is absorbance
x is the molar concentration of KMnO_4
y = 4.84E + 03x - 2.26E - 01
0.833 = 4.84 * 10⁺⁰³ x - 2.26 * 10⁻¹
1.059 = 4.84 * 10⁺⁰³ x
X = 0.000219 M
Hence, The molar Concentration of KMnO₄ is 0.000219 M
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For the first question, salt is soluble while sand is insoluble or not dissolvable in water. The salt should have vanished or melted, but the sand stayed noticeable or visible, making a dark brown solution probably with some sand particles caught on the walls of the container when the boiling water was put in to the mixture of salt and sand. The solubility of a chemical can be disturbed by temperature, and in the case of salt in water, the hot temperature of the boiling water enhanced the salt's capability to melt in it.
For the second question, the melted or dissolved salt should have easily made its way through the filter paper and into the second container, while the undissolved and muddy sand particles is caught on the filter paper. The size of the pores of the filter paper didn’t change. On the contrary, the size of the salt became smaller because it has been dissolved which is also the reason why it was able to go through the filter paper, while the size of the sand may have doubled or even tripled which made it harder to pass through.
