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Gnoma [55]
3 years ago
11

Sam records the mass of his evaporating dish as 6.488 g. He records the mass of the evaporating dish and the sample of hydrate a

s 19.72 g. After heating the sample in the evaporating dish to constant weight, the mass of them combined is 11.344 g. How many moles of water were removed from the sample by the heating process?
Chemistry
1 answer:
jok3333 [9.3K]3 years ago
5 0

Answer:

0.46 moles of water were removed from the sample by the heating process

Explanation:

Given

Mass of the evaporating dish = 6.488 grams

Mass of the sample of hydrate = 19.72 grams

Mass of the sample and dish after heating = 11.344 grams

The weight of water removed through evaporation

= 19.72 - 11.344 = 8.376 grams

Weight of one mole of water = 18 grams

Number of moles of water in 8.376 Grams of water is equal to

\frac{8.376}{18} = 0.46 moles

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Answer:

Sounds travels in transverse waves requires a medium to travel through

4 0
3 years ago
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A cylinder with a radius of 5.6 cm and a height of 10.7 cm has what volume? (You can look up the formula for the volume of a cyl
Keith_Richards [23]

Answer:

V=1054.2cm^{3}

Explanation:

1. First take the cylinder volume formula:

V=\pi.r^{2}.h

2. Then take the values for the radius and the height of the cylinder and replace them into the formula:

V=\pi*(5.6cm)^{2}*(10.7cm)

3. Solve the equation:

V=\pi*(31.36cm^{2})*(10.7cm)

V=\pi*335.5cm^{3}

V=1054.2cm^{3}

6 0
4 years ago
The liquid dispensed from a burette is called ___________.
11Alexandr11 [23.1K]

The liquid that is been dispensed during titration as regards this question is Titrant.

  • Titration can be regarded as  common laboratory method that is been carried out during quantitative chemical analysis.
  • This analysis helps to know the concentration of an identified analyte.
  • Burette can be regarded as laboratory apparatus.

It is used in the in measurements of variable amounts of liquid ,this apparatus helps in dispensation   of liquid, especially when performing titration.

  • The specifications is been done base on their volume, or resolution.
  • The liquid that comes out of this apparatus is regarded as Titrant, and this is gotten during titration process, which is usually carried out during volumetric analysis.

Therefore, burrete is used in volumetric analysis.

Learn more at:

brainly.com/question/2728613?referrer=searchResults

8 0
3 years ago
Read 2 more answers
The pressure of a gas is reduced from 1200.0 mmHg to 1.11842 atm as the volume of its container is increased by moving a piston
Artist 52 [7]

Answer:

          \large\boxed{T_2=786.\ºC}

Explanation:

Ideal gases follow the combined law of gases:

         P_1V_1/T_1=P_2V_2/T_2

Where,

P_1,V_1, and{\text{ }T_1\text{ are the pressure, temperature, and volume of the gas a state 1}

P_2,V_2, and{\text{ }T_2\text{ are the pressure, temperature, and volume of the gas a state 2}

  • Pressure is the absolute pressure and its units may be in any system, as long as they are the same for both states.

  • Also, volume may be in any units, as long as it they are the same for both states.

  • Temperature must be absolute temperature, whose unit is Kelvin.

Your data are:

  • P₁ = 1200.00 mmHg
  • P₂ = 1.11842 atm
  • V₁ = 85.0 mL
  • V₂ = 350.0 mL
  • T₂ = ?
  • T₁ = 90.0ºC

<u>1. Conversion of units:</u>

  • P₁ = 1200.00 mmHg × 1.00000 atm / 760.000 = 1.578947 mmHg
  • T₁ = 90.00ºC + 273.15 = 363.15K

<u>2. Solution</u>

  • Clearing T₂, from the combined gas equation you get:

      T_2=P_2V_2T_1/(P_1V_1)

  • Substituting the data:

         T_2=1.11842atm\times 350.0ml\times 363.15K/(1.578947atm\times 85.0ml)

          T_2=1,059K

  • Convert to celsius:

         T_2=1059-273.15=786.\ºC

8 0
3 years ago
If you perform the experiment described in investigation #15 by mixing 10 g of glue with 13 g of water and 8 g of sodium borate
Phantasy [73]

10 g of glue with 13 g of water ,

Mass ratio of the material can be calculated as:    

\frac{Mass of glue}{Mass of glue + Mass of water}  

\frac{10 g}{10 g + 13 g }  

\frac{10 g}{23 g }

8 g of sodium borate suspended in 11 g of water, mass ratio can be calculated as:

\frac{Mass of sodium borate}{Mass of sodium borate + Mass of water}  

\frac{ 8 g}{ 8 g + 11 g }  

\frac{8 g}{19 g }

3 0
3 years ago
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