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3 years ago

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 ?

1 answer:

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

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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