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arlik [135]
3 years ago
7

“What types of friction occur between your bike tires and the ground when you ride over cement, through a puddle, and when you a

pply your brakes?” Help..?
Chemistry
1 answer:
Kruka [31]3 years ago
4 0
The right answer is= <span>rolling, fluid, and sliding</span>
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Answer:

Erosion

Explanation:

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2 years ago
-Before we begin each flame test we will
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Answer:Poop

Explanation:poop

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Judging by the electron domain geometry, which type of hybrid orbitals does this molecule have? s sp sp2 sp3
Aleksandr-060686 [28]
The molecule have TETRAHEDRAL HYBRID ORBITAL.
SP3 hybridization involves the mixing of one orbital of S sub level and three orbitals of P sub level of the valence shell. All the orbitals possess equivalent energies and shapes. The SP3 orbital has 25% S character and 75% P character. S and P refers to the s and p sub shells.<span />
4 0
3 years ago
Read 2 more answers
Can you check this for me?
Aliun [14]
Hi, your answer is correct.
3 0
3 years ago
The molar mass is determined by measuring the freezing point depression of an aqueous solution. A freezing point of -5.20°C is r
Dima020 [189]

Answer:

The empirical formula is C2H4O3

The molecular formula is C4H8O6

The molar mass is 152 g/mol

Explanation:

The complete question is: An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57% C and 5.30% H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of -5.20°C is recorded for a solution made by dissolving 10.56 g of the compound in 25.0 g water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

Step 1: Data given

Mass % of Carbon = 31.57 %

Mass % of H = 5.30 %

Freezing point = -5.20 °C

10.56 grams of the compound dissolved in 25.0 grams of water

Kf water = 1.86 °C kg/mol

Step 2: Calculate moles of Carbon

Suppose 31.57% = 31.57 grams

moles C = mass C / Molar mass C

moles C = 31.57 grams / 12.0 g/mol = 2.63 moles

Step 3: Calculate moles of Hydrogen:

Moles H = 5.30 grams / 1.01 g/mol

moles H = 5.25 moles

Step 4: Calculate moles of Oxygen

Moles O = ( 100 - 31.57 - 5.30) / 16 g/mol

Moles O = 3.95 moles

Step 5: We divide by the smallest number of moles

C: 2.63 / 2.63 = 1 → 2

H: 5.25/2.63 = 2 → 4

O: 3.95/ 2.63 = 1.5 → 3

The empirical formula is C2H4O3

The molar mass of the empirical formula = 76 g/mol

Step 6: Calculate moles solute

Freezing point depression = 5.20 °C = m * 1.86

m = 5.20 / 1.86

m = 2.80 molal = 2.80 moles / kg

2.80 molal * 0.025 kg = 0.07 moles

Step 7: Calculate molar mass

Molar mass = mass / moles

Molar mass = 10.56 grams / 0.07 moles

Molar mass = 151 g/mol

Step 8: Calculate molecular formula

151 / 76 ≈  2

We have to multiply the empirical formula by 2

2*(C2H4O3) = C4H8O6

The molecular formula is C4H8O6

The molar mass is 152 g/mol

6 0
3 years ago
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