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4 years ago

I need help with this!!!

Chemistry

1 answer:

scZoUnD [109]4 years ago

5 0

11- Form of energy that can be reflected or emitted from objects through electrical or magnetic waves.

12-Energy that is caused by moving electric charges.

13-Energy stored in the bonds of chemical compounds.

I only know those. Sorry hoped I helped a little! :)

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When 5ml of HCl was added

777dan777 [17]

Answer:

A. There was still 140 ml of volume available for the reaction

Explanation:

According to Avogadro's law, we have that equal volumes of all gases contains equal number of molecules

According to the ideal gas law, we have;

The pressure exerted by a gas, P = n·R·T/V

Where;

n = The number of moles

T = The temperature of the gas

R = The universal gas constant

V = The volume of the gas

Therefore, given that the volumes and number of moles of the removed air and added HCl are the same, the pressure and therefore, the volume available for the reaction will remain the same

There will still be the same volume available for the reaction.

5 0

3 years ago

Which of the following examples illustrates a number that is correctly rounded to three significant figures?

Naily [24]

Answer:

c. 20.0332 g to 20,0 g

Explanation:

A significant figure is each of the digits of a number that are used to express it to the required degree of accuracy, starting from the first non-zero digit, with the exception of the trailing zeros.

Which of the following examples illustrates a number that is correctly rounded to three significant figures?

a. 109 526 g to 109 500 g. NO. The rounded number has 4 significant figures: 109 500.

b. 0.03954 g to 0.040 g. NO. The rounded number has 2 significant figures: 0.040.

c. 20.0332 g to 20.0 g. YES. The rounded number has 3 significant figures: 20.0.

d. 04.05438 g to 4.054 g. NO. The rounded number has 4 significant figures: 4.054.

e. 103.692 g to 103.7g. NO. The rounded number has 4 significant figures: 103.7.

8 0

3 years ago

Which is the correct name for p2o5? phosphorus dioxide phosphorus pentoxide diphosphorus pentoxide diphosphorus hexaoxide

Elenna [48]

Correct option:

The correct name for $P_{2}O_{5}$ is diphosphorus pentoxide.

Why $P_{2}O_{5}$ is called diphosphorus pentoxide?

$P_{2}O_{5}$ is commonly known as diphosphorus pentoxide.

Phosphorus pentoxide has an intriguing property in that $P_{2}O_{5}$ is actually its empirical formula, whereas $P_{4}O_{10}$ is its actual molecular formula.

However, the name of the chemical was obtained from its empirical formula rather than from its molecular formula. The official name for this substance is diphosphorus pentoxide.

Oxygen-containing binary compounds have "oxide" as their "last name." Phosphorus is the "first name."

We list each atom's numbers below:

The di- and Penta- prefixes are used to indicate the presence of two and five oxygen atoms, respectively, in the molecule.

Learn more about diphosphorus pentoxide here,

brainly.com/question/18237346

#SPJ4

6 0

2 years ago

How many grams of calcium phosphate (Ca3(PO4)2) re theoretically produced if we start with 3.40 moles of Ca(NO3)2 and 2.40moles

sattari [20]

1) Balance the chemical equation.

$3Ca(NO_3)_2+2Li_3PO_4\rightarrow6LiNO_3+Ca_3(PO_4)_2$

2) List the known and unknown quantities.

Reactant 1: Ca(NO3)2.

Amount of substance: 3.40 mol.

Reactant 2: Li3PO4.

Amount of substance: 2.40 mol.

Product: Ca3(PO4)2

Mass: unknown.

3) Which is the limiting reactant?

3.1-How many moles of Li3PO4 do we need to use all of the Ca(NO3)2?

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

$mol\text{ }Li_3PO_4=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{2\text{ }mol\text{ }Li_3PO_4}{3\text{ }mol\text{ }Ca(NO_3)_2}=2.2667\text{ }mol\text{ }Li_3PO_4$

We need 2.2667 mol Li3PO4 and we have 2.40 mol Li3PO4. We have enough Li3PO4. This is the excess reactant.

3.2-How many moles of Ca(NO3)2 do we need to use all of the Li3PO4?

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

$mol\text{ }Ca(NO_3)_2=2.40\text{ }mol\text{ }Li_3PO_4*\frac{3\text{ }mol\text{ }Ca(NO_3)_2}{2\text{ }mol\text{ }Li_3PO_4}=3.60\text{ }mol\text{ }Ca(NO_3)_2$

We need 3.60 mol Ca(NO3)2 and we have 3.40 mol Ca(NO3)2. We do not have enough Ca(NO3)2. This is the limiting reactant.

4) Moles of Ca3(PO4)2 produced from the limiting reactant.

We have 3.40 mol Ca(NO3)2 of the limiting reactant.

The molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 mol Ca(NO3)2: 1 mol Ca3(PO4)2.

$mol\text{ }Ca_3(PO_4)_2=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{1\text{ }mol\text{ }Ca_3(PO_4)_2}{3\text{ }mol\text{ }Ca(NO_3)_2}=1.1313\text{ }mol\text{ }Ca_3(PO_4)_2$

5) Mass of Ca3(PO4)2 produced.

The molar mass of Ca3(PO4)2 is 310.1767 g/mol.

$g\text{ }Ca_3(PO_4)_2=1.1333\text{ }mol\text{ }Ca_3(PO_4)_2*\frac{310.1767\text{ }g\text{ }Ca_3(PO_4)_2}{1\text{ }mol\text{ }Ca_3(PO_4)_2}$ $g\text{ }Ca_3(PO_4)_2=351.526\text{ }g\text{ }Ca_3(PO_4)_2$

The mass of Ca3(PO4)2 produced is 351 g Ca3(PO4)2.

Option D.

8 0

1 year ago

A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base

tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M