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3 years ago

Please, Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Chemistry

2 answers:

docker41 [41]3 years ago

5 0

Answer:1 chemical reactions, 2 properties,3 substances, 4light, 5 color, 6 precipitate, 7 gas, 8 temperature

Explanation: I think this is it I did this a week ago and that’s what I remember

ale4655 [162]3 years ago

3 0

Answer:

1 chemical reactions, 2 properties,3 substances, 4light, 5 color, 6 precipitate, 7 gas, 8 temperature

Explanation:

this a easy sciences

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In a certain experiment, 28.0 mL of 0.250 M HNO3and 53.0 mL of 0.320 M KOH are mixed. Calculate the number of moles of water for

Xelga [282]

Answer:

The number of moles of water formed in the resulting reaction is 6.03

[H+]: 37,2 M

[OH-]: 37,2 M

Explanation:

HNO3 + KOH ----> KNO3 + H2O

First, we must discover the limiting reagent and we need to find out the moles, we use for this.

Moles that are used = Molarity / volume

HNO3 : 0,250 mol/L / 0,028L = 8,93 moles

KOH : 0,320 mol/L / 0,053L = 6,03 moles

The ratio of the reagents by stoichiometry is 1 to 1, so the limiting reagent is KOH, if I need 1 mole of nitric per mole of KOH, for every 8.93 moles I will need the same. However I have only 6.03 moles of KOH

The ratio of the reagents/products by stoichiometry is 1 to 1 so if I need 1 mol of KOH to make 1 mol of Water, 6,03 moles of KOH are used to make 6,03 moles of H2O.

The equilibrium of water is this:

2H2O ⇄ H3O+ + OH-

2 moles of water are broken down into 1 mole of hydronium (H3O +) and 1 mole of hydroxyl (OH-)

6,03 moles of water are broken down into the half of those moles, so we have 3,015 moles of H3O+ and 3,015 moles of OH- but these moles are in 81,0 mL (the volume of the two solutions, 28 mL + 53 mL)

We must find out the moles in 1000 mL (1 L) so let's apply the rule of three.

81 mL ____ 3,015 moles

1000 mL ___ ( 1000 . 3,015) /81 = 37,2 M

7 0

3 years ago

The diameter of a biscuit is approximately 51 millimeters (mm). An atom of bismuth (Bi) is approximately 320. picometers (pm) in

astraxan [27]

Answer:

1.5e+8 atoms of Bismuth.

Explanation:

We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):

$\\ \frac{diameter\;biscuit}{diameter\;atom(Bi)}$

For this, it is necessary to know the values in meters for any of these diameters:

$\\ 1m = 10^{3}mm = 1e+3mm$

$\\ 1m = 10^{12}pm = 1e+12pm$

Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.

<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>

1 atom of Bismuth = 320pm in diameter.

$\\ 320pm*\frac{1m}{10^{12}pm} = 3.20*10^{-10}m$

<h3>Diameter of a biscuit in meters</h3>

$\\ 51mm*\frac{1}{10^{3}mm} = 51*10^{-3}m = 5.1*10^{-2}m$

<h3>Resulting Ratio</h3>

How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:

$\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1*10^{-2}m}{3.20*10^{-10}m}$

$\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}$

$\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{-2+10}$

$\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{8}=1.5e+8$

In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.

6 0

3 years ago

Define critical point. Group of answer choices The temperature and pressure below which a supercritical fluid exists. The temper

attashe74 [19]

Answer:

The temperature and pressure below which a supercritical fluid exists.

Explanation:

Critical point is a region on the phase diagram where fluid phases (liquids and gases) have the equal density.

This is caused by increased temperature and pressure of the fluid particles in a confided container. Supercritical fluids exist in a state above critical point.

I hope this explanation was clear.

8 0

3 years ago

Write the formulas for the following

vredina [299]

The formula of the given compounds are as follows:

a. copper (II) phosphate : Cu₃(PO₄)₂

b. phosphorus trichloride : PCl₃

c. potassium sulfite : K₂SO₃

d. strontium nitride : Sr₃N₂

e. nitrous acid : HNO₂

<h3>What is the formula of a compound?</h3>

The formula of a compound represents the compound using the symbol of the component elements in the compound showing the ratio in which the atoms of the elements combine in the compound.

The formula of the given compounds are as follows:

a. copper (II) phosphate : Cu₃(PO₄)₂

b. phosphorus trichloride : PCl₃

c. potassium sulfite : K₂SO₃

d. strontium nitride : Sr₃N₂

e. nitrous acid : HNO₂

In conclusion, the formula of a compound represents the compound with symbols.

Learn more about formula of compound at: brainly.com/question/13522594

#SPJ1

6 0

1 year ago

if it takes 54 ml of 0.100 M naoh to neutralize 125ml of an hcl solution, what is the concentration of the hcl

vichka [17]

Answer:

c = 0.0432moldm ^−3

Explanation:

The first step would be to find the molar ratio in the reaction. Now generally, one can simplify strong acid-strong base reaction by saying:

Acid+Base ->Salt+ Water

4 0

2 years ago