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UNO [17]
3 years ago
8

The element chlorine has an atomic weight of 35.5 and consists of two stable isotopes chlorine-35 and chlorine-37. The isotope c

hlorine-35 has a mass of 35.0 amu and a percent natural abundance of 75.5 %. The isotope chlorine-37 has a percent natural abundance of 24.5 %. What is the mass of chlorine-37? amu
Chemistry
2 answers:
mylen [45]3 years ago
6 0

Answer:

The answer is 37amu

Explanation:

We are given:

Chlorine-35: atomic mass  =35 amu  and percent abundance  =75.5% ≡ 0.755

Chlorine-37: atomic mass  = x amu  and percent abundance  =24.5% ≡ 0.25

From table, relative atomic mass of Chlorine is 35.5

⇒(0.755 × 35) + (24.5 × x) = 35.5

26.402 + 24.5x = 35.5

∴ x = 9.075 ÷ 24.5 = 0.37 ≡ 37.

∴ mass of Chlorine 37 = 37amu

ioda3 years ago
4 0

Answer:

The mass of Chlorine-37 is 37.04amu.

Explanation:

To calculate the atomic mass of a element with two isotopes we use:

(%abundance * mass)of chlorine-35 + (%abundance * mass) of chlorine-37 = atomic mass

Since the atomic mass has been given, we can equate the values as shown below and calculate:

(75.5/100 * 35) + (24.5/100 * Y) = 35.5

(0.755 * 35) + (24.5Y / 100) =35.5

26.425 + 24.5Y / 100 = 5.5

24.5Y/ 100 = 35.5 - 26.425

24.5Y/100 = 9.075

24.5Y = 9.075 * 100

24.5Y = 907.5

Y = 907.5 / 24.5

Y = 37.04amu

The mass of Chlorine-37 is 37.04amu.

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4.2g of cerium reacted with oxygen to form 5.16g of an oxide of cerium. Find
olganol [36]

Answer:

CeO₂

Explanation:

Hello!

In this case, since we are given the mass of both cerium and the cerium oxide, we can first compute the moles of cerium and the moles of oxygen as shown below:

n_{Ce}=4.2gCe*\frac{1molCe}{140.12gCe}=0.03molCe\\

m_O=5.16g-4.2g=0.96gO\\\\n_O=0.96g*\frac{1molO}{16.0gO} =0.06molO

Now, we simply divide each moles by 0.03 as the fewest moles in the formula to obtain the simplest formula (empirical formula) of this oxide:

Ce=\frac{0.03}{0.03}=1\\\\O =\frac{0.06}{0.03}=2

Thus, the formula turns out:

CeO_2

Regards!

6 0
3 years ago
Plz answer fast Name the four parts of a circuit.
nignag [31]

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A switch

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5 0
3 years ago
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How many grams of NaOH are produced from 20.0 grams of Na2CO3?
natita [175]

Answer:

Hope this helps!

Explanation:

Ans: 15.1 grams

Given reaction:

Na2CO3 + Ca(OH)2 → 2NaOH + CaCO3

Mass of Na2CO3 = 20.0 g

Molar mass of Na2CO3 = 105.985 g/mol

# moles of Na2CO3 = 20/105.985 = 0.1887 moles

Based on the reaction stoichiometry: 1 mole of Na2CO3 produces 2 moles of NaOH

# moles of NaOH produced = 0.1887*2 = 0.3774 moles

Molar mass of NaOH = 22.989 + 15.999 + 1.008 = 39.996 g/mol

Mass of NaOH produced = 0.3774*39.996 = 15.09 grams

4 0
3 years ago
In a single replacement reaction between 2.57 moles Aluminum and 3.59 moles of Hydrochloric acid, how many moles of Hydrogen can
Alenkasestr [34]

Answer:

1.795 mole of H2.

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2Al + 6HCl —> 2AlCl3 + 3H2

Step 2:

Determination of the limiting reactant.

From the balanced equation above,

2 moles of Al reacted with 6 moles

Therefore, 2.57 moles of Al will react with = (2.57 x 6)/2 = 7.71 moles of HCl.

From the calculation made above, it will require a higher amount of HCl than what was given to react completely with 2.57 moles of Al. Therefore, HCl is the limiting reactant and Al is the excess reactant.

Step 3:

Determination of the number of mole H2 produced from the reaction.

Here, we shall be using the limiting reactant because it will produce the maximum yield of the reaction since all of it were consumed by the reaction.

The limiting reactant is HCl and the amount of H2 produce can be obtained as follow:

From the balanced equation above,

6 moles of HCl reacted to produce 3 moles of H2.

Therefore, 3.59 moles of HCl will produce = (3.59 x 3)/6 = 1.795 mole of H2.

From the calculations made above, 1.795 mole of H2 is produced from the reaction.

5 0
3 years ago
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