Which is warmer -30C or -30 F?​

Answer:

5.0 × 10⁻⁶

Explanation:

Step 1: Write the balanced equation for the solution of chromium(III) iodate

Cr(IO₃)₃(s) ⇄ Cr³⁺(aq) + 3 IO₃⁻(aq)

Step 2: Calculate the solubility product constant (Ksp)

To relate Ksp and the solubility (S), we will make an ICE chart.

        Cr(IO₃)₃(s) ⇄ Cr³⁺(aq) + 3 IO₃⁻(aq)

I                               0                 0

C                             +S              +3S

E                                S               3S

The solubility product constant is:

Ksp = [Cr³⁺] × [IO₃⁻]³ = S × (3S)³ = 27 S⁴ = 27 × (2.07 × 10⁻²)⁴ = 5.0 × 10⁻⁶

Answer:

66.67%

Explanation:

From the given information:

mass of cyclohexane = 2.9949 grams

density of cyclohexane = 0.779 g/mL

Recall that:

Density = mass/volume

∴

Volume = mass/density

So, the volume of cyclohexane = 2.9949 g/ 0.779 g/mL

= 3.8445 mL

Also,

mass of propylbenzene = 1.6575 grams

density of propylbenzene = 0.862 g/mL

Volume of propylbenzene =  1.6575 g/ 0.862 g/mL

= 1.9229 mL

The volume % composition of cyclohexane from the mixture is:

[tex]= (\dfrac{v_{cyclohexane}}{v_{cyclohexane}+v_{propylbenzene}})\times 100[/tex]

[tex]= (\dfrac{3.8445}{3.8445+1.9229})\times 100[/tex]

[tex]= (\dfrac{3.8445}{5.7674})\times 100[/tex]

= 66.67%

Answer:

Solution temperature.

Explanation:

Hello there!

In this case, considering this question about chemical kinetics, it will be possible for us to analyze two perspectives:

1. Molecular: here, we infer that the solution temperature will provide more energy to the molecules in order to undergo more effective crashes which will make more products and therefore, increase the rate constant.

2. Mathematical: by means of the Arrhenius equation, it will be possible to tell that the increase in the temperature of the system, the negative of the exponent present in such equation will increase and therefore turn the rate constant bigger.

In such way, we infer the answer is solution temperature.

Regards!

Answer:

a measure of concentration equal to the gram equivalent weight per liter of solution.

Explanation:

Gram equivalent weight is the measure of the reactive capacity of a molecule. The solute's role in the reaction determines the solution's normality. Normality is also known as the equivalent concentration of a solution.

hope it helped

1.42liters, which is equivalent to 3pints, of blood is required for the surgery

1 US pint = 0.473 liters

3 × 0.473

= 1.419 liters of blood for the surgery

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The pressure of hydrogen gas is 607.7 mmHg

According of Dalton's law of Partial pressure, the total pressure of a mixture of gases is the sum of the partial pressures of the individual gases in the mixture.

We can now write;

The for hydrogen collected over water, we have a mixture of hydrogen gas and water vapour.

Total pressure = pressure of hydrogen gas + vapour pressure of water

Pressure of hydrogen gas = Total pressure - vapour pressure of water

Pressure of hydrogen gas =  636 mmHg - 28.3 mmHg

Pressure of hydrogen gas = 607.7 mmHg

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Answer:

F 64.6 percent of carbon may be

Explanation:

Add water, Na2CO3 dissolves, filter, PbCO3 stays in the paper and dissolved Na2CO3 goes through as the solution. Dry the PbCO3 and you have the dry solid.

OR

Add water to dissolve then filter to obtain PbCo3 as you're residue and Na2Co3 as the filtrate. Dry the insoluble PbCo3 between filter papers and you obtain solid PbCo3

Answer:

3.89 ×10^-5

Explanation:

Since they are gaseous reactants, we obtain the equilibrium constant from the given partial pressures;

p(H2O) = 0.0900 bar

p(H2) = 0.00100 bar

p(O2) = 0.00350 bar

The equation of the reaction is;2H2O(g)⇄2H2(g)+O2(g)

Kp= p(H2) . p(O2)/p(H2O)

Kp= 0.00100 × 0.00350/0.0900

Kp= 3.89 ×10^-5

The question is incomplete, the complete question is;

When determining the amount of an oxidant present by titration, you can use iodine and starch as an indicator.

First, the oxidant, like hypochlorite, oxidizes

Choose...

neutral iodine into iodide ion

iodide ion into neutral iodine

iodate polyatomic ion into iodide ion

When starch and iodine are both present, the solution is

Choose...

blue-black

brownish yellow

clear

During the titration, the titrant, like thiosulfate, reduces the

Choose...

iodide ion into iodate polyatomic ion

neutral iodine into iodide ion

iodide ion into neutral iodine

When the iodine has completely reacted at the endpoint of the titration, the solution should become

Choose...

clear

blue-black

brownish yellow

Answer:

1. iodide ion into neutral iodine

2. blue-black

3. neutral iodine into iodide ion

4. clear

Explanation:

Hypochlorite oxidizes the iodide ion to iodine molecule according to the reaction equation;

ClO-(aq) + 2H+(aq) + 2I-(aq) ---------> 6 I2(l) + Cl- (aq)+ H2O(l)

When iodine is added, the colour of the starch solution immediately changes to blue-black.

A reduction reaction occurs when the titrant, thiosulfate is added as follows;

I2 + 2S2O32- → 2I- + S4O62-

The solution at end point is found to become clear again.

Answer:

"[Temperature is a measurement of the average kinetic energy of the molecules in an object or a system. Kinetic energy is the energy that an object has because of its motion. The molecules in a substance have a range of kinetic energies because they don't all move at the same speed.]"

Answer:

Temperature is directly proportional to the average translational kinetic energy of molecules in an ideal gas

Explanation:

Answer:

a) 22 calories

b) 30 mg

Explanation:

Divide number of cal or mg of pot by 8 fl oz.

Answer:

d. 60.8 L

Explanation:

Step 1: Given data

Step 2: Calculate the work (W) done by the system

We will use the following expression.

ΔU = Q + W

W = ΔU - Q

W = -108.3 J - 53.1 J = -161.4 J

Step 3: Convert W to atm.L

We will use the conversion factor 1 atm.L = 101.325 J.

-161.4 J ×  1 atm.L/101.325 J = -1.593 atm.L

Step 4: Calculate the initial volume

First, we will use the following expression.

W = - P × ΔV

ΔV = - W / P

ΔV = - 1.593 atm.L / 0.677 atm = 2.35 L

The initial volume is:

V2 = V1 + ΔV

V1 = V2 - ΔV

V1 = 63.2 L - 2.35 L = 60.8 L

how many electrons does tin have?

Answer : 2, 8, 18, 18, 4

Therefore, total electrons = 50

Answer:

26.0L is the volume of the sample after the reaction

Explanation:

Based on the reaction, 1 mole of CH4 reacts with 1 mole of H2O to produce 1 mole of CO and 3 moles of H2.

That is, 1 mole of each reactant produce 4 moles of gases

As in the reaction, 0.1500 moles of CH4 and 0.1500 moles of H2O are added, 0.1500 moles of CO and 0.4500 moles of H2 are produced.

Before the reaction, the moles of gas are 0.3000 moles and after the reaction the moles are 0.6000 moles of gas.

Based on Avogadro's law, the moles of a gas are directly proportional to the volume under temperatura and pressure constant. The equation is:

V1/n1 = V2/n2

Where V is volume and n are moles of 1, initial state and 2, final state.

Replacing:

V1 = 13.0L

n1 = 0.3000 moles

V2 = ?

n2 = 0.6000 moles

13.0L*0.6000 moles / 0.3000 moles = V2

V2 = 26.0L is the volume of the sample after the reaction

Answer:

The options are incorrect.................

Explanation:

The Oxidation no. is +3

Answer:

Volume = 1222.5cm³

Explanation:

If the question is about the volume of the rectangle:

The volume of a rectangle is obtained by the multiplication of its 3 dimensions: Length, width, height.

In the problem, the length of the rectangle is 0.162m = 16.2cm

The width is 7.7cm

And the height is 9.8cm

The volume is:

Volume = 16.2cm*7.7cm*9.8cm

Answer:

The concentration of KOH is  0.06137 M

Explanation:

Step 1: Data given

Molar mass of KHP = 204.22 g/mol

Mass of KHP = 0.4877 grams

Volume of water = 50 mL

Volume of KOH solution = 38.91 mL

Step 2: The balanced equation

C8H5KO4 + KOH ⇒ C8H4K2O4+ H2O

Step 3: Calculate number of moles of KHP

Moles = Mass / molar mass

Moles KHP = 0.4877 grams / 204.22 g/mol

Moles KHP = 0.002388 moles

Step 4: Calculate moles of KOH

For 1 mol KHP we need 1 mol KOH to produce 1 mol C8H4K2O4 and 1 mol H2O

For 0.002388 moles KHP we need 0.002388 moles KOH

Step 5: Calculate the concentration of KOH

Concentration = moles / volume

Concentration of KOH = 0.002388 moles / 0.03891 L

Concentration of KOH = 0.06137 M

The concentration of KOH is  0.06137 M

Answer:

See Explanations

Explanation:

pH =-log[H₃O⁺] = -log[H⁺]

pOH = -log[OH⁻]

For weak acids [H⁺] = SqrRt(Ka·[Acid])

For weak bases [OH⁻] = SqrRt(Kb·[Base])

pH + pOH = 14

__________________________________________

A. Given 0.18M CH₃NH₂; Kb = (4.4 x 10⁻⁴)* => pH = 11.95

CH₃NH₂ + H₂O => CH₃NH₃OH ⇄ CH₃NH₃⁺ + OH⁻;

[OH⁻]  = SqrRt(Kb·[weak base]) = SqrRt(4.4 x 10⁻⁴ x 0.18)M = 8.97 x 10⁻³M

=> pOH = -log[OH⁻] = -log(8.93x10⁻³) = -(-2.05) = 2.05

=> pH = 14 - pOH = 14 - 2.05 = 11.95.

*Kb values for most ammonia derivatives in water can be found online by searching 'Kb-values for weak bases'. Kb-values for methyl amine and methylammonium chloride are both 4.4x10⁻⁴.

___________________________________________________

B. Given 0.18M CH₃NH₃Cl

In water ... CH₃NH₃Cl => CH₃NH₃⁺ + Cl⁻; Kb(CH₃NH₃Cl) = 4.4 x 10⁻⁴

Cl⁻ + H₂O => No Rxn (i.e.; no hydrolysis occurs) ... Cl⁻ does not react with H₂O.

Hydrolysis Reaction of Methylammonium Ion:

CH₃NH₃⁺ + H₂O => CH₃NH₄OH ⇄ CH₃NH₄⁺ + OH⁻

Ka' x Kb = Kw => Ka' = Kw/Kb = 10⁻¹⁴/4.4 x 10⁻⁴ = 2.27 x 10⁻¹¹                                   Ka' = [CH₃NH₄⁺][OH⁻]/[CH₃NH₄OH] = (x)(x)/(0.18M) = (x²/0.18M) = 2.27 x 10⁻¹¹ => x = [OH⁻] = SqrRt(2.27x10⁻¹¹ x 0.18)M = 2.02 x 10⁻⁶M => pOH = -log(2.02 x 10⁻⁶) = -(-5.69) = 5.69 => pH = 14 - pOH = 14 - 5.69 = 8.31.

*note => the general nature of halide interactions would increase acidity (lower pH) of the halogenated compound.

C. A mixture of 0.18M CH₃NH₂ and 0.18M CH₃NH₃Cl          

Mixture of 0.18M CH₃NH₂ + 0.18M CH₃NH₃Cl

In Water ...

=> 0.18M CH₃NH₃OH + 0.18M CH₃NH₃Cl

=> 0.18M CH₃NH₃⁺ + 0.1M OH⁻ + 0.18M CH₃NH₃⁺ + 0.18M Cl⁻

=> 0.36M CH₃NH₃⁺ + 0.18M OH⁻ + 0.18M Cl⁻

-----------------------------------------------------------

Ka'(CH₃NH₃⁺) x Kb(CH₃NH₂) = Kw => Ka'(CH₃NH₃⁺) = Kw/Kb(CH₃NH₂)

=> Ka'(CH₃NH₃⁺) = (10⁻¹⁴/4.4x10⁻⁴) = 2.27x10⁻¹¹

----------------------------------------------------------

From the 0.36M CH₃NH₃⁺

=>       CH₃NH₃⁺ + H₂O  ⇄ CH₃NH₄⁺ + OH⁻

C(eq)   0.36M        ----              x             x     (<= at equilibrium after mixing)

Ka'(CH₃NH₃⁺) = [CH₃NH₄⁺][OH⁻]/[CH₃NH₃⁺] = x²/(0.36M)

=> x = [OH⁻] = SqrRt(Ka'(CH₃NH₃⁺)·0.36M) = SqrRt(2.27x10⁻¹¹/0.36) = 0.0126M

=> Total [OH⁻] = 0.0126M + 0.18M = 0.1926M from hydrolysis process

=> final solution mix is therefore, 0.1926M in OH⁻ + 0.18M in Cl⁻

--------------------------------------------------------

∴pH = -log[H⁺] = -log(5.192x10⁻¹⁴) = -(-13.29) = 13.29 for solution mix

The acid and base dissociation constant and the 0.18 M of CH₃NH₂ and

CH₃NH₃Cl and the mixture give the following approximate values;

A. The pH value of the 0.18 M CH₃NH₂ is 11.93

B. The pH value of the 0.18 M CH₃NH₃Cl is 5.69

C. The pH value of the mixture is 10.644

A. 0.18 M CH₃NH₂

The solution is presented as follows;

CH₃NH₂ + H₂O → CH₃NH₃⁺ + OH⁻

Let x represent the number of moles of CH₃NH₃⁺ and OH⁻ produced, we

have;

The number of moles of CH₃NH₂ remaining = 0.18 - x

Which gives;

[tex]K_b = \mathbf{\dfrac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}}[/tex]

[tex]K_b[/tex] for CH₃NH₂ = 4.167 × 10⁻⁴

Therefore;

[tex]4.167 \times 10^{-4} = \mathbf{\dfrac{x \times x}{0.18 - x}}[/tex]

4.167 × 10⁻⁴ × (0.18 - x) = x²

4.167 × 10⁻⁴ × (0.18 - x) - x² = 0

Which gives;

x = [OH⁻] = 8.455 × 10⁻³

pH = 14 + log[OH⁻]

Which gives;

pH = 14 + log(8.455 × 10⁻³) ≈ 11.93

B.  0.18 M CH₃NH₃Cl

The solution is presented as follows;

CH₃NH₃⁺ → CH₃NH₂ + H⁺

Let x represent the number of moles of CH₃NH₂ and H⁺ produced,

respectively, we have;

The number of moles of CH₃NH₃⁺ remaining = 0.18 - x

Which gives;

[tex]K_a = \mathbf{\dfrac{[CH_3NH_2][H^+]}{[CH_3NH_3^+]}}[/tex]

Kₐ for CH₃NH₃Cl = 2.27 × 10⁻¹¹

Therefore;

[tex]2.27\times 10^{-11} = \dfrac{x \times x}{0.18 - x}[/tex]

2.27 × 10⁻¹¹ × (0.18 - x) = x²

2.27 × 10⁻¹¹ × (0.18 - x) - x² = 0

Which gives;

x = [H⁺] ≈ 2.02 × 10⁻⁶

pH = -log[H⁺]

Which gives;

pH = -log(2.02 × 10⁻⁶) ≈ 5.69

C. For the mixture of 0.18 M CH₃NH₂ and 0.18 M of CH₃NH₃Cl, we have;

Based on the Henderson-Hasselbalch equation, we have;

[tex]pH = \mathbf{ pKa + log\dfrac{[Conjugate \ base]}{[acid ]}}[/tex]

Which gives;

[tex]pH = -log\left(2.27 \times 10^{-11} \right)+ log\dfrac{0.18}{0.18} \approx \underline{10.644}[/tex]

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Answer:

3.8g of H2O are produced

Explanation:

The balanced reaction of the problem is:

CH4 + 2O2 → CO2 + 2H2O

Where 1 mole of CH4 reacts with 2 moles of O2

To solve this question we need to find, as first, the moles of each reactant in order to find limiting reactant. With limiting reactant we can find the moles of H2O produced and its mass as follows:

Moles CH4 - 16.04g/mol-

2.73g * (1mol/16.04g) = 0.170 moles CH4

Moles O2 -32g/mol-

6.7g (1mol/32g) = 0.209 moles O2

For a complete reaction of 0.170 moles of CH4 are needed:

0.170 moles CH4 * (2 mol O2 / 1mol CH4) = 0.340 moles O2

As there are just 0.209 moles of O2, oxygen is limiting reactant

The moles of water produced are:

0.209 moles O2 * (2mol H2O / 2mol O2) = 0.209 moles H2O

Mass water -Molar mass: 18.01g/mol-

0.209 moles H2O * (18.01g/mol) = 3.8g of H2O are produced

Answer:

[tex]\boxed {\boxed {\sf 0.566 \ mol \ Fe}}[/tex]

Explanation:

We are asked to convert a number of atoms to moles.  

We can convert atoms to moles using Avogadro's Number, which is 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this problem, the particles are atoms of iron (Fe). There are 6.022 ×10²³ atoms of iron in 1 mole of iron.  

We use dimensional analysis to convert atoms to moles. This involves setting up ratios. Use Avogadro's Number and the underlined information to make a ratio.

[tex]\frac {6.022 \times 10^{23}\ atoms \ Fe}{1 \ mol \ Fe}[/tex]

We are converting 3.41 × 10²³ atoms of iron to moles, so we multiply by this value.

[tex]3.41 \times 10^{23} \ atoms \ Fe *\frac {6.022 \times 10^{23}\ atoms \ Fe}{1 \ mol \ Fe}[/tex]

Flip the ratio. It stays equivalent, but it allows the units of atoms of iron to cancel.

[tex]3.41 \times 10^{23} \ atoms \ Fe *\frac{1 \ mol \ Fe} {6.022 \times 10^{23}\ atoms \ Fe}[/tex]

[tex]3.41 \times 10^{23}*\frac{1 \ mol \ Fe} {6.022 \times 10^{23}}[/tex]

[tex]\frac{3.41 \times 10^{23}} {6.022 \times 10^{23}} \ mol \ Fe[/tex]

[tex]0.5662570575\ mol \ Fe[/tex]

The original measure ment of iron atoms ( 3.41 × 10²³ ) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandths place. The 2 in the ten-thousandths place ( 0.5662570575) tells us to leave the 6 in the thousandths place.

[tex]0.566 \ mol \ Fe[/tex]

3.41 × 10²³ atoms of iron is equal to approximately 0.566 moles of iron.

Answer:

a) The energy transferred is 6.91 kJ

b) The internal energy is 4.90 kJ

c) The work done on the gas is - 2.01 kJ

Explanation:

Step 1: Data given

Number of moles of hydrogen gas = 2.00 moles

Pressure = constant

Temperature is heated from 294 K to 414 K

Molar heat capacity of hydrogen gas = 28.8 J/mol*K

Step 2: Calculate the energy transferred to the gas by heat.

Q = n* Cp * ΔT

⇒with Q =the energy transferred

⇒with n = the number of moles = 2.00 moles

⇒with Cp = the Molar heat capacity of hydrogen gas = 28.8 J/mol*K

⇒ with ΔT = Temperature 2 - Temperature 1 = 414 - 294 = 120K

Q = 2.00 * 28.8 * 120

Q = 6912 J = 6.91 kJ

Step 3: Calculate the increase in its internal energy.

ΔEint = n*Cv*ΔT

⇒with ΔEint = the increase in its internal energy.

⇒with n = the number of moles = 2.00 moles

⇒with Cv = The constant volume = 20.4 J/mol*K

⇒with  ΔT = Temperature 2 - Temperature 1 = 414 - 294 = 120K

ΔEint = 2.00 * 20.4 * 120

ΔEint =4896 J = 4.90 kJ

Step 4: Calculate the work done on the gas.

Work done on the gas = -Q + ΔEint

W = -6.91 kJ + 4.90 kJ

W = -2.01 kJ

Answer:

A. 1.9 × 103 g

(next one)

Which metric unit would be the best choice to report the result?  

A. kg

Answer:

1. 2

2. 1.9 × 10^3 g

3. kg

Explanation:

Answer:

AgNO3 + NaOH = AgOH + NaNO3.

Explanation:

Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.

Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.However, the equation balanced here is the initial reaction which produces AgOH and NaNO3.

Answer:

[H₃O⁺] = [H⁺] = 10^-pH = 10⁻¹°⁸⁶ = 0.0138M in [H⁺]  

Explanation:

By definition pH = -log[H⁺] => [H⁺] = 10^-pH = 10⁻¹°⁸⁶ = 0.0138M in [H⁺]

Using your calculator ... I am using a TI-30XA scientific calculator.

Depending on the calculator, the answer may show as 0.0138, or  1.38x10⁻², or 1.38E-2 (=1.38 x 10⁻²). It is the user's job to insert dimensional units into answer of interest =>  0.0138M, or  1.38 x 10⁻²M, or 1.38E-2M.

1.38E-2 which is 1.38 x 10⁻².

Answer:

Vanlency of hydrogen - 11

Electrons of hydrogen - 1

Answer:

The answer is: 1

Hope this helps :) <3

Explanation:

Answer:

d

Explanation:

Answer:

12

Explanation:

2+8+2=12

atomic no is the No of protons

Answer:

Atomic number is 12.

Explanation:

Atomic number = electrons in filled shells + outermost electrons

= 2 + 8 +2

= 12

Answer:

See explanation

Explanation:

The particular reactants in the Fischer esterification reaction were not stated.

Generally, a Fischer esterification is a reaction that proceeds as follows;

RCOOH + R'OH ⇄RCOOR' + H2O

This reaction occurs in the presence of an acid catalyst.

We can shift the equilibrium of this reaction towards the products side in two ways;

I) use of a large excess of either of the reactants

ii) removal of one of the products as it is formed.

Any of these methods shifts the equilibrium of the Fischer esterification reaction towards the products side.