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Artist 52 [7]
3 years ago
10

Calculate the mass of 0.00456 moles of (NH4)2SO4

Chemistry
1 answer:
Anna [14]3 years ago
7 0
To find the mass you need to find the weight of a mol of the molecules by adding up the atomic mass.

N =  14.007 g/mol
H = 1.008 g/mol
S = 32.065 g/mol
O = 16 g/mol

2(14.007) + 8(1.008) + 32.065 + 4(16) = 132.143 g/mol

Now you know how much an entire mol weight you multiply it by how much you actually have 

0.00456 * 132.143 = 0.603 g
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Draw a sketch of a cross section of a graduated cylinder illustrating the proper method for reading the volume
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Answer:

in reading volume - you read from the bottom of the meniscus, which is the curve formed from the liquid in the graduated cylinder. Most graduated cylinders are in ml, so measure in the most accurate reading.

Explanation:

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What is the purpose of an adaptation
aleksandrvk [35]
An adaptation is a mutation, or genetic change, that helps an organism, such as a plant or animal, survive in its environment.
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Write the first and second ionization equations for H2SeO3 please.
IRISSAK [1]
As a diprotic acid, the H₂SeO₃ can ionize by step. First step is H₂SeO₃ =(reversible reaction) H⁺ + HSeO₃⁻. And second step is HSeO₃⁻ =(reversible reaction) H⁺ +SeO₃ ²⁻.
3 0
3 years ago
A 2.4 L balloon holds 3.7 mol He. If 1.6 mol He are added to the balloon, what is the new volume?
Elanso [62]
For this question, lets apply Avagadro's law
when Pressure and temperature are constant, the volume occupied is directly proportional to the number of moles of gases.
\frac{V}{n} = k
where V-volume, n-number of moles and k - constant 
Therefore at 2 instances 
\frac{V1}{n1} =  \frac{V2}{n2}
where V1 and n1 are for 1st instance 
and V2 and n2 are for 2nd instance 
therefore 
\frac{V1}{n1} = \frac{V2}{n2}
V1 = 2.4 L
n1 = 3.7 mol
n2 = 3.7 + 1.6 = 5.3 mol
since more He moles are added at the 2nd instance its the sum of the moles.
V2 needs to be calculated \frac{2.4}{3.7}  = \frac{V2}{5.3}
V2 = 2.4 x 5.3 / 3.7
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Answer is 1st option 3.4 L


6 0
3 years ago
A compound that contains both potassium and oxygen formed when potassium metal was burned in oxygen gas. the mass of the compoun
Artemon [7]
Answer:
             3.2 g of O₂

Solution:
             This reaction is for the formation of Potassium Superoxide, The reaction is as follow,

                                           K  +  O₂     →     KO₂

First let us confirm that either the given amount of Potassium produces the given amount of Potassium oxide or not,
So,
As,
                39.098 g (1 mole) K produced  =  71.098 g of K₂O
So,
                          3.91 g of K will produce  =  X g of K₂O

Solving for X,
                      X  =  (3.91 g × 71.098 g) ÷ 39.098

                      X  =  7.11 g of K₂O

Hence, it is confirmed that we have selected the right equation,
So,
As,
                     39.098 g of K required  =  32 g of O₂
So,
                     3.91 g of K will require  =  X g of O₂

Solving for X,
                     X  =  (3.91 g × 32 g) ÷ 39.098 g

                     X  =  3.2 g of O₂
4 0
3 years ago
Read 2 more answers
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