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scoundrel [369]
2 years ago
7

What volume would be occupied by 8.1 g of a substance with a density of 1.65041 g/cm3 ? answer in units of cm3 ?

Chemistry
1 answer:
ladessa [460]2 years ago
7 0
Hey there!

Mass = 8.1 g

density = 1.65041 g/cm³

Volume = ?

Therefore:

D = m / V

<span>1.65041  = 8.1 / V

V = 8.1 / </span><span>1.65041 g

V = 4.907 cm</span>³
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Sam follows the directions on a box of pancake mix. The box says that it makes 10 pancakes, but he is able to make 8 pancakes fr
Mice21 [21]

Given that the pancake mix can be used for the preparation of 10 pancakes. The other ingredients added to the pancake mix would be eggs, cheese, milk, salt and baking soda. If all of thee ingredients are added to the pan cake mix in required proportions, we can make 10 pan cakes. So, the theoretical yield will be 10 pancakes. But in real Sam was able to prepare only 8 pancakes. That means the number of pancakes produced was limited by the amount of other ingredients. So, the actual yield will be 8 pancakes.

Hence the correct answer B) Theoretical yield

8 0
2 years ago
Read 2 more answers
Write a ground state electron configuration for each neutral atom
Gre4nikov [31]

Answer:

Pb[lead] [Xe]4f^145d^106s^26p^2

U[uranium] 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

This notation can be written in core notation or noble gas notation by replacing the

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

with the noble gas [Rn].

[Rn]7s25f4

N[nitrogen] The full electron configuration for nitrogen is 1s^2 2s^2 2p^3.

Ti[titanium] Ti2+:[Ar]3d^2

Ti:1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2

1s^2 2s^2 2p^6 3s^2 3p^5 = 17 electrons

(1) electron gain will result to a

negative charge (−), and

(2) electron loss will result to a positive charge (+),

1s^2 2s^2 2p^6 3s^2 3p^6 = 18 electrons

Hg[mercury] You should then find its atomic number is 80. It has a Xe core, so in shorthand notation, you can include [Xe]instead of

1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6,

for 54 electrons. For the 6th row of the periodic table, we introduce the 4f orbitals, and proceed to atoms having occupied 5d orbitals. We, as usual, have the ns orbitals, and n=? for the 6th period?

Mercury has a regular electron configuration. It becomes:

[Xe]4f145d106s2

Explanation:

socratic.org helped me! I'm really sorry if this is wrong!

6 0
2 years ago
Which is the correct Lewis dot structure for BeH2 ?
kherson [118]
H• •Be• •H --> H:Be:H
Since there are 4 valence electrons in total, Beryllium has 2 and Hydrogen has 2. You would put the Be in the middle because there is only 1 of them.
7 0
2 years ago
rate of a certain reaction is given by the following rate law: rate Use this information to answer the questions below. What is
Sunny_sXe [5.5K]

Complete Question

The  rate of a certain reaction is given by the following rate law:

            rate =  k [H_2][I_2]

rate Use this information to answer the questions below.

What is the reaction order in H_2?

What is the reaction order in I_2?

What is overall reaction order?

At a certain concentration of H2 and I2, the initial rate of reaction is 2.0 x 104 M / s. What would the initial rate of the reaction be if the concentration of H2 were doubled? Round your answer to significant digits. The rate of the reaction is measured to be 52.0 M / s when [H2] = 1.8 M and [I2] = 0.82 M. Calculate the value of the rate constant. Round your answer to significant digits.

Answer:

The reaction order in H_2 is  n =  1

The reaction order in I_2 is  m = 1

The  overall reaction order z =  2

When the hydrogen is double the the initial rate is   rate_n  =  4.0*10^{-4} M/s

The rate constant is   k = 35.23 \  M^{-1} s^{-1}

Explanation:

From the question we are told that

   The rate law is  rate =  k [H_2][I_2]

   The rate of reaction is rate =  2.0 *10^{4} M /s

Let the reaction order for H_2 be  n and for I_2  be  m

From the given rate law the concentration of H_2 is raised to the power of 1 and this is same with I_2 so their reaction order is  n=m=1

   The overall reaction order is  

               z  = n +m

               z  =1 +1

               z  =2

At  rate =  2.0 *10^{4} M /s

        2.0*10^{4}  = k  [H_2] [I_2] ---(1)

= >    k  = \frac{2.0*10^{4}}{[H_2] [I_2]  }

given that the concentration of hydrogen is doubled we have that

            rate  = k [2H_2] [I_2] ----(2)

=>      k = \frac{rate_n  }{ [2H_2] [I_2]}

 So equating the two k

           \frac{2.0*10^{4}}{[H_2] [I_2]  } = \frac{rate_n  }{ [2H_2] [I_2]}

    =>    rate_n  =  4.0*10^{-4} M/s

So when

      rate_x =  52.0 M/s

        [H_2] = 1.8 M

         [I_2] =  0.82 \ M

We have

      52 .0 =  k(1.8)* (0.82)

     k = \frac{52 .0}{(1.8)* (0.82)}

      k = 35.23 M^{-2} s^{-1}

     

     

3 0
2 years ago
( Endocytosis / Exocytosis ) is the movement of substances out of a cell by vesicular transport.
tiny-mole [99]

Answer:

Exocytosis

Explanation:

Some molecules are simply too big to move via a transport protein or the plasma membrane. To carry these macromolecules in or out of the cell, cells employ two more active transport pathways. Macromolecules or big particles are transported across the plasma membrane via Vesicles transport or other cytoplasmic structures. They are of two types, Endocytosis and Exocytosis

From the given information, Exocytosis is the right answer.

It is the process of vesicles combining with the plasma membrane thereby releasing their contents to the exterior of the cell. When a cell creates components for export, such as proteins, or when it gets rid of a waste product or a toxin, exocytosis occurs. Exocytosis is the process by which newly generated membrane proteins and membrane lipids are transported on top of the plasma membrane.

4 0
3 years ago
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