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Answer:

stratosphere

Explanation:

contains a high concentration of ozone in relation to other parts of the atmosphere, although still small in relation to other gases in the stratosphere.

Answer:

stratosphere

Explanation: Most atmospheric ozone is concentrated in a layer in the stratosphere, about 9 to 18 miles (15 to 30 km) above the Earth's surface. Ozone is a molecule that contains three oxygen atoms.

Answer:

Explanation:

Here you go!! Hope this helps.

Answer:

true

Explanation:

I'm not sure why cause I dont know how to explain but it's TRUE

Answer:

True  

Explanation:

The particles moving faster in a substance the hotter it gets.

Answer:

The answer is D - Atom

The smallest unit of an element that can exist either alone or in combination with other such particles of the same or different elements is the atom.

An atom is defined as the smallest unit of matter which forms an element. Every form of matter whether solid,liquid , gas consists of atoms . Each atom has a nucleus which is composed of protons and neutrons and shells in which the electrons revolve.

The protons are positively charged and neutrons are neutral and hence the nucleus is positively charged. The electrons which revolve around the nucleus are negatively charged and hence the atom as a whole is neutral and stable due to presence of oppositely charged particles.

Atoms of the same element are similar as they have number of sub- atomic particles which on combination do not alter the chemical properties of the substances.

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Answer:

C3Br6

Explanation:

C= (1 X 12.011) = 12.011

Br= (2 X 79.904)= 159.808

159.808+12.011 = 171.819

515.46 divided by 171.819 = 3.00

so you mulitpy CBr2 by 3 which gives you C3Br6

Answer:

B

Explanation:

I did the question before and got it right.

Answer:

3.65 mol O₂

Explanation:

Step 1: RxN

2H₂O₂ → 2H₂O + O₂

Step 2: Define

Given - 7.30 mol H₂O₂

Solve - x mol O₂

Step 3: Stoichiometry

[tex]7.30 \hspace{3} mol \hspace{3} H_2O_2(\frac{1 \hspace{3} mol \hspace{3} O_2}{2 \hspace{3} mol \hspace{3} H_2O_2} )[/tex] = 3.65 mol O₂

Answer:

during sublimation

Explanation:

just took the test

Answer:

I'm on the exact same queston

Answer:

The test variable (independent variable) controls the outcome variable (dependent variable)

Explanation:

its right on study island

Answer:

The Group 2 metals become more reactive towards the water as you go down the Group.

Explanation:

These all react with cold water with increasing vigour to give the metal hydroxide and hydrogen. ... You get less precipitate as you go down the Group because more of the hydroxide dissolves in the water. Summary of the trend in reactivity.

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Answer:

molar mass of unknown gas = 1.987 g/mol

Explanation:

First, the number of moles of the unknown gas is found

Using the ideal gas equation: PV = nRT

P = 1.00 atm, V = 5.00 L, T = 298.15 K, R = 0.082 L.atm.mol⁻¹K⁻¹

n = PV/RT

n = (1.00 atm * 5.00 L)/(298.15 K *0.082 L.atm.mol⁻¹K⁻¹)

n = 0.2109 moles

Molar mass = mass/ number of moles

molar mass = 0.419 g/ 0.2109 mols

molar mass of unknown gas = 1.987 g/mol

The molar mass of unknown gas by using ideal gas equation = 1.987 g/mol.

This equation gives the relation between pressure, volume, temperature as given below:

[tex]PV = nRT[/tex]

P = 1.00 atm, V = 5.00 L, T = 298.15 K, R = 0.082 L.atm.mol⁻¹K⁻¹

Substitute the above values in the above equation as follows:

n = (1.00 atm * 5.00 L)/(298.15 K *0.082 L.atm.mol⁻¹K⁻¹)

n = 0.2109 moles

[tex]Molar mass = mass/ number of moles[/tex]

Calculate molar mass by using the above equation,

molar mass = 0.419 g/ 0.2109 moles

The molar mass of unknown gas = 1.987 g/mol

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Answer:

A chemical formula tells us the number of atoms of each element that is in a compound. It contains the symbols of the atoms for the elements present in the compound as well as how many there are for each element in the form of subscripts.

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Answer:

[tex]Total = 50.6\ moles[/tex]

Explanation:

Given

[tex]Propane = C_3H_8[/tex]

Represent Carbon with C and Hydrogen with H

[tex]C = 13.8[/tex]

Required

Determine the total moles

First, we need to represent propane as a ratio

[tex]C_3H_8[/tex] implies

[tex]C:H = 3:8[/tex]

So, we're to first solve for H when [tex]C = 13.8[/tex]

Substitute 13.8 for C

[tex]13.8 : H = 3 : 8[/tex]

Convert to fraction

[tex]\frac{13.8}{H} = \frac{3}{8}[/tex]

Cross Multiply

[tex]3 * H = 13.8 * 8[/tex]

[tex]3 H = 110.4[/tex]

Solve for H

[tex]H = 110.4/3[/tex]

[tex]H = 36.8[/tex]

So, when

[tex]C = 13.8[/tex]

[tex]H = 36.8[/tex]

[tex]Total = C + H[/tex]

[tex]Total = 13.8 + 36.8[/tex]

[tex]Total = 50.6\ moles[/tex]

Answer:

Chromatography involves solvent separation on a solid medium.

Distillation takes advantage of differences in boiling points.

Evaporation removes a liquid from a solution to leave a solid material.

Filtration separates solids of different sizes.

Explanation:

Answer:

Explanation:

Half life of radioactive materials do not depend upon the mass of the material . It only depends upon the nature of radioactive materials . The half life of 20 g is 232 seconds . That means 20 gram will be reduced to 10 gram in 232 seconds .

Half life of 80 gram is also 232 seconds . So , 80 gram will be reduced to 40 gram in 232 second .

Answer:

atom

Explanation:

the atom is the smallest form.

Answer:

k Nishant

Explanation:

i don't know sorry but u can search in google

Answer:

[tex]Kc=\frac{[SF_6]^8}{[F_2]^2^4}[/tex]

Explanation:

Hello.

In this case, for the undergoing chemical reaction:

[tex]S_8(s) + 24F_2(g) \rightleftharpoons 8SF_6(g)[/tex]

We consider the law of mass action in order to write the equilibrium expression yet we do not include S8 as it is solid and make sure we power each gaseous species to its corresponding stoichiometric coeffient (24 for F2 and 8 for SF6), thus we obtain:

[tex]Kc=\frac{[SF_6]^8}{[F_2]^2^4}[/tex]

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Answer:

Explanation:

Kb of  (CH₃)₃N is 7.4 x 10⁻⁵

initial concentration of (CH₃)₃N   a   is .48 M

(CH₃)₃N    +   H₂O =  (CH₃)₃NH⁺  +  OH⁻

a - x                                     x               x  

x² / (a - x )  = Kb

x is far less than a so a - x can be replaced by a .

x² / a   = Kb

x²  = a x Kb = .48 x 7.4 x 10⁻⁵ = 3.55 x 10⁻⁵ = 35.5 x 10⁻⁶

x = 5.96 x 10⁻³

pOH = - log ( 5.96 x 10⁻³ )

= 3 - log 5.96

= 3 - .775

= 2.225

Answer:

you can see the empirical formula at the pic

The empirical formula for compound (a) is NiI2, (b) is BaBr2 and (c) is BeS.

Empirical formula of a compound is defined as the simplest whole number ratio of atoms present in a compound.

(a) 1.245 g Ni : 5.381 g I

Mole of Ni ; Mole of I = 1.245/59 : 5.381/127 = 0.02 : 0.04 = 1:2

So the formula is NiI2

(b) 2.677 g Ba : 3.115 g Br

Mole of Ba : Mole of Br = 2.677/137 : 3.115/60 = 0.019 : 0.038

= 0.02 : 0.04 = 1:2

So the formula is BaBr2

(c) 2.128 g Be : 7.557 g S

Mole of Be : Mole of S = 2.128/9 : 7.557/32 = 0.2 : 0.2 = 1:1

So the formula is BeS

Thus, empirical formula for compound (a) is NiI2, (b) is BaBr2 and (c) is BeS.

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Answer:

The correct answer is 0.67 g H₂

Explanation:

Isopropyl alcohol (C₃H₇OH) can decompose to give acetone (C₂H₆OH) and hydrogen gas (H₂) according to the following chemical equation:

C₃H₇OH (g) ⇒ C₂H₆CO(g) + H₂(g)

We can calculate the initial mass of isopropyl alcohol from the density and volume data:

density = m/V = 0.785 g/mL

⇒ m = density x V = 0.785 g/mL x 25.6 mL = 20.096 g C₃H₇OH

According to the chemical equation 1 mol of C₃H₇OH gives 1 mol H₂. The molar mass of C₃H₇OH is:

molar mass C₃H₇OH = (12 g/mol x 3) + (1 g/mol x 7) + 16 g/mol + 1 g/mol = 60 g/mol

molar mass H₂ = 1 g/mol x 2 = 2 g/mol

So, we obtain: 2 g H₂ from 60 g C₃H₇OH. We multiply this stoichiometric ratio (2 g H₂/60 g C₃H₇OH) by the initial mass of C₃H₇OH to obtain the mass of H₂ is formed:

20.096 g C₃H₇OH x (2 g H₂/60 g C₃H₇OH) = 0.6698 g ≅ 0.67 g H₂

Answer:

b is the correct answer

do not trust answer one

Explanation:

Answer:

seafloor spreading

Explanation:

i took the test

Answer:

Explanation:

In my opinion the answer should be SO2

Answer:

a should be answer i think.

Answer:

266Lr

Thirteen isotopes of lawrencium are currently known; the most stable is 266Lr with a half-life of 11 hours, but the shorter-lived 260Lr (half-life 2.7 minutes) is most commonly used in chemistry because it can be produced on a larger scale.

Explanation:

hopefully that helps you