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FromTheMoon [43]

3 years ago

13

A graduated cylinder contains 17.5 ml of water. When a metal cube is placed onto the cylinder, it’s water level rises to 20.3 ml

. What is the volume of the cube cm3

1 answer:

MrRa [10]3 years ago

4 0

Answer:

20.3-17.5=2.8ml 1ml=1cm3 vol= 2.8cm3

Explanation:

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At 35 C, a sample of gas has a volume of 256 ml and a pressure of 720.torr. What would the volume

Natalija [7]

Answer: Volume would be 196.15 mL if the temperature were changed to $22^{o}C$ and the pressure to 1.25 atmospheres.

Explanation:

Given: $T_{1} = 35^{o}C = (35 + 273) K = 308 K$ , $V_{1}$ = 256 mL,

$P_{1}$ = 720 torr (1 torr = 0.00131579 atm) = 0.947368 atm

$T_{1} = 22^{o}C = (22 + 273) K = 295 K$ , $P_{2} = 1.25 atm$

Formula used to calculate volume is as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substitute the values into above formula as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{1 atm \times 256 mL}{308 K} = \frac{1.25 atm \times V_{2}}{295 K}\\V_{2} = 196.15 mL$

Thus, we can conclude that the volume would be 196.15 mL if the temperature were changed to $22^{o}C$ and the pressure to 1.25 atmospheres.

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2 years ago

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For the reaction Na2CO3+Ca(NO3)2⟶CaCO3+2NaNO3 how many grams of calcium carbonate, CaCO3, are produced from 79.3 g of sodium car

Alexus [3.1K]

Answer:

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.

Explanation:

The balanced reaction is:

Na₂CO₃ + Ca(NO₃)₂ ⟶ CaCO₃ + 2 NaNO₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

Na₂CO₃: 1 mole
Ca(NO₃)₂: 1 mole
CaCO₃: 1 mole
NaNO₃: 2 mole

Being the molar mass of the compounds:

Na₂CO₃: 106 g/mole
Ca(NO₃)₂: 164 g/mole
CaCO₃: 100 g/mole
NaNO₃: 85 g/mole

then by stoichiometry the following quantities of mass participate in the reaction:

Na₂CO₃: 1 mole* 106 g/mole= 106 g
Ca(NO₃)₂: 1 mole* 164 g/mole= 164 g
CaCO₃: 1 mole* 100 g/mole= 100 g
NaNO₃: 2 mole* 85 g/mole= 170 g

You can apply the following rule of three: if by stoichiometry 106 grams of Na₂CO₃ produce 100 grams of CaCO₃, 79.3 grams of Na₂CO₃ produce how much mass of CaCO₃?

$mass of CaCO_{3} =\frac{79.3 grams of Na_{2} CO_{3} *100 grams of of CaCO_{3}}{106 grams of Na_{2} CO_{3}}$

mass of CaCO₃= 74.81 grams

<u><em>74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.</em></u>

6 0

3 years ago

Select the correct answer.

Elina [12.6K]

Answer:

CU

Explanation:

Cu is the correct answer if I'm correct

4 0

2 years ago