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PilotLPTM [1.2K]
3 years ago
5

In a 5.609g of H20, how many molecules H20 arc present?

Chemistry
1 answer:
nignag [31]3 years ago
8 0

Answer:

1.88 x 10²³molecules of H₂O

Explanation:

Given parameters:

Mass of H₂O  = 5.609g

Unknown:

Number of molecules of  H₂O  = ?

Solution:

To solve this problem, we need to find the number of moles of given compound first.

 Number of moles  = \frac{mass}{molar mass}  

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

 Number of moles  = \frac{5.609}{18}   = 0.31mole

From mole concept;

            1 mole of a substance contains 6.02 x 10²³molecules

       0.31 mole of H₂O will therefore contain 0.31 x 6.02 x 10²³molecules

                                             = 1.88 x 10²³molecules of H₂O

       

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Answer:

magnesium

Explanation:

magnesium is in Group 2, in the periodic table. this means that it has 2 valence electrons. the less valence electrons an element or atom has, the more reactive. Selenium has 6 valence electrons. as a result, Mg is more reactive

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3 years ago
How many grams are in 8.3 moles of CaCl2?
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Answer:

(8.3×40)+(8.3×71)

921.3grames

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3 years ago
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Which two elements make up the greatest percentages by mass in Earth’s crust?
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<h2><u>Answer</u> :</h2>

The most appropriate option is :

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Since, they are the most abundant elements found in Earth's Crust.

  • Oxygen (46%)

  • Silicon (28%)

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2 years ago
QUESTION 5<br> Match the symbol with the type of symbiotic relationship it represents
marshall27 [118]

Answer:

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Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s
bazaltina [42]

<u>Answer:</u> The time taken by the reaction is 84.5 seconds

<u>Explanation:</u>

The equation used to calculate half life for first order kinetics:

k=\frac{0.693}{t_{1/2}}

where,

t_{1/2} = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

k=\frac{0.693}{9}=0.077s^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}     ......(1)

where,

k = rate constant  = 0.077s^{-1}

t = time taken for decay process = 50.7 sec

[A_o] = initial amount of the reactant = ?

[A] = amount left after decay process =  0.0741 M

Putting values in equation 1, we get:

0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}

[A_o]=3.67M

Now, calculating the time taken by using equation 1:

[A]=0.0055M

k=0.077s^{-1}

[A_o]=3.67M

Putting values in equation 1, we get:

0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s

Hence, the time taken by the reaction is 84.5 seconds

6 0
3 years ago
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