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Bogdan [553]
3 years ago
7

A student needs to prepare 50.0 mL of a 1.30 M aqueous H2O2 solution. Calculate the volume of 5.0 M H2O2 stock solution that sho

uld be used to prepare the solution.
Chemistry
1 answer:
lisov135 [29]3 years ago
7 0

Answer:

13.0mL

Explanation:

So this is a classic M1V1=M2V2 problem, where M is molarity and V is volume and the subsequent numbers represent the two sets of condition (1 being before dilution and 2 being after dilution)

So M1 is going to be 5.0M because it is our initial molarity and V1 is what we are trying to find since we are trying to find how much of initial volume should be diluted.

M2 is 1.30M since it is what molarity is after dilution and M2 is what volume is after dilution which is 50mL.

So M1V1=M2V2 (becomes an algebra problem)

5*V1=1.30*50

V1= 13.0mL

Now this answer should make sense since to dilute something with large molarity to small you only need very few mL than the final volume because you add water to dilute it.

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The mass number of an atom is 15, and its atomic number is 7. The atom probably has... 8 neutrons in the nucleus. about as much
Artist 52 [7]

Answer: Option (a) is the correct answer.

Explanation:

Atomic number is the sum of only total number of protons present in an element. Whereas mass number is the sum of total number of both protons and neutrons present in an element.

For example, given atom has mass number as 15 and its atomic number is 7.

Therefore, number of neutrons present in it will be calculated as follows.

                    Mass number = no. of protons + no. of neutrons

                         15 = 7 + no. of neutrons

                    no. of neutrons = 15 - 7

                                               = 8

Thus, we can conclude that the given atom contains 8 neutrons in the nucleus.

3 0
3 years ago
A student dissolves of glucose in of a solvent with a density of . The student notices that the volume of the solvent does not c
nikitadnepr [17]

Answer:

0.052 M

0.059 m

Explanation:

There is some missing info. I think this is the complete question.

<em>A student dissolves 4.6 g of glucose in 500 mL of a solvent with a density of 0.87 g/mL. The student notices that the volume of the solvent does not change when the glucose dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to 2 significant digits.</em>

Step 1: Calculate the moles of glucose (solute)

The molar mass of glucose is 180.16 g/mol.

4.6 g × 1 mol/180.16 g = 0.026 mol

Step 2: Calculate the molarity of the solution

0.026 moles of glucose are dissolved in 500 mL (0.500 L) of solution. We will use the definition of molarity.

M = moles of solute / liters of solution

M = 0.026 mol / 0.500 L = 0.052 M

Step 3: Calculate the mass corresponding to 500 mL of the solvent

The solvent has a density of 0.87 g/mL.

500 mL × 0.87 g/mL = 435 g = 0.44 kg

Step 4: Calculate the molality of the solution

We will use the definition of molality.

m = moles of solute / kilograms of solvent

m = 0.026 mol / 0.44 kg = 0.059 m

4 0
3 years ago
Three 5-l flasks, fixed with pressure gauges and small valves, each contains 4 g of gas at 273 k. flask a contains h2, flask b c
Varvara68 [4.7K]
First, please check the missing part in your question in the attachment.
a) So first, the Rank of pressure:
according to this formula PV = nRT and when n = m/Mw
PV = m/Mw * R*T
when we have the same mass m and the same V volume so P will proportional with the mole weight M as when the M is smaller the pressure will be greater 
when Mw of H2(A) = 2 g / Mw of He (B) = 4 g and Mw of CH4(C) = 16 g
∴ Pressure :
 (A) > (B) > (c)

B) The rank of average molecular kinetic energy:
when K = 3/2 KB T
when K is the average kinetic energy per molecule of gas 
and KB is Boltzmann's constant
and T is the temperature (K)
So from this equation, we can know that K only depends on T value, and when we have the T constant here for A, B, and C So the rank of K will be like the following:
∴ A = B = C
C) the rank of diffusion rate after the valve is opened:
according to this formula:
R2/R1 = √M1/M2
from this equation, we can see that diffusion is proportional to the reciprocal of the molecular mass M so,
when Mw H2 (A) = 2 g & Mw He(B) = 4 g & CH4 (C) = 16 g
∴ the rank of diffusion:
A > B > C

D) The rank of the Total kinetic energy of the molecules:
when we have the Mw different so it will make the no.of molecules differs as when the Mw is low the no.of molecules will be hight, and when the average molecular kinetic energy equals. so the total kinetic energy will depend on no. of molecules 
∵ Mw A < Mw B < Mw C 
∴no .of molecules of A > B >C
∴ the rank of total kinetic energy is:
A > B > C

e) the rank of density:

when ρ = m/ v 
and m is the mass & v is the volume and we have both is the same for A, B, and C
so the density also will be the same, ∴ the rank of the density is:
A = B = C

F) the rank of the collision frequency:
as the no.of molecules increase the collision frequency increase and depend also on the velocity and it's here the same.
∴ Collision frequency will only depend on the no.of molecules
we have no.of molecules of A > B > C as Mw A < B < C 
∴the rank of the collision frequency is:
A > B > C 

 



7 0
3 years ago
Which of these elements is this group?
sergey [27]

Answer:

Strontium

Explanation:

In the periodic table, an element with two (2) valence electrons is found on group 2. Group 2 is a group of the periodic table that harbors element called ALKALINE EARTH METALS. As the name implies, they are metals that possess shiny and solid characteristics at room temperature.

Group 2 elements include the following: Beryllium (Be), Magnesium (Mg), Calcium (Ca), Strontium (Sr), barium (Ba), and radium (Ra). Based on the descriptive information in this question, the element being described is a GROUP 2 element. Based on the elements in the option, only STRONTIUM (Sr) is a group 2 element.

3 0
3 years ago
I NEED HELP ASAP!!!!<br><br><br><br> ITS DUE IN A FEW MINUTES!!!!
OleMash [197]

Answer:

The Sun and the planets were born from a cloud of gas and dust called the solar nebula 4.6 billion years ago. The collapse of the solar nebula was most likely triggered by a shock wave from a nearby supernova explosion. The Sun formed in the center, with the planets surrounding it in a thin disk.

Explanation:

6 0
2 years ago
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