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sukhopar [10]
2 years ago
10

What are the freezing and boiling points of water on the celsius temperature table

Chemistry
2 answers:
KengaRu [80]2 years ago
4 0
There are 100 degrees between the freezing (0°) and boiling points (100°) of water on the Celsius scale and 180 degrees between the similar points (32° and 212°) on the Fahrenheit scale.


Problem: Two scientists are doing an experiment designed to identify the boiling point
Answer: 250°F is the higher temperature by 2°F
Evgesh-ka [11]2 years ago
4 0

Answer:

freezing point is 0°C and boiling point is 100°C

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Find change H for the reaction as it occurred in the calorimeter.
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Delta H = q / mass * delta temperature
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Consider molecular, complete ionic, and net ionic equations.
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Explanation:

(a) The molecular equations shows the equation in which all the species of the reactants and the products are in molecules and the net charge is zero.

The complete ionic equations shows the equation in which all the species of the reactants and the products are in dissociated form and are represented as ions.

The net ionic equations shows the equation in which all the species of the reactants and the products are in dissociated form and do not show the spectator ions which are same in the reactants and the products.

(b) If there is no spectator ions in the reaction, then the complete and the net ionic equations would be identical.

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, glucose is converted into what three products? science
Ivenika [448]

Answer:

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6 0
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Read 2 more answers
Aluminum has a density of 2.7 g/cm3, how much space in cm3 would 81 grams of aluminum occupy? Show steps to answering this equat
Evgesh-ka [11]

Answer:

30 cm³

Explanation:

Step 1: Given data

  • Density of aluminum (ρ): 2.7 g/cm³
  • Mass of aluminum (m): 81 g
  • Volume occupied by aluminum (V): ?

Step 2: Calculate the volume occupied by aluminum

The density of aluminum is equal to its mass divided by its volume.

ρ = m/V

V = m/ρ

V = 81 g / 2.7 g/cm³

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4 0
3 years ago
Calculate the mass of CO2 that can be produced if the reaction of 54.0 g of propane and sufficient oxygen has a 64.0% yield.
Oduvanchick [21]

Answer:

103.9 g

Explanation:

  • C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

First <u>we convert 54.0 g of propane (C₃H₈) into moles</u>, using its <em>molar mass</em>:

  • 54.0 g ÷ 44 g/mol = 1.23 mol C₃H₈

Then we <u>convert 1.23 moles of C₃H₈ into moles of CO₂</u>, using the <em>stoichiometric coefficients</em>:

  • 1.23 mol C₃H₈ * \frac{3molCO_2}{1molC_3H_8} = 3.69 mol CO₂

We <u>convert 3.69 moles of CO₂ into grams</u>, using its <em>molar mass</em>:

  • 3.69 mol CO₂ * 44 g/mol = 162.36 g

And <u>apply the given yield</u>:

  • 162.36 g * 64.0/100 = 103.9 g
7 0
3 years ago
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