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Virty [35]
2 years ago
11

What is density? How do you find it?

Chemistry
1 answer:
Assoli18 [71]2 years ago
5 0

Density is the measure of a mass per unit volume of a material substance

density= Mass/volume

The unit of density is g/cm^3

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Dimas [21]

Answer:

2 is the valancy of alkine

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3 years ago
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All the following are oxidation–reduction reactions except:________
Blizzard [7]

Answer:

e. Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g)

Explanation:

All the following are oxidation–reduction reactions except:________

a. H₂(g) + F₂(g) → 2HF(g).  Redox. H is oxidized and F is reduced.

b. Ca(s) + H₂(g) → CaH₂(s).  Redox. Ca is oxidized and H is reduced.

c. 2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g).  Redox. K is oxidized and H is reduced.

d. 6Li(s) + N₂(g) → 2Li₃N(s).  Redox. Li is oxidized and N is reduced.

e. Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g). Not redox. All the elements have the same oxidation number

5 0
3 years ago
Which metal pan should the chef choose because it undergoes the largest temperature change for a given input of energy?
hoa [83]

Answer:

If the chef is in a hurry we would assume he would want the pan to heat up so copper

Explanation:

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Write the equation for the dissociation of hcl when hcl is dissolved in water.
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HCl -> H+ + Cl- (monoprotic acid)

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2 years ago
Acetone has a boiling point of 56.5 celcius. How many grams of the acetone vapor would occupy the 250 mL Erlenmeyer flask at 57
vodomira [7]

Answer:

0.515 g

Explanation:

<em>Acetone (C₃H₆O) has a boiling point of 56.5 °C. How many grams of the acetone vapor would occupy the 250 mL Erlenmeyer flask at 57 °C and 730 mmHg?</em>

<em />

Step 1: Given data

Temperature (T): 57°C

Pressure (P): 730 mmHg

Volume (V): 250 mL

Step 2: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15 = 57°C + 273.15 = 330 K

Step 3: Convert "P" to atm

We will use the conversion factor 1 atm = 760 mmHg.

730 mmHg × (1 atm/760 mmHg) = 0.961 atm

Step 4: Convert "V" to L

We will use the conversion factor 1 L = 1,000 mL.

250 mL × (1 L/1,000 mL) = 0.250 L

Step 5: Calculate the moles (n) of acetone

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 0.961 atm × 0.250 L/(0.0821 atm.L/mol.K) × 330 K

n = 8.87 × 10⁻³ mol

Step 6: Calculate the mass corresponding to 8.87 × 10⁻³ moles of acetone

The molar mass of acetone is 58.08 g/mol.

8.87 × 10⁻³ mol × 58.08 g/mol = 0.515 g

8 0
3 years ago
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