Which number represetns a coefficient?
2
3
4
7

Answer:

k Nishant

Explanation:

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

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:

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:

look at the explanation

Explanation:

The air that you breathe is a homogeneous mixture of oxygen, nitrogen, argon, and carbon dioxide, along with other elements in smaller amounts. Because each layer of the Earth's atmosphere has a different density, each layer of air is its own homogeneous mixture.

Hope this helps. :)

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]

Best regards!

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:

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

Find more information about ideal gas equation here,

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Answer : The number of moles of precipitate, [tex]PbBr_2[/tex] formed will be 0.01 moles.

Explanation : Given,

Concentration of NaBr = 0.200 M

Volume of solution = 100.0 mL = 0.1 L      (1 L = 1000 mL)

First we have to calculate the moles of NaBr.

[tex]\text{Moles of NaBr}=\text{Concentration of NaBr}\times \text{Volume of solution in L}[/tex]

[tex]\text{Moles of NaBr}=0.200M\times 0.1L=0.02mol[/tex]

Now we have to calculate the moles of precipitate, [tex]PbBr_2[/tex] formed.

The balanced chemical reaction is:

[tex]2NaBr(aq)+Pb(NO_3)_2(aq)\rightarrow PbBr_2(s)+2NaNO_3(aq)[/tex]

From the balanced chemical reaction we conclude that:

As, 2 moles of NaBr react to give 1 mole of [tex]PbBr_2[/tex]

So, 0.02 moles of NaBr react to give [tex]\frac{0.02}{2}=0.01[/tex] mole of [tex]PbBr_2[/tex]

Therefore, the number of moles of precipitate, [tex]PbBr_2[/tex] formed will be 0.01 moles.

The number of mole of the precipitate (i.e PbBr₂) formed when 100 mL of 0.2 M NaBr react with excess Pb(NO₃)₂ is 0.01 mole

Volume = 100 mL = 100 / 1000 = 0.1 L

Molarity of NaBr = 0.2 M

Mole = Molarity x Volume

Mole of NaBr = 0.2 × 0.1

2NaBr(aq) + Pb(NO₃)₂(aq) → PbBr₂(s) + 2NaNO₃ (aq)

From the balanced equation above,

2 moles of NaBr reacted to produce 1 mole of PbBr₂.

Therefore,

0.02 mole of NaBr will react to produce = [tex]\frac{0.02}{2} \\\\[/tex] = 0.01 mole of PbBr₂.

Thus, the number of mole of the precipitate (i.e PbBr₂) produced from the reaction is 0.01 mole

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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:

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:

Explanation:

Here you go!! Hope this helps.

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:

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:

K and ba

Explanation:

Answer: K and Ba

Explanation:

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:

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:

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:

b is the correct answer

do not trust answer one

Explanation:

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:

during sublimation

Explanation:

just took the test

Answer:

Explanation:

Materials are clothes and heavy rope

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.

To learn more about empirical formula, refer to the link below:

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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

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:

B

Explanation:

I did the question before and got it right.

Answer:

19. Option B. ⁰₋₁B

20. Option D. ²¹⁰₈₄Po

Explanation:

19. ²²⁸₈₈Ra —> ²²⁸₈₉Ac + ʸₓZ

Thus, we can determine ʸₓZ as follow:

228 = 228 + y

Collect like terms

228 – 228 = y

y = 0

88 = 89 + x

Collect like terms

88 – 89 = x

x = –1

Thus,

ʸ ₓZ => ⁰₋₁Z => ⁰₋₁B

²²⁸₈₈Ra —> ²²⁸₈₉Ac + ʸₓZ

²²⁸₈₈Ra —> ²²⁸₈₉Ac + ⁰₋₁B

20. ᵘᵥX —> ²⁰⁶₈₂Pb + ⁴₂He

Thus, we can determine ᵘᵥX as follow:

u = 206 + 4

u = 210

v = 82 + 2

v = 84

Thus,

ᵘᵥX => ²¹⁰₈₄X => ²¹⁰₈₄Po

ᵘᵥX —> ²⁰⁶₈₂Pb + ⁴₂He

²¹⁰₈₄Po —> ²⁰⁶₈₂Pb + ⁴₂He

Answer:

Explanation:

In my opinion the answer should be SO2

Answer:

a should be answer i think.