The number 0.0007270 is larger than the number 5.7 × 10–3.

Answer:

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Answer:

There will be very little of BrOCl BrCl

Explanation:

Based on the equilibrium:

Br2(g) + OCl2(g) ⇄ BrOCl(g) + BrCl(g)

The equilibrium constant, Kc, is:

Kc = 1.58x10⁻⁵ = [BrOCl] [BrCl] / [Br2] [OCl2]

As Kc is <<< 1, in equilibrium, the concentration of products will remain lower regard to the concentration of the reactants. That means, right answer is;

Answer:

n = 3, l = 2, ml =-2

Explanation:

Quantum numbers are a set of values which can be used to describe the energy and position of an electron in space.

There are four sets of quantum numbers;

1) principal quantum number

2) orbital quantum number

3) spin quantum number

4) magnetic quantum number.

The values of orbital quantum number include; -l to +l;

The set of quantum numbers without error is ; n = 3, l = 2, ml =-2

Answer:

Chloride ion is a strong nucleophile and bromide is a good leaving group. The major product of treating (S)-2-bromobutane with NaCl in CH3C(O)CH3 (acetone) is _________. (1S,2R)-1-chloro-2-bromobutane cis-2-butene (1R,2S)-1-chloro-2-bromobutane (S)-2-chlorobutane trans-2-butene (R)-2-chlorobutane

Explanation:

The reaction of (S)-2-bromobutane with NaCl in CH3C(O)CH3 (acetone) forms the following product:

The answer is (R)-2-chlorobutane.

The reaction take splace through [tex]S_{N} _2[/tex] mechansim and inversion in configuration happens.

Answer:

0.111 g

Explanation:

1 g = 1000 mg

Doctor ordered the following concentration of Claforan:

C = 1 g/100 mL x 1000 mg/1 g = 10 mg/mL

If we add 2 g iof Claforan, we obtain:

2 g Claforn ---- 180 mg/mL Claforan

To reach a concentration equal to C (10 mg/mL), we need:

10 mg/mL Claforan x 2 g Claforn/(180 mg/mL Claforan) = 0.111 g Claforn

Therefore, we have to add 0,111 g (111 mg) of Claforn to the bag of 100 ml D5W to obtain the ordered concentration of 10 mg/mL Claforan.  

Answer: The total partial pressure of the solution is 131.37 torr.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of glucose = 46.8 g

Molar mass of glucose = 180 g/mol

Plugging values in equation 1:

[tex]\text{Moles of glucose}=\frac{46.8g}{180g/mol}=0.26 mol[/tex]

Given mass of methanol = 117 g

Molar mass of methanol = 32 g/mol

Plugging values in equation 1:

[tex]\text{Moles of methanol}=\frac{117g}{32g/mol}=3.66 mol[/tex]

Mole fraction is defined as the moles of a component present in the total moles of a solution. It is given by the equation:

[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex] .....(2)

where n is the number of moles

Putting values in equation 2:

[tex]\chi_{methanol}=\frac{3.66}{0.26+3.66}=0.934[/tex]

Raoult's law is the law used to calculate the partial pressure of the individual gases present in the mixture. The equation for Raoult's law follows:

[tex]p_A=\chi_A\times p_T[/tex] .....(3)

where [tex]p_A[/tex] is the partial pressure of component A in the mixture and [tex]p_T[/tex] is the total partial pressure of the mixture

We are given:

[tex]p_{methanol}=122.7torr\\\chi_{methanol}=0.934[/tex]

Putting values in equation 3, we get:

[tex]122.7torr=0.066\times p_T\\\\p_T=\frac{122.7torr}{0.934}=131.37torr[/tex]

Hence, the total partial pressure of the solution is 131.37 torr.

Answer: There are [tex]1.23 \times 10^{22}[/tex] atoms present in 37.1 mg of tantalum.

Explanation:

Given: Mass of single tantalum atom = [tex]3.01 \times 10^{-22} g[/tex]

Mass of tantalum atoms = 37.1 mg (1 mg = 0.001 g) = 0.0371 g

Therefore, number of tantalum atoms present in 0.0371 grams is calculated as follows.

[tex]No. of atoms = \frac{0.0371 g}{3.01 \times 10^{-22}}\\= 1.23 \times 10^{20}[/tex]

Thus, we can conclude that there are [tex]1.23 \times 10^{22}[/tex] atoms present in 37.1 mg of tantalum.

There are [tex]1.23\times 10^{20}[/tex] atoms of tantalum in 37.1 mg of tantalum.

Explanation:

Given:

Mass of single atom of tantalum =[tex]3.01\times 10^{-22} g[/tex]

To find:

The number of atoms of tantalum in 37.1 milligrams.

Solution:

Mass of tantalum = 37.1 mg

[tex]1 mg = 0.001 g\\37.1 mg=37.1\times 0.001 g=0.0371 g[/tex]

The number of atoms in 0.0371 grams of tantalum = N

Mass of a single atom of tantalum = [tex]3.01\times 10^{-22} g[/tex]

Then a mass of N atoms of tantalum will be:

[tex]0.0371 g=N\times 3.01\times 10^{-22} g\\N=\frac{0.0371 g}{ 3.01\times 10^{-22} g}\\=1.23\times 10^{20}[/tex]

There are [tex]1.23\times 10^{20}[/tex] atoms of tantalum in 37.1 mg of tantalum.

Learn more about the unitary method here:

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Pyruvate Kinase

Pyruvate Kinase performs a substrate level phosphorylation on ADP to generate an ATP and pyruvate, the final product of glycolysis.

PK dificiency is transmitted in an autosomal recessive disorder in which both alleles must contain the mutated gene, PK-LR.

Hope it helps you! ＼(^ᴥ^)／

Answer:

Fill in the blanks with each titration term with its definition.

a. Solution of an unknown concentration that has another solution slowly added to it ________________

b. Process of slowly adding a solution to react with another solution and determine the concentration of one of the solutions based on the reaction between them ______________

c. A reagent added to the analyte solution that changes color when the reaction is complete ______________

d. Glassware that allows a solution to be precisely and slowly added to another solution _____________

e. Solution of known concentration that is slowly added to a solution of unknown concentration ________________

f. When the required amount of one solution has been added to the second solution to complete the reaction ____________

Explanation:

a. Solution of an unknown concentration that has another solution slowly added to it is called analyte.

b. Process of slowly adding a solution to react with another solution and determine the concentration of one of the solutions based on the reaction between them is called titration.

c. A reagent added to the analyte solution that changes color when the reaction is complete is called an indicator.

d. Glassware that allows a solution to be precisely and slowly added to another solution is called a pipette.

e. Solution of known concentration that is slowly added to a solution of unknown concentration is called titrant.

f. When the required amount of one solution has been added to the second solution to complete the reaction is called neutralization.

Answer:

so 0.15 moles X 22.4 dm3/mole=3.36 dm3. Next we find the moles of hexane combusted, and then the moles of CO2. Finally, we find the volume of CO2 using the fact that at STP, 1 mole of gas = 22.4 dm3.

Answer:

green gemstone

Explanation:

hope this helps someone

Answer:

CF₂

Explanation:

Let's assume we have 100 g of the gas. If that were the case we'd have

Now we convert both masses into moles, using their respective molar mass:

We can express those results as C₂F₄.

To determine the empirical formula we reduce those coefficients to the lowest possible integers, leaving us with CF₂.

Answer:

Huh!?

Explanation:

explain me please

Answer:

See explanation

Explanation:

First we divide the percentage by mass of each element by it's relative atomic mass then we divide the quotients obtained by the lowest ratio obtained in the first step.

C- 76.57/12, H= 6.43/1, O = 17.00/16

C- 6.38/1.06, H= 6.43/1.06, O= 1.06/1.06

C- 6, H- 6, O- 1

Empirical formula: C6H6O

[(12 ×6) + (6 × 1) + (16 × 1)]n=94.11

[72 + 6 +16]n = 94.11

n = 94.11/94

n= 1

Molecular formula = C6H6O

2)

C- 53.30/12, H- 11.19/1, O- 35.51/16

C- 4.44/2.22, H- 11.19/2.22, O- 2.22/2.22

C- 2, H- 5, O- 1

Empirical formula: C2H5O

[(2×12) + (5× 1) + (1×16)]n = 90.12

[24 + 5 + 16] n = 90.12

n= 90.12/45

n= 2

Molecular formula = C4H10O2

Answer:

Ni

Explanation:

An active metal is a highly reactive metal. Active metals are found high up in the activity series.

Active metals react with other metals that are lower than them in the activity thereby displacing the lower metals from a solution of their salts. This is what may have happened in the other two reactions.

Ni is the most active metal listed in the question since it can react a compounds with Pb(NO3)2(aq) to liberate Pb metal.

Answer:

0.465

Explanation:

To find the volume of a substance, divide the mass by the density.

M/D = V

10.0 / 21.5 = 0.4651163

Then round to 3 significant figures: and the density is 0.465

Answer:

[tex]{ \boxed{ \bf{vapour \: density = 2 \times molecular \: mass}}} \\{ \tt{ PV= (\frac{m}{ m_{r}}) RT}} \\ { \tt{3 \times 5.6 = \frac{11}{m _{r}} \times 0.0831 \times 273}} \\ { \tt{m _{r} = 14.85 \: g}} \\ \\ { \bf{vapour \: density = 2 \times m _{r}}} \\ = 2 \times 14.85 \\ = 29.7 \: { \tt{g {dm}^{ - 3} }}[/tex]

Answer:

H2

Explanation:

Answer:

[tex]{ \bf{H _{2} }} \\ { \tt{hydrogen \: gas}}[/tex]

Answer:

2040 kJ

Explanation:

Step 1: Calculate the energy provided by 21 g of protein

17 kJ are provided per gram of protein.

21 g × 17 kJ/g = 357 kJ

Step 2: Calculate the energy provided by 59 g of carbohydrate

17 kJ are provided per gram of carbohydrate.

59 g × 17 kJ/g = 1003 kJ

Step 3: Calculate the energy provided by 18 g of fat

38 kJ are provided per gram of fat.

18 g × 38 kJ/g = 684 kJ

Step 4: Calculate the total energy provided by the dinner

357 kJ + 1003 kJ + 684 kJ = 2044 kJ ≈ 2040 kJ

Answer:

The answer would be B, putting thermal energy into something means you're adding heat into it.

Answer:

heat rate= 1281W

length = 15.8m

Explanation:

we have this data to answer this question with

Tmi = 85 degrees

Tmo = 35 degrees

Ts = 25 dgrees

flow rate = 25 degrees

using engine oil property from table a-5

Tm = Tmo - TMi/2 = 333k

u =0.522x10⁻²

k = 0.26

pr = 51.3

cp = 2562 J/kg.k

mcp(Tmo-Tmi) =

0.01 x 2562(35-85)

= 1281 W

we find the change in Tim

= [(35-25)-(85-25)]/ln[(35-25)/(85-25)]

= -50/ln0.167

= -50/-1.78976

= 27.9°c

we finf the required reynold number

4x0.01/πx0.003x0.522x10⁻²

= 0.04/0.00004921

= 812.8

= 813

we find approximate correlation

NuD = hd/k

NuD = 3.66

3.66 = 0.003D/0.26

cross multiply

0.003D = 3.66x0.26

D = 3.66x0.26/0.003

= 317.2

As = 1281/317x27.9

= 0.145

As = πDL

L = As/πD

= 0.145/π0.003

= 0.145/0.009429

L = 15.378

Explanation:

here is the answer. Feel free to ask for more chem help

Answer:

cell

chloroplast and cell wall

nucleus

life processes

cell membrane

shape and size

vacuole

Hope it helps

Answer:

The entropy change of carbon dioxide = 0.719 kJ/k

Explanation:

Given:

1.5 m - 3 insulated rigid tank contains 2.7 kg of carbon dioxide at 100 kPa

The objective is to determine the entropy change of carbon dioxide

Formula used:

ΔS=

Solution:

On considering,

[tex]C_{P} =0.846 kJ/kg K\\C_V=0.657 kJ/kg k\\[/tex]

ΔS=[tex]mc_{v} lu\frac{p_{2} }{P_{1} }[/tex]

On substituting the values,

ΔS=[tex]2.7*0.657lu\frac{150}{100}[/tex]

ΔS=0.719 kJ/k

The entropy change is "0.719 kJ/K".

Given values are:

Mass of tank,

Pressure,

Rised pressure,

Assumption of constant specific heat is,

As we know the formula,

→ [tex]\Delta S = mC_v \ ln(\frac{P_2}{P_1} )[/tex]

         [tex]= (2.7)(0.657) \ ln (\frac{150}{100} )[/tex]

         [tex]= 1.7739\times 0.4055[/tex]

         [tex]= 0.7193 \ kJ/K[/tex]

Thus above answer is right.

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Answer:

0.796 M

Explanation:

Step 1: Given data

Gravimetric concentration (Cg): 4.78 g%g

Density of the solution (ρ): 1.00 g/mL

Step 2: Calculate the volumetric concentration of the solution (Cv)

We will use the following expression.

Cv = Cg × ρ

Cv = 4.78 g%g × 1.00 g/mL = 4.78 g%mL

Step 3: Calculate the molarity of the solution (M)

The volumetric concentration is 4.78 g%mL, that is, there are 4.78 g of acetic acid per 100 mL of solution. We can calculate the molarity using the following expression.

M = mass solute / molar mass solute × liters of solution

M = 4.78 g / 60.05 g/mol × 0.1 L = 0.796 M

Answer:

d is the answer have a good one

Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, [tex]CH_{4}[/tex].

Explanation:

Given: Mass of methane = 146.6 g

As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.

[tex]Moles = \frac{mass}{molar mass}\\= \frac{146.6 g}{16.04 g/mol}\\= 9.14 mol[/tex]

The given reaction equation is as follows.

[tex]C + 2H_{2} \rightarrow CH_{4}[/tex]

This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.

[tex]Moles of H_{2} = \frac{9.14}{2}\\= 4.57 mol[/tex]

Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, [tex]CH_{4}[/tex].