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1 year ago

What volume in milliliters of 6.0 M NaOH is needed to prepare 175mL of 0.20 M NaOH by dilution?

Chemistry

1 answer:

Ilya [14]1 year ago

4 0

Answer:

V¹N²= V²N²

here V¹= ?

N¹= 6.00

V²= 175ml

M²= 0.2M

So V¹= (V²N²)/N² = (175 x 0.2)/6

V¹ = 5.83 ml

Explanation:

Therefore diluting 5.83 ml of 6.00M NaOH to 175 m l ,we get 0.2M Solution.

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1. Which class of compounds contains at least one element from Group 17 of the Periodic Table ? A) aldehyde B ) amine C) ester D

blondinia [14]

Halide

hope it helps

4 0

1 year ago

Fire a single electron towards the hydrogen atom. Describe what happens in a step by step fashion. [N.B. - It may be helpful to

scZoUnD [109]

Answer:

The electrons will be added by the hydrogen.

Explanation:

If we fire a single electron towards the hydrogen atom, the hydrogen atoms added the electron to its shell by applying force of attraction and becomes stable as well as non reactive in nature because the hydrogen attains the electronic configuration of helium which is a noble gas and have completed its outermost shell. The proton that is present in the nucleus attracts this electron and compel it to add in the electron.

7 0

1 year ago

A sample that weighs 103.75 g is a mixture of 30% helium atoms and 70% krypton atoms. How many particles are present in the samp

aalyn [17]

1) we calculate the molar mass of He (helium) and Kr (Krypton).
atomic mass (He)=4 u
atomic mas (Kr)=83.8 u

Therefore the molar mass will be:
molar mass(He)=4 g/mol
molar mass(Kr)=83.8 g/mol.

1) We can find the next equation:
mass=molar mass x number of moles.

x=number of moles of helium
y=number of moles of helium.

(4 g/mol) x +(83.8 g/mol)y=103.75 g
Therefore, we have the next equation:

(1)

4x+83.8y=103.75

2) We can find other equation:

We have 30% helium atoms and 70% Kryptum atoms, therefore we have 30% Helium moles and 70% of Krypton moles.

1 mol is always 6.022 * 10²³ atoms or molecules, (in this case atoms).

Then:
x=number of moles of helium
y=number of moles of helium.
(x+y)=number of moles of our sample.

x=30% of (x+y)

Therefore, we have this other equation:
(2)

x=0.3(x+y)

With the equations(1) and (2), we have the next system of equations:

4x+83.8y=103.75

x=0.3(x+y) ⇒ x=0.3x+0.3y ⇒ x-0.3x=0.3y ⇒ 0.7 x=0.3y ⇒ x=0.3y/0.7
⇒x=3y/7

We solve this system of equations by substitution method.
x=3y/7

4(3y/7)+83.8y=103.75
lower common multiple)7
12y+586.6y=726.25
598.6y=726.25
y=1.21

x=3y/7=3(1.21)/7=0.52

We have 0.52 moles of helium and 1.21 moles of Krypton.

1 mol=6.022 * 10²³ atoms

Total number of particles=(6.022 *10²³ atoms /1 mol) (number of moles of He+ number of moles of Kr).

Total number of particles=6.022 * 10²³ (0.52+1.21)=6.022 * 10²³ (1.73)=
=1.044 * 10²⁴ atoms.

Answer: The sample have 1.044 * 10²⁴ atoms.

6 0

2 years ago

Read 2 more answers

A 7.591-9 gaseous mixture contains methane (CH4) and butane

mestny [16]

Answer:

65.71%

Explanation:

First, we can write the mass of the mixture, thus:

7.519g = X + Y (1)

Where X is the mass of methane and Y the mass of butane

Also, the reactions of combustion are:

CH₄ + 2O₂ → CO₂ + 2H₂O

2 moles of oxygen react per mole of methane

C₄H₁₀ + 13/2 O₂ → 4CO₂ + 5H₂O

13/2 moles of oxygen react per mole of methane

That means, in therms of moles of oxygen we can write:

0.9050 moles = 2X/16.04 + 13/2Y/ 58.12

0.9050 = 0.12469X + 0.11184Y (2)

Where 16.04 and 58.12 are molar masses of methane and butane

That is because if X is the mass of methane:

X g Methane * (1mol / 16.04g) = Moles methane

Moles methane * (2 moles Oxygen / mole methane) = Moles oxygen

Replacing (1) in (2):

0.9050 = 0.12469X + 0.11184 (7.519 - X)

0.9050 = 0.12469X + 0.841 - 0.11184X

0.0641 = 0.01285X

X = 4.988g = Mass of methane.

And mass percent of methane is:

4.988g / 7.591g * 100

7 0

1 year ago

Why would it be better to be an r-selected species if the water resources in an area were to become more limited over a short pe

Gnoma [55]

An r-selected species reproduces much faster than K-selected species.

r-selected species focuses on maturing and reproducing quickly. r-selected species will probably reproduce when the water supply is there for the short period of time; thus, increasing the chance of the r-selected species of surviving.

K-selected species, on the other hand, focus on raising their young and reproduce later. Since the K-selected species take long to mature before reproducing, water supply may run out before they have a chance of fully maturing; thus, K-selected species have a lower chance of survival.

Hope this helps.
If you need anything more, feel free to comment! Have an awesome day! :)

~Collinjun0827, Junior Moderator

6 0

1 year ago

Read 2 more answers