please help! what is the correct answer to this picture

Please Help! What Is The Correct Answer To This Picture

Answers

Answer 1

Answer:

i think its c [everything is so blury]

Explanation:


Related Questions

a. Compound A and compound B are constitutional isomers with molecular formula C3H7Cl. When compound A is treated with sodium methoxide, a substitution reaction predominates. When compound B is treated with sodium methoxide, an elimination rection predominates. Propose structures A and B.

b. An unknown compound with molecular formula C6H13Cl is treated with sodium ethoxide to produce 2,3-dimethyl-2-butene as the major product. Identify the structure of the unknown compound.

Answers

Answer:

história phkfk

Explanation:

guiooupigjdytrss

A student named a particular compound 2-ethyl-3-methyl-2-butene. Assuming that the student's choice actually corresponded to the correct distribution of the double bond and the substituents, what is the correct IUPAC name for this compound

Answers

Answer:

2-ethyl-3-methylbut-2-ene

Explanation:

The whole idea of IUPAC nomenclature is to devise a universally accepted system of writing the name of a compound from its structure.

According to IUPAC nomenclature, the root of the compound is the longest carbon chain. The substituents are named in alphabetical order and in such a way as to give each one the lowest number. The position of the functional group is indicated accordingly.

For the compound in question, its correct IUPAC name is 2-ethyl-3-methylbut-2-ene.

Lab 2: paper chromatography of organic dyes
Picture of questions below.

Answers

Answer:

The three primary colors used when mixing dyes or paints are red, yellow, and blue. Other colors are often a mixture of these three colors. Try running a chromatography test again with non-primary-color markers, like purple, brown, and orange.

Explanation:

Mixtures that are suitable for separation by chromatography include inks, dyes and colouring agents in food. ... As the solvent soaks up the paper, it carries the mixtures with it. Different components of the mixture will move at different rates. This separates the mixture out.

why is platinum metal preferred to other metals for the flame test​

Answers

Answer:

Platinum is especially good for this because it is unreactive, and does not produce a color in the flame which will mask the presence of other metals.

Hope this answer is right!

Answer:

Hey mate, here is your answer

1. Platinum doesn't impart any color to the flame.

2. It is not oxidised under the high temperature of the flame from a bunsen burner.

3. It is almost chemically inert. Even at high temperatures, it remains unattacked by free radicals / acid radicals.

Therefore, platinum wire is crucial for a flame test. Also, a platinum wire should be thoroughly cleaned before using it for a new flame test.

A platinum wire is cleaned by dipping it into concentrated HNO3 and then placing it in the non luminous part of the bunsen flame. Otherwise, the perviously tested radicals will impart color to the flame, which may cause confusion.

Explanation:

Hope it helps you

pasagot po please!!

science po ito pasagot po matino need ko po!!​

Answers

Explanation:

Efficiency = (output/input)×100%

70% = output/(800 W)

which means

output = 0.70×(800 W) = 560 W

A solution of acetic acid that has a concentration of 0.10 moles per liter has a pH of 2.87. What is the likely pH of a 0.10 mole per liter solution of the conjugate base sodium acetate?
A. 8.97
B. 1.00
C. 2.87
D. 4.74
E. 13.00

Answers

Answer: The correct option is A) 8.97

Explanation:

To calculate the [tex]K_b[/tex] of a reaction, we use the equation:

[tex]K_a\times K_b=10^{-14}[/tex]

where,

[tex]K_a[/tex] = acid dissociation constant of acetic acid = [tex]1.86\times 10^{-5}[/tex]

[tex]K_b[/tex] = base dissociation constant

Putting values in above equation, we get:

[tex](1.86\times 10^{-5})\times K_b=10^{-14}\\\\K_b=\frac{10^{-14}}{1.86\times 10^{-5}}=5.37\times 10^{-10}[/tex]

To calculate the hydroxide ion concentration of conjugate base, we use the equation:

[tex][OH^-]=\sqrt{K_b\times \text{[Conjugate base]}}[/tex]

where,

[Conjugate base] = 0.10 M

Putting values in above equation, we get:

[tex][OH^-]=\sqrt{(5.37\times 10^{-10})\times 0.1}[/tex]

[tex][OH^-]=7.33\times 10^{-6}[/tex]

To calculate the pOH of the solution, we use the equation:

[tex]pOH=-\log [OH^-][/tex]

[tex]pOH=-\log (7.33\times 10^{-6})[/tex]

[tex]pOH=5.03[/tex]

To calculate the pH of the solution, we use the equation:

pH + pOH = 14

pH + 5.03 = 14

pH = (14 - 5.03) = 8.97

Hence, the correct option is A) 8.97

atomic number of element is 15 write a formula of an oxide​

Answers

Answer:

Atomic Number. 15=phosphorus

Valency=3

So, Oxide=P203

A solution of KMnO4 has an absorbance of 0.539 when measured in the colorimeter. Determine the concentration of the KMnO4 given the following data for a calibration plot.

Concentration of KMNO4 (M) Absorbance
0.0150 0.081
0.0300 0.159
0.0450 0.260
0.0600 0.334

Answers

Answer:

Concentration of unknown solution is 0.0416 M

Explanation:

As we know  

Absorbance is equal to the product of molar absorptivity of KMnO4 m, path length and concentration

From the given set of graphical data, it is clear that the absorbance vs concentration is a straight line.  

From the graph, we can obtain-  

Y = 5.73 X – 0.0065

Absorbance = 0.232

0.232 = 5.73 X – 0.0065

X = 0.0416  

Concentration of unknown solution is 0.0416 M

The seagulls on the beach -​

Answers

Seagulls like beaches because they can find food

A 14.570 g sample of CaCl2 was added to 12.285 g of K2CO3 and mixed in water. A 3.494 g yield of CaCO3 was obtained.
What is the limiting reagent?
-CaCO3
-K2CO3
-CaCl2

Calculate the percent yield of CaCO3.
yield of CaCO3= %

Answers

Answer:

Limiting reagent is the potassium carbonate.

Percent yield of calcium carbonate is: 39.3 %

Explanation:

The reaction is:

CaCl₂ + K₂CO₃ → CaCO₃ + 2KCl

Formula for percent yield is:

(Produced yield / Thoeretical yield) . 100

Firstly we determine the moles of each reactant, in order to say what is the limiting reagent: ratio is 1:1.

1 mol of chloride need 1 mol of carbonate.

14.570 g . 1 mol /110.98 g = 0.131 moles of CaCl₂

12.285 g . 1 mol / 138.2g = 0.0889 moles of carbonate.

Limiting reagent is carbonate. For 0.131 moles of CaCl₂ we need the same amount of carbonate and we have less moles.

Ratio is also 1:1, with calcium carbonate.

1 mol of potassium carbonate produces 1 mol of calcium carbonate

then, 0.0889 moles will produce the same amount of CaCO₃

We convert moles to mass: 0.0889 mol . 100.08g /mol = 8.89 g

That's the theoretical yield; to find the percent yield:

(3.494 g / 8.89g) . 100 = 39.3%

 

Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH.
pH= 2.89

Answers

Answer: The value of [tex][H_{3}O^{+}][/tex] is 0.0012 M and [tex][OH^{-}][/tex] is [tex]1.02 \times 10^{-14}[/tex].

Explanation:

pH is the negative logarithm of concentration of hydrogen ion.

It is given that pH is 2.89. So, the value of concentration of hydrogen ions is calculated as follows.

[tex]pH = - log [H^{+}]\\2.89 = - log [H^{+}]\\conc. H^{+} = 0.0012 M[/tex]

The relation between pH and pOH value is as follows.

pH + pOH = 14

0.0012 + pOH = 14

pOH = 14 - 0.0012 = 13.99

Now, pOH is the negative logarithm of concentration of hydroxide ions.

Hence, [tex][OH^{-}][/tex] is calculated as follows.

[tex]pOH = - log [OH^{-}]\\13.99 = - log [OH^{-}]\\conc. OH^{-} = 1.02 \times 10^{-14} M[/tex]

Thus, we can conclude that the value of [tex][H_{3}O^{+}][/tex] is 0.0012 M and [tex][OH^{-}][/tex] is [tex]1.02 \times 10^{-14}[/tex].

What is the quantity of
heat required to raise the
temperature of 500 g of
iron by 2°C?
The specific heat capacity
of iron is 500 J/(kg °C)

Answers

Answer:

The quantity of  heat required to raise the  temperature of 500 g of  iron by 2°C is 500 J.

Explanation:

Calorimetry is responsible for measuring the amount of heat generated or lost in certain physical or chemical processes.

The sensible heat of a body is the amount of heat received or transferred by a body when undergoing a temperature variation (Δt) without there being a change in physical state (solid, liquid or gaseous).

Its mathematical expression is the fundamental equation of calorimetry. This is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this  case:

Q= ?c= 500 [tex]\frac{J}{kg*C}[/tex]m= 500 g= 0.500 kgΔT= 2 C

Replacing:

Q= 500 [tex]\frac{J}{kg*C}[/tex] *0.500 kg*2 C

Solving:

Q= 500 J

The quantity of  heat required to raise the  temperature of 500 g of  iron by 2°C is 500 J.

heating, the particle _______________ increases as more __________ __________ is added​

Answers

Answer: what are the choices?!.

Explanation:

what does PH scale measure

Answers

pH is a measure of how acidic/basic water is

Explanation:

pH is a measure of how acidic/basic water is. The range goes from 0 - 14, with 7 being neutral. pHs of less than 7 indicate acidity, whereas a pH of greater than 7 indicates a base.

assume in a different experiment, you prepare a mixture containing 10.0 M FeSCN2+, 1.0 M H+, 0.1 MFe3+ and 0.1 M HSCN. Is the initial mixture at equilibrium? If not, in what direction must the reactionproceed to reach equilibrium? (Hint: You will need to use the value of Kc you determined in the lab

Answers

Answer:

The mixture is not in equilibrium, the reaction will shift to the left.

Explanation:

Based on the equilibrium:

Fe³⁺+ HSCN ⇄ FeSCN²⁺ + H⁺

kc = 30 = [FeSCN²⁺] [H⁺] / [Fe³⁺] [HSCN]

Where [] are concentrations at equilibrium. The reaction is in equilibrium when  the ratio of concentrations = kc

Q is the same expression than kc but with [] that are not in equilibrium

Replacing:

Q = [10.0M] [1.0M] / [0.1M] [0.1M]

Q = 1000

As Q > kc, the reaction will shift to the left in order to produce Fe³⁺ and HSCN untill Q = Kc

A 8.29g sample of calcium sulfide was decomposed into its constituent elements, producing 4.61g of calcium and 3.68g of sulfur. Which of the statements are consistent with the law of constant composition (definite proportions)?

a. Every sample of calcium sulfide will have 44.4% mass of calcium.
b. Every sample of calcium sulfide will have 2.86 g of calcium.
c. The mass ratio of Ca to S in every sample of calcium sulfide is 1.25.
d. The ratio of calcium to sulfur will vary based on how the sample was prepared.
e. The mass percentage of calcium plus the mass percentage of sulfur in every sample of calcium sulfide equals 100%.

Answers

Answer:

d,e

Explanation:

Consider the following reaction at 298 K.
2 SO2(g) + O2(g) → 2 SO3(g)
An equilibrium mixture contains O2(g) and SO3(g) at partial pressures of 0.43 atm and 2.6 atm, respectively. Using data from Appendix 4, determine the equilibrium partial pressure of SO2 in the mixture.
______atm.

Answers

Answer and Explanation:

The reaction is in the gas phase, so the equilibrium constant is expressed in terms of the partial pressures (P) of the products and reactants, as follows:

[tex]Kp = \frac{P^{2}_{SO_{3} } }{P_{SO_{2}} ^{2}P_{O_{2}} }[/tex]

We have the following data:

P(SO₃) = 2.6 atm

P(O₂) = 0.43 atm

We need Kp for this reaction. We can assume that in Appendix 4 we found that Kp = 7 x 10²⁴.

Then, we introduce the data in the equilibrium constant expression to calculate the partial pressure f SO₂ (PSO₂), as follows:

[tex]P_{SO_{2} } = \sqrt{\frac{P_{SO_{3} } ^{2} }{Kp P_{O_{2} } } } = \sqrt{\frac{(2.6 atm)^{2} }{(7 x 10^{24)}(0.43 atm) } } = 1.5 x 10^{-12} atm[/tex]

Therefore, the partial pressure of SO₂ is 1.5 x 10⁻¹² atm (for the given Kp).

For each molecule, specify the polarity of the bonds and the overall polarity of the molecule.

a. BeCl2
b. H2O
c. O3

Answers

Each Be–Cl bond is polar because the two atoms have different electronegativities. The number of outer atoms (2) and lone pairs on the central atom (0) indicate that this molecule has a linear geometry. The bonds in a linear molecule are symmetric, and so their dipoles cancel out.

Each O–H bond is polar because the two atoms have different electronegativities. The number of outer atoms (2) and lone pairs on the central atom (2) indicate that this molecule has a bent geometry. The bonds in a bent molecule are asymmetric, and so their dipoles do not cancel out. In addition, the asymmetric arrangement of the lone pairs on O further contribute to the dipole of this molecule.

An O–O or O=O bond is nonpolar because the two atoms have the same electronegativity. Because there is no overall polarity in O2, the molecule is nonpolar.

Given the following values for the change in enthalpy (deltaH) and entropy (deltaS), which of the following processes can occur at 298 K without violating the second Law of Thermodynamics?
(a) DeltaH = -84 kj mol-2 (-20 kcal mol-1), DeltaS = +125j mol-2K-1)(+30 cal mol-1 K-1)
(b) DeltaH = -84 kj mol-2 (-20 kcal mol-1), DeltaS = -125j mol-2K-1)(-30 cal mol-1 K-1)
(c) DeltaH = +84 kj mol-2 (+20 kcal mol-1), DeltaS = +125j mol-2K-1)(+30 cal mol-1 K-1)
(d) DeltaH = +84 kj mol-2 (+20 kcal mol-1), DeltaS = +125j mol-2K-1)(-30 cal mol-1 K-1)

Answers

Answer:

Option A and B

Explanation:

(a) DeltaH = -84 kj mol-2 (-20 kcal mol-1), DeltaS = +125j mol-2K-1)(+30 cal mol-1 K-1)

Delta G = Delta H – T * DS

Substituting the given values, we get –  

Delta G = -84 -298 *(125/1000) = -121.25  KJ/mol

Delta G is negative hence the process is spontaneous and will not violate the second law of thermodynamics

(b) DeltaH = -84 kj mol-2 (-20 kcal mol-1), DeltaS = -125j mol-2K-1)(-30 cal mol-1 K-1)

Delta G =-84 -298 *(-125/1000) = -46.75 KJ/mol

Delta G is negative hence the process is spontaneous and will not violate the second law of thermodynamics

(c) DeltaH = +84 kj mol-2 (+20 kcal mol-1), DeltaS = +125j mol-2K-1)(+30 cal mol-1 K-1)

Delta G = 84 -298 *(125/1000) = +46.75 KJ/mol

Delta G is positive hence the process is non-spontaneous and will violate the second law of thermodynamics

(d) DeltaH = +84 kj mol-2 (+20 kcal mol-1), DeltaS = +125j mol-2K-1)(-30 cal mol-1 K-1)

Delta G = 84 -298 *(-125/1000) = + 121.25  KJ/mol

Delta G is positive hence the process is non-spontaneous and will violate the second law of thermodynamics

How many grams of KNO3 can dissolve in 100g of water at 20°C?

Answers

Answer:

30 grams of KNO3 can be dissolved.

Explanation:

Hello there!

In this case, since the solubility is defined as the maximum amount of solute that can be dissolved in a certain amount of solvent, usually 100 grams of water as function of the temperature, we will need to recall the graph for the solubility of KNO3 as shown on the attached file.

Thus, by identifying the curve for KNO3, we realize that at a temperature of 20 °C, the solubility is about 30 grams; which means 30 grams can be dissolved in 100 grams of water at 20 °C.

Regards!

phương pháp VI PHÂN ĐỒ THỊ để xác định bậc phản ứng

Answers

Answer:

mwlooy kagabi jal

64 JAHA VI PHÂN KAY

Dugongs are animals that live in the ocean and eat underwater grasses. The sun is shining on the shallow ocean water where the grasses and dugongs live.

What is happening to the carbon in the water around the grasses and the dugongs? Is carbon moving into the water, moving out of the water, or both?

Answers

Answer:

please mark as brainliest

Explanation:

The sun is shining on the shallow ocean water where the grasses and dugongs live. What is happening to the carbon in the water around the grasses and the dugongs? Is carbon moving into the water, moving out of the water, or both? Carbon is not moving into the water; it is only moving out of the water.

Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide.

4HCl(aq)+MnO2(s)⟶MnCl2(aq)+2H2O(l)+Cl2(g)

A sample of 43.1g MnO2 is added to a solution containing 42.9g HCl.

a. What is the limiting agent?
b. What is the theoretical yield of Cl2?
c. If the yield of the reaction is 72.9%, what is the actual yield of chlorine?

Answers

Answer:

A. HCl is the limiting reactant.

B. Theoretical yield of Cl₂ is 20.9 g.

C. Actual yield of Cl₂ = 15.2 g.

Explanation:

The balanced equation for the reaction is given below:

4HCl + MnO₂ –> MnCl₂ + 2H₂O + Cl₂

Next, we shall determine the masses of HCl and MnO₂ that reacted and the mass of Cl₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of HCl = 1 + 35.5

= 36.5 g/mol

Mass of HCl from the balanced equation = 4 × 36.5 = 146 g

Molar mass of MnO₂ = 55 + (2×16)

= 55 + 32

= 87 g/mol

Mass of MnO₂ from the balanced equation = 1 × 87 = 87 g

Molar mass of Cl₂ = 2 × 35.5

= 71 g/mol

Mass of Cl₂ from the balanced equation = 1 × 71 = 71 g

SUMMARY:

From the balanced equation above,

146 g of HCl reacted with 87 g of MnO₂ to produce 71 g of Cl₂.

A. Determination of the limiting reactant.

From the balanced equation above,

146 g of HCl reacted with 87 g of MnO₂.

Therefore, 42.9 g of HCl will react with = (42.9 × 87)/146 = 25.6 g of MnO₂.

From the calculation made above, we can see clearly that only 25.6 g out of 43.1 g of MnO₂ given was needed to react completely with 42.9 g of HCl.

Therefore, HCl is the limiting reactant.

B. Determination of theoretical yield of Cl₂.

Here, the limiting reactant will be used.

From the balanced equation above,

146 g of HCl reacted to produce 71 g of Cl₂.

Therefore, 42.9 g of HCl will react to produce = (42.9 × 71)/146 = 20.9 g of Cl₂.

Thus, the theoretical yield of Cl₂ is 20.9 g.

C. Determination of the actual yield of Cl₂.

Theoretical yield of Cl₂ = 20.9 g

Percentage yield of Cl₂ = 72.9%

Actual yield of Cl₂ =?

Percentage yield = Actual yield / Theoretical yield × 100

72.9% = Actual yield / 20.9

Cross multiply

Actual yield = 72.9% × 20.9

Actual yield = 72.9/100 × 20.9

Actual yield = 0.729 × 20.9

Actual yield of Cl₂ = 15.2 g

carbon dioxide gas evolve during the fermentation of sugar which was collected at 22.5°C and 0.945 ATM after perfect strangers in the volume was found to be 25.0 ML how many grams of carbon dioxide were collected

Answers

Answer:

0.043 grams

Explanation:

We can find the mass of carbon dioxide as follows:

[tex] m = n*M [/tex]

Where:

n: is the number of moles

M: is the molar mass = 44.01 g/mol

First, we need to calculate the number of moles. We can use the Ideal gas equation:

[tex] PV = nRT [/tex]

Where:

P: is the pressure = 0.945 atm

V: is the volume = 25.0 mL  

R: is the gas constant = 0.082 L*atm/(K*mol)

T: is the tempearture = 22.5 °C

[tex]n = \frac{PV}{RT} = \frac{0.945 atm*25 mL*\frac{1 L}{1000 mL}}{0.082 L*atm/K*mol*(22.5 + 273) K} = 9.75 \cdot 10^{-4} moles[/tex]

Hence, the mass is:

[tex]m = 9.75 \cdot 10^{-4} moles*44.01 g/mol = 0.043 g[/tex]

Therefore, 0.043 grams were collected.

                         

I hope it helps you!

What is the concentration of s solution that contains 55 mL of alcohol per 145 mL solution?​

Answers

Answer:

37.9% v/v

Explanation:

Since both the alcohol and solution are presumed to be liquid, this concentration can be expressed as a volume concentration (or % v/v):

volume concentration = volume of solute / volume of solution

[tex]\% v/v = 55/145= 0.379[/tex]

which is a characteristic of oxygen​

Answers

Answer:

Characteristics of oxygen are as follows:

In normal conditions oxygen is a colourless, odourless and insipid gas.It condensates in a light blue liquid.Oxygen is part of a small group of gasses literally paramagnetic and it's the most paramagnetic of all.Liquid oxygen is also slightly paramagnetic.

The product of an organic reaction is analyzed by column chromatography using silica as the stationary phase and toluene as the mobile phase.

a. True
b. False

Answers

Answer:

The product of an organic reaction is analyzed by column chromatography using silica as the stationary phase and toluene as the mobile phase.

Explanation:

The given statement is true.

In chromatography silica gel is used as the predominant stationary phase.

Since silica gel is a good adsorbent.

It is a polar adsorbent.

In order to remove polar components, silica gel is used as the stationary phase.

Answer is a.true.

Which is a property of barium (Ba)?
O A. It rarely reacts with other elements.
O B. It is brittle as a solid.
O c. It is very reactive.
O D. It does not conduct electricity.
Plzzzzz helppppp!!!

Answers

Answer:

a it rarely reach with other elements

The carbon-carbon bonds in benzene are:
O a) Easily broken in chemical reactions
b) A hybrid between double bonds and single bonds
c)
Identical to the carbon-carbon bonds in cyclohexene
d) Identical to the carbon-carbon bonds in cyclohexane
please hurry

Answers

Answer:

a

Explanation:

Easily broken in chemical reactions

1. How many atoms of chlorine are present in 1.70x1023 molecules Cl2?​

Answers

Explanation:

the answer is in the image above

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