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stealth61 [152]
3 years ago
11

23g of of sodium reacted with 35.5g of chlorine. calculate the mass of the sodium chloride compound formed

Chemistry
1 answer:
Marrrta [24]3 years ago
3 0

Answer:

the answer is 58.5g

Explanation:

23+35.5=58.5g

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Elements ending in the electron configurations ns^1 are highly reactive metals . What family does these elements belong to ?
myrzilka [38]

Answer:

<h3>A . Alkali metals</h3>

Explanation:

The highlighted elements of the periodic table belong to the alkali metal element family. The alkali metals are recognized as a group and family of elements. These elements are metals. Sodium and potassium are examples of elements in this family.

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6 0
3 years ago
The density of water is 1.00 g/ml at 4°c. how many water molecules are present in 2.36 ml of water at this temperature?
spin [16.1K]

Mass of 1 ml of water 1 g as density of water 1.00 g/ml at 4° C. So, mass of 2.36 ml of water \frac{1}{2.36} g= 0.423 g.

Molecular mass of water is 18 g which indicates 18 g water contains 6.023 X 10^{23} number of water molecules. So, 0.423 g of water contains \frac{6.023 X 10^{23} X 0.423 }{18}  g = 0.141 X 10^{23} number of water molecules= 1.41 X 10^{22} number of water molecules.

5 0
3 years ago
Density is anything that has mass and takes up space.<br><br> A. True<br><br> B. False
sergij07 [2.7K]
True because it is correct
4 0
3 years ago
Read 2 more answers
Calculate the empirical formula for each natural flavor based on its elemental mass percent composition.
algol [13]
The complete question is;
Calculate the empirical formula for each of the following naturalflavors based on their elemental mass percent composition.

Q1)
methyl butyrate (component of apple taste andsmell):  C -58.80 %  H- 9.87 % 
O -31.33.%Express your answer as a chemical formula.


Q2)
vanillin (responsible for the taste and smellof vanilla):  C - 63.15%  H- 5.30 % 
O - 31.55%Express your answer as a chemical formula.

Q1)
empirical formula is the simplest ratio of whole number of elements in the compound. as the percentages have been given, lets calculate for 100 g of compound 
                                          C                         H                         O
mass                             58.80 g                  9.87 g                   31.33
molar mass                   12 g/mol                 1 g/mol                 16 g/mol
number of moles           58.80/12                9.87/1                    31.33/16
                                      = 4.9                      =9.87                     = 1.95
then divide number of moles by least number of moles - 1.95 in this case
                                      4.9/1.95 = 2.51      9.87/1.95 = 5.06    1.95/1.95= 1
next multiply by 2 to get numbers that can be rounded off to whole numbers
                                       2.51x2 = 5.02        5.06x2 = 10.12      1x2 = 2
when rounded off to the nearest whole number 
C - 5
 H - 10
 O - 2
therefore empirical formula is C₅H₁₀O₂

Q2) for this too since elemental composition has been given in percentages lets calculate for 100 g of compound 
                                          C                         H                         O
mass                               63.15 g                5.30 g                 31.55 g
molar mass                     12 g/mol              1 g/mol                16 g/mol
number of moles             63.15/12             5.30/1                  31.55/16
                                        =5.26                  =5.30                   =1.97
divide the number of moles by the least number of moles - 1.97
                                        5.26/1.97            5.30/1.97               1.97/1.97 
                                        =2.67                   = 2.69                      = 1
multiply each by 3 to get numbers that can be rounded off to whole numbers
                                        2.67x3 = 8.01     2.69x3 = 8.07         1x3 = 3
rounded off to the nearest whole numbers 
C - 8
H - 8
O - 3
empirical formula = C₈H₈O₃
7 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 10.085 g at 20ºC. If the d
Tanzania [10]
To calculate percent errorsubtract the accepted value from the experimental value.Take the absolute value of step 1.Divide that answer by the accepted value.Multiply that answer by 100
so% error = [|(10.085 g/ 10 ml) - 0.9975| / 0.9975] x 100
% error = 1.1 %
5 0
3 years ago
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