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Andrei [34K]
4 years ago
13

A 50 ml volumetric flask is calibrated to the _______ and will hold exactly _________ when filled to the calibration line: 0.01

ml, 100.00 ml 0.01 ml, 50.00 ml 0.01 ml. 50.000 ml 0.01 ml, 50.0 ml 0.1 ml, 50.00 ml 0.1 ml, 50.0 ml 0.1 ml, 50 ml
Chemistry
2 answers:
tensa zangetsu [6.8K]4 years ago
7 0
<h2>Answer : Option B) 0.01 mL, 50.00 mL</h2><h3>Explanation :</h3>

A 50  mL volumetric flask is calibrated to the <u>0.01 mL</u> and will hold exactly <u>50.00 mL </u> when filled to calibration line.

Every volumetric flask is calibrated with 0.01 mL so that when the liquid is filled till the calibration line the flask should hold 50.00 mL fluid in it. It gives precise and accurate measurements for the experiments. Calibration is a must for every glassware which ensures less experimental errors.

artcher [175]4 years ago
5 0

Answer;

0.1 mL, 50.00 mL

A 50 ml volumetric flask is calibrated to the 0.1 mL and will hold exactly 50.00 mL when filled to the calibration line.

Explanation;

A volumetric flask is a piece of laboratory glassware, a type of laboratory flask, calibrated to contain a precise volume at a particular temperature. Volumetric flasks are used for precise dilutions and preparation of standard solutions.

Calibration is a comparison between a known measurement (the standard) and the measurement using your instrument. The accuracy of the standard should be ten times the accuracy of the measuring device being tested.

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How much heat energy, in kilojoules, is required to convert 72.0 gg of ice at −−18.0 ∘C∘C to water at 25.0 ∘C∘C ? Express your a
Ghella [55]

Answer: The enthalpy change is 34.3 kJ

Explanation:

The conversions involved in this process are :

(1):H_2O(s)(-18^0C)\rightarrow H_2O(s)(0^0C)\\\\(2):H_2O(s)(0^0C)\rightarrow H_2O(l)(0^0C)\\\\(3):H_2O(l)(0^0C)\rightarrow H_2O(l)(25^0C)

Now we have to calculate the enthalpy change.

\Delta H=[m\times c_{s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{l}\times (T_{final}-T_{initial})]

where,

\Delta H = enthalpy change = ?

m = mass of water = 72.0  g

c_{s} = specific heat of ice = 2.09J/g^0C

c_{l} = specific heat of liquid water = 4.184J/g^0C

n = number of moles of water = \frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{72.0g}{18g/mole}=4.00moles

\Delta H_{fusion} = enthalpy change for fusion = 6010 J/mole

Now put all the given values in the above expression, we get

\Delta H=[72.0g\times 2.09J/g^0C\times (0-(-18)^0C]+4.00mole\times 6010J/mole+[72.0g\times 4.184J/g^)C\times (25-0)^0C]\Delta H=34279.8J=34.3kJ        (1 KJ = 1000 J)

Therefore, the enthalpy change is 34.3 kJ

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3 years ago
Which of these mixtures will MOST LIKELY result in soil? A) sand, clay, and heat B) sand, clay, and water C) sand, clay, heat, a
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Answer:

I think it should be C.

Explanation:

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4 years ago
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Hello there,


Your correct answer would be 8.<span> There are </span>8 valence electrons<span> available for the Lewis structure for NH</span>3<span>.

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4 0
3 years ago
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If you have 30.O g of hydrogen gas burned in excess oxygen how many moles of water can you make
vladimir1956 [14]

Answer:

15 moles

Explanation:

Data given:

mass of hydrogen (H₂) = 30.0 g

amount of oxygen (O₂) = excess

moles of water = ?

Solution:

First we look to the reaction in which hydrogen react with oxygen and make (H₂O)

Reaction:

              2H₂  + O₂  -----------> 2H₂O

Now look at the reaction for mole ratio

             2H₂  + O₂  -----------> 2H₂O

             2 mole                       2 mole

So it is 2:2 mole ratio of hydrogen to water

As we Know

molar mass of H₂  = 2(1) = 2 g/mol

molar mass of H₂O = 2(1) + 16 = 18 g/mol

Now convert moles to gram

                  2H₂         +       O₂        ----------->    2H₂O

          2 mole (2 g/mol)                                 2 mole (18 g/mol)

                    4 g                                                     36 g

So,

we come to know that 4 g of hydrogen gives 36 g of water then how many grams of water will be produce by 30 grams of hydrogen.

Apply unity formula

                       4 g of H₂ ≅ 36 g of H₂O

                        30 g of H₂ ≅ X of H₂O

Do cross multiplication

                  X of H₂O =  30 g x 36 g / 4 g

                  X of H₂O =  270 g

Now convert grams of H₂O into moles

               No. of moles = mass in grams/molar mass

Put values in above formula

               No. of moles = 270 g / 18 (g/mol)

               No. of moles = 15 mol

so 30 gram of hydrogen produce 15 mol of water.

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4 years ago
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