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Dafna1 [17]
3 years ago
10

The mass of a solid substance is 21.112 g. If the volume of the solid substance is 19.5 cm3, calculate the density of the substa

nce (in g/cm3) with the correct number of significant digits. The density is g/cm3.
Chemistry
1 answer:
Ostrovityanka [42]3 years ago
6 0

Answer:

ρ = 1.08 g/cm³

Explanation:

Step 1: Given data

Mass of the substance (m): 21.112 g

Volume of the substance (V): 19.5 cm³

Step 2: Calculate the density of the substance

The density (ρ) of a substance is equal to its mass divided by its volume.

ρ = m / V

ρ = 21.112 g / 19.5 cm³

ρ = 1.08 g/cm³

The density of the substance is 1.08 g/cm³.

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Joseph drives 25miles North in 15 minutes then he drives 40 miles east in 30 minutes, then he uses cruise control option for 1 h
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Answer: 1.25 miles per minute

Explanation:

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Total time taken = ( 15 minutes + 30 minutes + 1 hour + 15 minutes) = 120 minutes

Since 60 minutes = 1 hour, the total time taken is 120 minutes

Now, apply Average speed = (Total distance / Time taken)

= (150 miles / 120 minutes)

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Thus, Joseph drove with an average speed of 1.25 miles per minute.

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3 years ago
How many sig figs does 0.00002 have?
DiKsa [7]
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N2H4 + N2O4 --> N2 + H2O
ahrayia [7]

Answer:

  1. The limiting reagent is N2O4
  2. 14,09g

Explanation:

  • First, we adjust the reaction.

2N_{2} H_{4} + N_{2} O_{4} ⇄6N_{2} +  4H_{2}O

  • Second, we assume that the participating moles are equal to the stoichiometric ratios because we do not know the amounts of the reagents.

We can determinate what is the limiting reagent comparing of product amounts which can be formed from each reactant.

Using N_{2} H_{4} to form H_{2}O

               molH_{2} O = 1mol N_{2} H_{4} } . \frac{4 mol H_{2} O}{2mol N_{2} H_{4} }. \frac{18\frac{g}{mol}  H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}H_{4}  }{32,04\frac{g}{mol}  N_{2} H_{4} }

                                           molH_{2} O = 1, 125 mol

Using N_{2} O_{4} to form H_{2} O

              molH_{2} O = 1mol N_{2} O_{4} } . \frac{4 mol H_{2} O}{1mol N_{2} O_{4} }. \frac{18\frac{g}{mol}  H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}O_{4}  }{92\frac{g}{mol}  N_{2} O_{4} }

                                           molH_{2} O = 0,783 mol

The limiting reagent is N2O4, because can produce only 0, 783 mol of H2O.

This is the minimum measure can be formed of each product.

∴                          MassOfH_{2}O = 0,783mol . 18\frac{g}{mol}

                                      MassOfH_{2}O = 14,09g

5 0
3 years ago
3 NO2(g) + H2O(ℓ) −→
gulaghasi [49]
The balanced equation for the above reaction is as follows;
3NO₂ + H₂O --> 2HNO₃ + NO
stoichiometry of NO₂ to NO is 3:1
molar volume is where 1 mol of any gas occupies a volume of 22.4 L
volume of gas is directly proportional to number of moles of gas.
therefore stoichiometry can be applied for volume as well.
volume ratio of NO₂ to NO is 3:1
volume of NO₂ reacted - 854 L
therefore volume of NO formed - 854 L /3 = 285 L
volume of NO formed - 285 L
7 0
3 years ago
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