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Papessa [141]
3 years ago
8

How many moles of cacl2 are in 10.5g of cacl2?

Chemistry
1 answer:
Arte-miy333 [17]3 years ago
5 0

Answer:

0.09460823181719888

Explanation:

Round to whatever decimal point is needed for your answer.

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A beaker contains 42.0 mL of carbon tetrachloride ( CCl4 , density is 1.59 g/mL). Determine how many molecules of carbon tetrach
Alex

Answer:

Number of molecules = 2.61  x 10²³

Explanation:

Given:

Volume of tetra-chloride = 42 ml

Density = 1.59 g/ml

Find:

Number of molecules

Computation:

Mass = Volume of tetra-chloride x Density

Mass = 42 x 1.59

Mass = 66.78

Molecular mass of tetra-chloride = 154 g/mol

Moles = 66.78 / 154

Moles = 0.4337

0.4337 mole = 0.4337 x Avogadro number

0.4337 mole = 0.4337 x 6.022 x 10²³

Number of molecules = 2.61  x 10²³

3 0
3 years ago
How many protons and neutrons are in the nucleus of isotope with mass of 68.926 amu?
Sidana [21]
There are 30 protons and 39 neutrons in the nucleus.

This must me the isotope of an element with an atomic mass close to 69 u.
The only candidates are Zn and Ga.
Zn has a zinc-69 isotope with mass 68.926 u.
Ga has a gallium -69 isotope with mass 68.925 u.
The isotope is probably _{30} ^{69}Zn.
It has 30 protons and 39 neutrons.
4 0
3 years ago
When a 10 ml graduated cylinder is filled to the 10 ml mark, the mass of the water was measured to be 9.955 g. if the density of
Lady_Fox [76]

Given, the density of water is  0.9975 g/ml. Density of water is mass of water per unit volume. Mass of 1 ml of water supposed to be  0.9975 g from density of water. So, mass of 10 ml of water is (0.9975 X 10) g= 9.975 g. From graduated cylinder, mass of 10 ml water is measured to be 9.955 g. So, error for mass of 10 ml water= (9.975-9.955)=0.02 g. Percentage of error for 10 ml water is \frac{0.02}{10} X 100 = 0.2. Error in the mass  for the 10 ml of water is 0.2 %.

5 0
3 years ago
In a paper chromatography chamber, which example could be the mobile phase?
Anika [276]

Answer:alcohol

Explanation:

5 0
3 years ago
Determine the % yield when 7.80 grams of benzene (c6h6) burns in oxygen gas to form 3.00 grams of co2 gas and water vapor.
gogolik [260]
Benzene reacts with O₂ to produce CO₂ and H₂O, i.e.

                           C₆H₆  +  7.5 O₂    →    6 CO₂  +  3 H₂O

According to equation,

    78.11 g (1 mole) C₆H₆ reacts to produce  =  264 g (6 moles) of CO₂

Hence,

      7.80 g C₆H₆ when reacted will produce  =  X g of CO₂

Solving for X,
                                X  =  (7.80 g × 264 g) ÷ 78.11 g

                                X  =  26.36 g of CO₂

Theoretical Yield:
                             26.36 g 
of CO₂ produced is theoretical yield which shows 100% reaction between benzene and oxygen.

Actual Yield:
                   According to statement the actual amount of CO₂ produced is 3.0 g of CO₂.

%age Yield:
      
                %age Yield  =  (Actual Yield ÷ Theoretical Yield) × 100

Putting Values,

                %age Yield  =  (3.00 g ÷ 26.36 g) × 100

                %age Yield  =  11.38 %
3 0
4 years ago
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