You have run an experiment studying the effects of the molecular weight of a compound on the rate of diffusion in agar. Compound X has a molecular weight of 25.3 g/mol and compound Y has a molecular mass of 156.2 g/mol. On two separate agar plates, 0.1 g of each substance were transferred and allowed to diffuse for 2 hours. The results below were obtained. Use this information about this situation to answer the following three questions.
a. What was the independent variable in this experiment?
b. What was the independent variable in this experiment?
c. List two controls that were held constant for this experiment or that you would hold constant for this experiment

Answers

Answer 1

Answer:

A) Molecular weight

B) Rate of diffusion

C) Agar plates

Time of diffusion

Explanation:

A) The independent variable is the molecular weight because it is the variable that the researcher changes at will to aid his research. It doesn't depend on other variables.

B) The dependent variable is "Rate of diffusion because it depends on the molecular weight of the compound.

C) A control variable is one that would be held constant throughout the research.

In this case, the agar plates and the time of 2 hours diffusion remain the same throughout.


Related Questions

Classify each amino acid by the chemical properties of its side chain (R group) at pH 7 . Select the amino acid that fits best in each category. Each amino acid will be selected only once. Which amino acid has a positively charged R group

Answers

Answer: The amino acid that has a positively charged R group are LYSINE and ARGININE.

Explanation:

AMINO ACIDS are the basic structural units of proteins. Each amino acid contains an amino group ( -NH2) and a carboxyl group (-COOH) in its molecule. The carbon atom of amino acid to which the functional groups are attached is know as the alpha-carbon. In neutral solution, amino acids are mainly in the form of dipolar ions. Amino acids can be prepared through hydrolysis of protein by boiling with dilute hydrochloric acid.

Amino acids can be classified according to the chemical properties of its side chain (R group) at pH 7, these include:

--> Amino acids with POSITIVELY CHARGED R group

--> Amino acids with negatively charged R group

--> Amino acids with neutral polar R group

--> Amino acids with nonpolar aliphatic R group.

Amino acids with POSITIVELY CHARGED R group are those amino acids that has side chains which contain nitrogen and resemble ammonia, which is a base. Their pKa's are high enough that they tend to bind protons, gaining a positive charge in the process. Example include LYSINE and ARGININE.

Write a balanced equation for the complete combustion of 2,3,3-trimethylpentane. Use the molecular formula for the alkane (C before H) and the smallest possible integer coefficients.

Answers

Answer:

C8H18 + 25/2O2 ----> 8CO2 + 9H2O

Explanation:

2,3,3-trimethylpentane has the molecular formula C8H18.

The general formula for the combustion of an alkane is;

CnH2n+2 + 3n+1/2O2 ----->nCO2 + (n+1)H2O

In writing a balanced chemical reaction equation, the number of atoms of each element on the left hand side of the reaction equation must be the same as the number of atoms of the same element on the right hand side of the reaction equation.

For C8H18, the balanced chemical reaction equation for combustion is;

C8H18 + 25/2O2 ----> 8CO2 + 9H2O

The placement of carbonyl group of a kerose sugar is at second-carbon only (b) fir + carbon only () first and fast carbon (d) list carbon only Chirulit,​

Answers

Answer:

(d)

Explanation:

Carbonyl group can be the placement of kerosene sugar

Calculate the mass of water produced when 9.57 g of butane reacts with excess oxygen.
Express your answer to three significant figures and include the appropriate units.

Answers

Answer:

14.9 g

Explanation:

Step 1: Write the balanced equation

C₄H₁₀ + 6.5 O₂ ⇒ 4 CO₂ + 5 H₂O

Step 2: Calculate the moles corresponding to 9.57 g of C₄H₁₀

The molar mass of C₄H₁₀ is 58.12 g/mol.

9.57 g × 1 mol/58.12 g = 0.165 mol

Step 3: Calculate the moles of H₂O produced from 0.165 moles of C₄H₁₀

0.165 mol C₄H₁₀ × 5 mol H₂O/1 mol C₄H₁₀ = 0.825 mol H₂O

Step 4: Calculate the mass corresponding to 0.825 mol of H₂O

The molar mass of H₂O is 18.02 g/mol.

0.825 mol × 18.02 g/mol = 14.9 g

separete the ALKALI from the following bases :

NH4OH(ammonium nitrate)

CuO(copper oxide)

Zn(OH)2 (zinc hydroxide)

MgO(magnesium oxide)

Na2O(sodium oxide)

NaOH(sodium hydroxide)

CoO(cobalt oxide)

Mg(OH)2(magnesium hydroxide)

LIOH(lithium hydroxide)

help me with this i will surely mark u as Brainliest

plss help!!!

Answers

Answer:

Ammonium hydroxide, NH₄OH

Magnesium hydroxide, Mg(OH)₂

Sodium hydroxide, NaOH

Lithium hydroxide, LiOH

Explanation:

A base is a substance which neutralizes acids to produce salt and water. Bases are hydroxide or oxides of metals. Bases may be soluble or insoluble in water. Bases generally have a bitter taste and turn red litmus paper or indicator red.

Alkalis are bases which are soluble in water. They form the hydroxide of the alkali metals or alkaline earth metals in solution and they ionize to produce hydroxide ions. They are slippery to touch and turn red litmus blue being bases.

Therefore, all alkalis are bases but not all bases are alkalis. Insoluble bases are not alkalis.

From the given chemical compounds the alkalis present in the list are:

Ammonium hydroxide, NH₄OH; since it is soluble in water and produces hydroxide ions

Magnesium hydroxide, Mg(OH)₂; since it is slightly soluble in water and produces hydroxide ions

Sodium hydroxide, NaOH; since it is soluble in water and produces hydroxide ions

Lithium hydroxide, LiOH; since it is soluble in water and produces hydroxide ions

CuO(copper oxide) is a base but not an alkali as it does not produce hydroxide ions.

Zn(OH)2 (zinc hydroxide) is amphoteric and is insoluble

MgO(magnesium oxide) is a base but not an alkali as it does not produce hydroxide ions.

Na2O(sodium oxide) is a base but not an alkali as it does not produce hydroxide ions.

CoO(cobalt oxide) is a base but not an alkali as it does not produce hydroxide ions.

Which best describes how the total mass of the substances that go into
photosynthesis compares to the mass of substances that are present
afterward?
O A. The mass increases because the molecules that are produced are
larger than those that are used.
B. The mass increases because some light energy changes into
mass.
O C. The mass stays the same because the total number of atoms
does not change
O D. The mass decreases because plants destroy some of the atoms
during photosynthesis.

Answers

Answer:

C. The mass stays the same because the total number of atoms does not change

Explanation:

According to the law of conservation of matter/mass, matter cannot be created nor destroyed, hence, the amount of matter in the reactants must be the same amount in the products.

Using the photosynthetic reaction as a case study, carbon dioxide (CO2) and water (H2O) are the compounds that go into the reaction (reactants) while glucose and oxygen (O2) are the products of the reaction.

Using the law of conservation of matter to explain, the total mass of both the reactants and products stays the same because the total number of atoms does not change i.e. if 6 atoms of Carbon starts the reaction, 6 atoms of carbon will end it.

mẫu khi thêm NH4NO3 vào đem nung để nguội lại thêm NH4NO3 có tác dụng gì?

Answers

Answer:

Adding ammonium nitrate to water turns the mixture cold and is a good example of an endothermic chemical reaction!

Identify the conjugate acid/base pairs in each of the following equations:
(a) H2S + NH3 ⇔ NH4+ + HS-
Pair 1: H2S and
Pair 2: NH3 and
(b) HSO4- + NH3 ⇔ SO42- + NH4+
Pair 1: HSO4- and​
Pair 2: NH3 and
(c) HBr + CH3O- ⇔ Br- + CH3OH
Pair 1: HBr and
Pair 2: CH3O- and
(d) HNO3 + H2O → NO3- + H3O+
Pair 1: HNO3 and​
Pair 2: H2O and

Answers

Answer:

(a) Pair 1: H₂S and HS⁻

    Pair 2: NH₃ and NH₄⁺

(b) Pair 1: HSO₄⁻ and SO₄⁻

    Pair 2: NH₃ and NH₄⁺

(c) Pair 1: HBr and Br⁻

    Pair 2: CH₃O⁻ and CH₃OH

(d)  Pair 1: HNO₃ and NO₃⁻

     Pair 2: H₃O⁺

Explanation:

When an acid loses its proton (H⁺), a conjugate base is produced.

When a base accepts a proton (H⁺), it forms a conjugate acid.

(a) H₂S is an acid. When it loses a proton, it forms the conjugate base HS⁻.

    NH₃ is a base. When NH₃ gains a proton, it forms the conjugate acid NH₄⁺

(b) The acid HSO₄⁻ loses a H⁺ ion and forms the conjugate base SO₄²⁻.

     The base NH₃ accepts a H⁺ ion to form the conjugate acid NH₄⁺.

(c) HBr is an acid. When loses the H⁺ ion, it forms the conjugate base Br⁻.

   CH₃O⁻ accepts a H⁺ ion to form the conjugate acid CH₃OH.

(d) HNO₃ loses a proton to form the conjugate base NO₃⁻.

    H₂O gains a proton to form the conjugate acid H₃O⁺.

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.

The rate of the reaction is 1.6*10-2 M/s when the concentration of A is 0.15 M. Calculate the rate constant if the reaction is (a) first order in A and (b) second order in A.

Answers

Answer:

[tex]k_1=0.107s^{-1} \\\\k_2=0.711M^{-1}s^{-1}[/tex]

Explanation:

Hello there!

In this case, according to the given information and the attached picture in which we can see the units of the rate constant, it turns out possible for us to realize the two called rate laws are:

[tex]r=k[A]\\\\r=k[A]^2[/tex]

The former is first-order and the latter second-order; in such a way, we solve for the rate constant in both cases to obtain the following:

[tex]k=\frac{r}{[A]}=\frac{1.6x10^{-2}M/s}{0.15M}=0.107s^{-1} \\\\k=\frac{r}{[A]^2}=\frac{1.6x10^{-2}M/s}{(0.15M)^2}=0.711M^{-1}s^{-1}[/tex]

Regards!

Sợ sánh kỹ thuật trích ly rắn lỏng với kỹ thuật trích ly lỏng
lỏng

Answers

Answer:

which language is this please tell me

He predicted an element with an atomic weight between 65 (zinc) and 75 (arsenic) with a valence similar to aluminum that he named ekaboron. This element was discovered in 1879. What is this element:?

Answers

Answer:

Scandium

Explanation:

Mendeleev played an important role in the development of the modern periodic table. His periodic table was filled with gaps. He said that these gaps were elements that were yet to be discovered. He rightly predicted many elements which have now been discovered and fitted in their proper places in the periodic table.

He used the prefix ''eka'' to refer to elements whose properties were alike but were yet to be discovered at that time.

The compound named ekaboron which he predicted to have an atomic weight between 65 (zinc) and 75 (arsenic) with a valence similar to aluminum was later discovered in 1879 and properly named scandium.

How many milliliters of a 6.00 M NaCl solution are needed to make 250.0 milliliters of a 0.500 M NaCl solution?​

Answers

Answer:

20,8ml  NaCl 6M

Explanation:

C1V1 = C2V2

A 100.0-g sample of water at 27.0oC is poured into a 71.0-g sample of water at 89.0oC. What will be the final temperature of the water? (Specific heat capacity of water = 4.184 J/goC.)

Answers

Answer: The final temperature will be [tex]52.74^oC[/tex]

Explanation:

Calculating the heat released or absorbed for the process:

[tex]q=m\times C\times (T_2-T_1)[/tex]

In a system, the total amount of heat released is equal to the total amount of heat absorbed.

[tex]q_1=-q_2[/tex]

OR

[tex]m_1\times C\times (T_f-T_1)=-m_2\times C\times (T_f-T_2)[/tex] ......(1)

where,

C = heat capacity of water = [tex]4.184J/g^oC[/tex]

[tex]m_1[/tex] = mass of water of sample 1 = 100.0 g

[tex]m_2[/tex] = mass of water of sample 2 = 71.0 g

[tex]T_f[/tex] = final temperature of the system = ?

[tex]T_1[/tex] = initial temperature of water of sample 1 = [tex]27^oC[/tex]

[tex]T_2[/tex] = initial temperature of the water of sample 2 = [tex]89.0^oC[/tex]

Putting values in equation 1, we get:

[tex]100.0\times 4.184\times (T_f-27)=-71.0\times 4.184\times (T_f-89)\\\\171T_f=9019\\\\T_f=\frac{9019}{171}=52.74^oC[/tex]

Hence, the final temperature will be [tex]52.74^oC[/tex]

.What shape does each galaxy have?

Answers

Answer:

as in shapes

Explanation:

4 different shapes, spiral, elliptical, lenticular, and irregular. Spiral galaxies are one of the most familiar galaxy shapes.

What are the two types of addition compounds

Answers

Answer:

electrophilic addition and

nucleophilic addition.

Answer:

the two types of addition compoundsare:

1.electrophilic addition

2. nucleophilic addition.

Classify each structure according to its functional class.
Compound A contains a carbonyl bonded to two alkyl groups.
Compound B contains an oxygen bonded to two alkyl groups.
Compound C contains a carbonyl bonded to propyl and N H C H 3.
Compound D is a nitrogen bonded to three alkyl groups.
Classify structure A according to its functional class.
Classify structure B according to its functional class.
Classify structure C according to its functional class.
Classify structure D according to its functional class.

Answers

Answer:

Classify each structure according to its functional class.

Compound A contains a carbonyl bonded to two alkyl groups.

Compound B contains an oxygen bonded to two alkyl groups.

Compound C contains a carbonyl bonded to propyl and N H C H 3.

Compound D is a nitrogen bonded to three alkyl groups.

Explanation:

Compound A contains a carbonyl bonded to two alkyl groups.

-C=O group is called a carbonyl group.

If it is present between two alkyl groups then, it is a ketone.

Compound B contains oxygen bonded to two alkyl groups.

Compound B is an example of an ether molecule.

Compound C contains a carbonyl bonded to propyl and N H C H 3.

Compound C is C3H7-CO-NHCH3 which is an amide molecule.

Compound D is nitrogen bonded to three alkyl groups.

This is an example of a tertiary amine group.

Select the net ionic equation for the reaction that occurs when magnesium sulfate and nickel(II) nitrate are mixed.
a. Ni2+(aq) + SO4^2- → NISO2 (s) + O2 (g).
b. Mg2+(aq) + 2NO3 (aq) → Mg(NO3)2(s).
c. Mg2+(aq) + NO3- → MgNO3 (s).
d. Mg2+(aq) + SO4^2- (aq) + Ni2+ (aq) + 2NO3- → Mg2+ (aq) + 2NO3 (aq) + NISO4 (s).
e. Ni2+(aq) + SO4^2- (aq) → NISO4 (s).
f. No reaction occurs.

Answers

Answer:

No reaction occurs.

Explanation:

The molecular reaction is as follows;

MgSO4(aq) + Ni(NO3)2(aq) ----> Mg(NO3)2(aq) + NiSO4(aq)

We can see from the reaction above that the both products of the reaction are soluble. Recall that a double replacement reaction often yields one insoluble product which separates as a precipitate.

This reaction does not occur since the two products that ought to be obtained are soluble in water.

Balance each of the following equations. Then, drag and drop each equation to match the coefficient of H2O in the balanced chemical equation. A coefficient for water may be used once, more than once, or not at all. Drag and drop your selection from the following list to complete the answer:
C2H5OH + O2 + CO2 + H2O NH3 + O2 + NO2 + H20 C3H2 + O2 + CO2 + H2O H2SO4 + NaOH → Na2SO4 + H20 NO2 + H2O → HNO3 + NO

Answers

Answer:

C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O

2 NH₃ + 3.5 O₂ → 2 NO₂ + 3 H₂O

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

3 NO₂ + H₂O → 2 HNO₃ + NO

Explanation:

We will balance the equation using the trial and error method.

C₂H₅OH + O₂ → CO₂ + H₂O

1) We balance C atoms by multiplying CO₂ by 2 and H atoms by multiplying H₂O by 3.

C₂H₅OH + O₂ → 2 CO₂ + 3 H₂O

2) We balance O atoms by multiplying O₂ by 3.

C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O

NH₃ + O₂ → NO₂ + H₂O

1) We balance H atoms by multiplying NH₃ by 2 and H₂O by 3.

2 NH₃ + O₂ → NO₂ + 3 H₂O

2) We balance N atoms by multiplying NO₂ by 2.

2 NH₃ + O₂ → 2 NO₂ + 3 H₂O

3) We balance O atoms by multiplying O₂ by 3.5

2 NH₃ + 3.5 O₂ → 2 NO₂ + 3 H₂O

C₃H₈ + O₂ → CO₂ + H₂O

1) We balance C atoms by multiplying CO₂ by 3 and H atoms by multiplying H₂O by 4.

C₃H₈ + O₂ → 3 CO₂ + 4 H₂O

2) We balance O atoms by multiplying O₂ by 5.

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

H₂SO₄ + NaOH → Na₂SO₄ + H₂O

1) We balance Na atoms by multiplying NaOH by 2.

H₂SO₄ + 2 NaOH → Na₂SO₄ + H₂O

2) We balance H and O atoms by multiplying H₂O by 2.

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

NO₂ + H₂O → HNO₃ + NO

1) We balance H atoms by multiplying HNO₃ by 2.

NO₂ + H₂O → 2 HNO₃ + NO

2) We balance N atoms by multiplying NO₂ by 3.

3 NO₂ + H₂O → 2 HNO₃ + NO

please answer all if you can!

Answers

Answer:1.b 2.A (i think it might be c) 3 (i have no clue) 4.(no clue) Sorry i only got 1 and 2 witch i think are right its been a while sense i have done this....

Explanation:

____________ can increase the presence of 5HT in the terminal button or synaptic cleft.

Answers

Answer:

Selective serotonin reuptake inhibitors (SSRIs)

Explanation:

A synaptic cleft is a space that separates two neurons thereby forming a junction between two or more neurons. The synaptic cleft helps in the transfer of nerve impulse from one neuron to the other.

5-HT is found in the enteric nervous system located in the gastrointestinal tract and it helps in modulating cognition, memory, sleep, and numerous physiological processes.

Selective serotonin reuptake inhibitors (SSRIs), such as fluoxetine and citalopram are used to increase the level of 5-HT in the synaptic cleft by inhibiting its reuptake into the presynaptic terminal.

Identify the item that does not have kinetic energy.
A. A worm crawling
B. A falling leaf
C. An airplane flying
D. A butterfly sitting on a twig

Answers

Answer:

D

Explanation:

there is no other answer choice that does not move. This is a fool-proof question because all the other answer choices contain movement except the butterfly resting

For a particular chemical reaction the rate​ (g/hr) at which one of the reactants changes is proportional to the amount of that reactant present. If y represents the amount of that reactant at time​ t, StartFraction dy Over dt EndFraction equals minus0.7y. If there were 70 grams of the reactant when the process started ​(tequals ​0), how many grams will remain after 4 ​hours?

Answers

Answer:

Amount of reactant after four hours = 4,26 grams

Explanation:

Suppose y denotes the amount of reactant at the time (t)

The given function:

[tex]\dfrac{dy}{dt} = -0.7 y[/tex]

[tex]\dfrac{dy}{y} = -0.7 dt[/tex]

Taking integral on both sides

㏑(y) = -0.7t + c¹

[tex]e^{In(y)}= e^{-0.7t + c^1}[/tex]

[tex]y(t) = Ce ^{-0.7t}[/tex]

At t = 0 ; y (t) = 70

[tex]70 = Ce^{-0.7(0)}[/tex]

C = 70

As such; [tex]\mathtt{y(t) = 70 e^{-0.7*t}}[/tex]

After four hours, the amount of the reactant is:

[tex]\mathtt{y(t) = 70 e^{-0.7*4}}[/tex]

[tex]\mathtt{y(t) = 70 e^{-2.8}}[/tex]

[tex]\mathtt{y(t) = 4.26}[/tex]

Amount of reactant after four hours = 4,26 grams

Fe có tác dụng với HCL không

Answers

Fe có tác dụng với Hcl

Fe + 2hcl -> fecl2 +h2

Answer:

Explanatio:

Which event is an example of melting?
A. Wax drips down the side of a lit candle.
B. Perspiration dries on a person's skin.
C. Rain turns to ice pellets.
D. A mirror fogs up when someone takes a hot shower.
I’m just curious tbh

Answers

Answer:

A. Wax drips down the side of a lot candle.

Explanation:

The chemical change from solid to liquid. This is a combustion reaction, so carbon dioxide gas and water vapour is also produced but you can't see them

Answer:

A. Wax drips down the side of a lot candle.

Explanation:

Which of the following colors has the highest energy? O A. Red O B. Green O C. Blue O D. Yellow​

Answers

Answer:

C. Blue

Explanation:

This is because, Blue color highest frequency of energy after Violet and Indigo.

please help! what is the correct answer to this picture

Answers

Answer:

i think its c [everything is so blury]

Explanation:


Spell out the full name of the compounds
Help plz

Answers

Answer:

propanal

Explanation:

hope this helps :)

A company manufacturing KMnO4 wants to obtain the highest yield possible. Two of their research scientists are working on a technique to increase the yield.

Both scientists started with 50.0 g of manganese oxide.

What is the theoretical yield of potassium permanganate when starting with 50.0 g MnO2?

The equation for the production of potassium permanganate is as follows:

2 MnO2 + 4 KOH + O2 → 2 KMnO4 + 2 KOH + H2

Answers

Answer:

The theoretical yield potassium permanganate, KMnO₄ when starting with 50.0 g MnO₂ is 90.8 g

Explanation:

Molar mass of MnO₂ = (55 + 2 × 16) = 87.0 g/mol

Molar mass of KMnO₄ = (39 + 55 + 4 × 16) = 158 g/mol

Moles of MnO₂ in 50 g = reacting mass / molar mass

where reacting mass = 50 g

Moles of MnO₂ in 50 g = 50 g /87 g/mol = 0.575 moles

The equation for the production of potassium permanganate is as follows:

2 MnO2 + 4 KOH + O2 → 2 KMnO4 + 2 KOH + H2

From the equation of the reaction above, 2 moles of MnO₂ produces 2 moles of KmNO₄. The mole ratio of MnO₂ to KMnO₄ is 1 : 1

Therefore, 0.575 moles of MnO₂ will produce theoretically 0.575 moles of KMnO₄

Mass of 0.575 moles of KMnO₄ = number of moles × molar mass

Mass of 0.575 moles of KMnO₄ = 0.575 moles × 158 g/mol = 90.8 g of KMnO₄

Therefore, the theoretical yield potassium permanganate when starting with 50.0 g MnO₂ is 90.8 g

why Mg(OH)2 is soluble in HCL​

Answers

Answer:

While Mg(OH)2 is practically insoluble, a certain amount of Mg(OH)2 dissociates into ions when put in water. ... As HCl is added to the beaker containing milk of magnesia, the H+ ions from the HCl react with the OH– ions (those that are actually in solution from the Mg(OH)2) according to Equation 3 below.

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