Which compound would you expect to have the lowest boiling point?
a)CH3-CH2-CH2-CH2-CH2-CH3
b)CH3-CH2-OCH2-CH2-CH3
c)CH3-CH2-CH2-CH2-CH2-OH
d)CH3-CH2-OCH(CH3)2
e)CH3-OCH2-CH2-CH2-CH3

Answer:

Silver.

Explanation:

Hello!

In this case, we can consider the specific heat as the property that we can analyze in order to answer to this question. In such a way, as the specific heat is known as the energy required to modify the temperature of 1 g of the substance by 1 °C, since the masses of all the substances are the same, we can that their specific heats are respectively 0.240, 0.444 and 4.184 J/(g°C), from the equation:

[tex]Q=mCp\Delta T[/tex]

We can see that the higher the specific heat (Cp) the lower the change in temperature considering their inversely proportional relationship. However, as 100 J of energy is applied to all the substances, we can see that silver will exhibit the largest temperature change because a higher change is needed to fit with the provided energy.

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Answer: it is covalent and there are 2 hydrogen molecules and 1 oxygen molecule.

Explanation: it just is

Answer:

The rate at which heat is added in the boiler is 8128.610 kilowatts.

Explanation:

A boiler is a device that works at steady state, whose function is pressurizing and heating water. All gravitational electric and magnetic effects can be neglected. From First Law of Thermodynamics the boiler is characterized by the following model:

[tex]\dot Q_{in} + \dot H_{in}-\dot H_{out} + \dot K_{in} - \dot K_{out}= 0[/tex] (Eq. 1)

Where:

[tex]\dot Q_{in}[/tex] - Heat transfer rate, measured in kilowatts.

[tex]\dot H_{in}[/tex], [tex]\dot H_{out}[/tex] - Inlet and outlet fluid enthalpy rates, measured in kilowatts.

[tex]\dot K_{in}[/tex], [tex]\dot K_{out}[/tex] - Inlet and outlet kinetic energy rate, measured in kilowatts.

And by the Principle of Mass Conservation we have that:

[tex]\dot m_{in}-\dot m_{out} = 0[/tex] (Eq. 2)

Where [tex]\dot m_{in}[/tex] and [tex]\dot m_{out}[/tex] are mass flows at entrance and exit of the boiler, measured in kilograms per second.

Then, we clear the heat transfer rate and expand (Eq. 1) by definitions of enthalpy and kinetic energy, as well as (Eq. 2):

[tex]\dot Q_{in} = \dot H_{out}-\dot H_{in}+\dot K_{out}-\dot K_{in}[/tex]

[tex]\dot Q_{in} = \dot m \cdot \left[(h_{out}-h_{in}\right)+\frac{1}{2}\cdot (v_{out}^{2}-v_{in}^{2})\left][/tex] (Eq. 3)

Speeds of fluid entering and exiting the boiler ([tex]v[/tex]), measured in meters per second, are found by applying the following formula:

[tex]\dot m = \left(\frac{\pi}{4} \right)\cdot \frac{D^{2}\cdot v}{\nu}[/tex]

[tex]v = \frac{4\cdot \dot m\cdot \nu}{\pi\cdot D^{2}}[/tex] (Eq. 4)

Where:

[tex]\nu[/tex] - Specific volume of fluid, measured in cubic meters per kilogram.

[tex]D[/tex] - Inner diameter of pipe, measured in meters.

Entrance ([tex]\dot m = 2.778\,\frac{kg}{s}[/tex], [tex]\nu_{in} = 0.001093\,\frac{m^{3}}{kg}[/tex] and [tex]D_{in} = 0.08\,m[/tex])

[tex]v_{in} = \frac{4\cdot \left(2.778\,\frac{kg}{m^{3}} \right)\cdot \left(0.001093\,\frac{m^{3}}{kg} \right)}{\pi\cdot (0.08\,m)^{2}}[/tex]

[tex]v_{in}\approx 0.604\,\frac{m}{s}[/tex]

Specific enthalpy at given conditions is:

[tex]h_{in} = 640.09\,\frac{kJ}{kg}[/tex] (From Saturated Water tables)

Exit ([tex]\dot m = 2.778\,\frac{kg}{s}[/tex], [tex]\nu_{out} = 0.61731\,\frac{m^{3}}{kg}[/tex] and [tex]D_{out} = 0.30\,m[/tex])

[tex]v_{out} = \frac{4\cdot \left(2.778\,\frac{kg}{m^{3}} \right)\cdot \left(0.61731\,\frac{m^{3}}{kg} \right)}{\pi\cdot (0.30\,m)^{2}}[/tex]

[tex]v_{out} \approx 24.261\,\frac{m}{s}[/tex]

Specific enthalpy at given conditions is:

[tex]h_{out} = 3272.4\,\frac{kJ}{kg}[/tex] (From Superheated Steam tables)

If we know that [tex]\dot m = 2.778\,\frac{kg}{s}[/tex], [tex]h_{in} = 640.09\,\frac{kJ}{kg}[/tex], [tex]h_{out} = 3272.4\,\frac{kJ}{kg}[/tex], [tex]v_{in}\approx 0.604\,\frac{m}{s}[/tex], [tex]v_{out} \approx 24.261\,\frac{m}{s}[/tex], the heat transfer rate is:

[tex]\dot Q = \left(2.778\,\frac{kg}{s} \right)\cdot \left\{ \left(3272.4\,\frac{kJ}{kg}-640.09\,\frac{kJ}{kg} \right)+\frac{1}{2}\cdot \left[\left(24.261\,\frac{m}{s} \right)^{2}-\left(0.604\,\frac{m}{s} \right)^{2}\right] \right\}[/tex]

[tex]\dot Q = 8128.610\,kW[/tex]

The rate at which heat is added in the boiler is 8128.610 kilowatts.

It is critical to watch or observe what is happening with every step of your experiment

Answer:

The answers of the questions are found below

Explanation:

1.) higher than

2.) solution and the liquid taste

3.) hard

4.) have varying solubilities

-Just took it on edge and got them all right

-Hope you do well <3 :)

The complete statement about ionic compounds is:

The melting point of an ionic compound is likely to be higher than a  molecular compound.

Ionic compounds are those compounds that are formed by holding ions with an electrostatic force, called ionic bonding. The compounds contain negatively charged ions called anions and positively charged ions called cations. The anions and cations connect together to form a  neutral compound.

The compounds are bound with an electrostatic force, so their melting point will be higher than the molecular compounds.

Thus, the melting point of an ionic compound is likely to be higher than a molecular compound.

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Answer:

Elements in the same group of the periodic table have the same number of valence electrons. These are the electrons in their outer energy level that can be involved in chemical reactions. ... All the elements in group 1 have just one valence electron. This makes them very reactive.

Explanation:

Answer:

a) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = pKa?

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

Ka = 1.8*10^-5

ph = pK = -log(1.8*10^-5) = 4.74

Step 2:

Use the Henderson-Hasselbalch equation.

pH = pKa + log(A-/HA)

4.74 = 4.74 + log(A-/HA)

0 =  log(A-/HA)

A-/HA = 1

Consider X = moles of NaOH added (and moles of A- formed)

Remaining moles of HA = 1.8 - X

moles of A- = X

HA = 1.8 - X

X/(1.8-X) = 1

X =0.9

We have to add 0.9 mole NaOH to 1.0 L of 1.8 M HC2H3O2

To control we can do the following equation:

4.74 = 4.74 + log(0.9/0.9) = 4.74

b) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 4.00?

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

Ka = 1.8*10^-5

ph = 4

Step 2:

Use the Henderson-Hasselbalch equation.

pH = pKa + log(A-/HA)

4 = 4.74 + log(A-/HA)

-0.74 =  log(A-/HA)

A-/HA = 0.182

Consider X = moles of NaOH added (and moles of A- formed)

Remaining moles of HA = 1.8 - X

moles of A- = X

HA = 1.8 - X

X/(1.8-X) = 0.182

X =0.277

We have to add 0.277 mole NaOH to 1.0 L of 1.8 M HC2H3O2

To control we can do the following equation:

4 = 4.74 + log(0.277/1.523)

c) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 5.00

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

Ka = 1.8*10^-5

ph = 5

Step 2:

Use the Henderson-Hasselbalch equation.

pH = pKa + log(A-/HA)

5 = 4.74 + log(A-/HA)

0.26 =  log(A-/HA)

A-/HA = 1.82

Consider X = moles of NaOH added (and moles of A- formed)

Remaining moles of HA = 1.8 - X

moles of A- = X

HA = 1.8 - X

X/(1.8-X) = 1.82

X =1.16

We have to add 1.16 mole NaOH to 1.0 L of 1.8 M HC2H3O2

To control we can do the following equation:

5 = 4.74 + log(1.16/0.64) = 5

Explanation:

Answer:

a. Phosphoric Acid

b. Acetic Acid

c. Hypochlorous Acid

Explanation:

A buffer works when the pH of this one is in pKa ± 1. That means, to find which buffer system works in some pH you need to find pKa:

pKa = -log Ka

pKa Acetic acid:

-log1.8x10⁻⁵ = 4.74

pKa phosphoric acid:

-log7.5x10⁻³ = 2.12

pKa hypochlorous acid:

-log3.5x10⁻⁸ = 7.46

a. For a pH of 2.8 the best choice is phophoric acid because its effective range is: 1.12 - 3.12 and 2.8 is between these values.

b. pH 4.5. Acetic acid. effective between pH's 3.74 - 5.74

c. pH 7.5. Hypochlorous acid that works between 6.46 and 8.46

Moles of Sr²⁺ = 0.3375

Molarity is a way to express the concentration of the solution

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

[tex]\large{\boxed {\bold {M ~ = ~ \frac {n} {V}}}[/tex]

Where

M = Molarity

n = Number of moles of solute

V = Volume of solution

So to find the number of moles can be expressed as

[tex]\boxed{\bold{n=V\times M}}[/tex]

Strontium phosphide (Sr₃P₂)

Sr₃P₂ ⇒ 3Sr ²⁺ + 2P³⁻

[tex]\tt 0.45~M \times 0.25~L=0.1125[/tex]

mol ratio Sr₃P₂ : Sr ²⁺ = 1 : 3

[tex]\tt \dfrac{3}{1}\times 0.1125=0.3375[/tex]

Answer:

True

Explanation:

According to some calculations, the Earth is losing 50,000 metric tons of mass every single year, even though an extra 40,000 metric tons of space dust converge onto the Earth's gravity well, it's still losing weight.

Answer:

true

Explanation:

Answer:

11.023 pounds

Explanation:

Answer:

a

Explanation:

Answer:

The acceleration will be .5 and the force will be 37.5

Have a nice day!   :)

Answer:

Hello. The answer is 37.5N.

Explanation:

If you look carfully ، every time that you add one weight the force changes 2.5N so you can do this:

[tex](2.5 \times 5) + 25 = 37.5[/tex]

Answer:

it is Calcium (Ca)

4th period, 2nd group, 2 valence electrons

Answer:

Matter changes to a plasma state

Explanation:

Plasma is superheated matter, so when matter is heated to the point that electrons are stripped off of atoms forming ionized gas.  Palsma comprises over 99% of the observable universe.  Plasma is visible in the form of stars, nebulas, and sometimes even auroras.

Answer:

C. Matter changes to a plasma state.

Explanation:

Correct on Edge 2022!!!

Good luck everyone, you got this! Have a great day!

Answer:

0.0132 g CuSO₄

General Formulas and Concepts:

[tex]\text{Molarity}=\frac{\text{moles of solute}}{\text{liters of solution}}[/tex]

Explanation:

Step 1: Define variables

3.29 mL

4.00 M CuSO₄

x g CuSO₄

Step 2: Define conversions

1000 mL = 1 L

Step 3: Solve

Step 4: Simplify

We are given 3 sig figs.

0.01316 g CuSO₄ ≈ 0.0132 g CuSO₄

Answer:

See explanation

Explanation:

a) -About 1-2 drops of the unknown is dissolved in approximately 1 mL of analytical grade acetone standard solution. The test solution is added and shaken thoroughly.  If a positive test is not immediately observed, the set should be allowed to stand for 1-2 minutes. A positive result means the appearance of a green color.

b) The functional groups that can be distinguished by the test are alcohols and aldehydes.

Aldehydes give a positive test to chromic acid but ketones do not.

Primary and secondary alcohols give a positive test to chromic acid but tertiary alcohols do not.

c) Chromic Acid Test involves Cr in the +6 oxidation state. A positive test implies the reduction of orange Cr^6+ to green chromium Cr^3+.

d) The compounds were not shown but this image attached from lumen learning summarizes the reaction mechanism of chromic acid test.

This question is incomplete, the complete question is;

The flask contains 10.0mL of HCl and a few drops of phenolphthalein indicator.

The buret contains 0.320 M NaOH,

It requires 11.2 mL of the NaOH solution to reach the end point of the titration

What is the initial concentration of HCI?

Answer:  

initial concentration of HCl is 0.3584 M

Explanation:

Given that;

the reaction between HCl and NaOH is 1:1 molar

HCl(aq) + NaOH(aq) -------> NaCl(aq) + H2O(l)

we know that molarity is the number of moles of solute per liter of solution

so moles of NaOH consumed

= ( 0.320 mol / 1000ml) × 11.2ml

= 0.003584 mol

now

0.003584 moles of NaOH reacts with 0.003584 moles of HCl

moles of HCl present in the 10.0ml solution = 0.003584 mol

therefore the initial concentration of HCl

= ( 0.003584 mol / 10.0ml )× 1000ml

= 0.3584 M

the initial concentration of HCl is 0.3584 M

Answer:

D. Territory

Explanation:

Territory is defined as an area that is claimed by one or more individuals.

Territory is often used as a form of ownership of land or water mass between two parties. It is important as it gives each party a knowledge of their designated areas.

This validates it being an area that is claimed by one or more parties as organisms use that to protect their interest of food, water and mates for their survival.

Answer:extrinisic properties

i really need a brainliest

Answer:

The molar mass of Na (sodium) is 23.0 g/moles

Explanation:

Answer:

the molar mass of Na (sodium) is 23.0 g/mol.

Explanation:

Answer:D. Both of their ions will have the same negative charge

Explanation:

Answer:

1. 14 moles of NH3 were produced.

2. 7 moles of CO2 were produced.

3. 7 moles of H2O were produced.

Explanation:

We'll begin by writing a balanced equation for the decomposition of ammonium carbonate, (NH4)2CO3.

This is illustrated below:

(NH4)2CO3(s) —> NH3(g) + CO2(g) + H2O(g)

There are 2 atoms of N on the left side and 1 atom on the right side. It can be balance by putting 2 in front of NH3 as shown below:

(NH4)2CO3(s) —> 2NH3(g) + CO2(g) + H2O(g)

Now the equation is balanced.

From the balanced equation above,

1 mole of (NH4)2CO3 decompose to produce 2 moles of NH3, 1 mole of CO2 and 1 mole of H2O.

1. Determination of the number of mole of NH3 produced.

From the balanced equation above,

1 mole of (NH4)2CO3 decompose to produce 2 moles of NH3.

Therefore, 7 moles of (NH4)2CO3 will decompose to produce = 7 × 2 = 14 moles of NH3.

Therefore, 14 moles of NH3 were produced.

2. Determination of the number of mole CO2 produced.

From the balanced equation above,

1 mole of (NH4)2CO3 decompose to produce 1 mole of CO2.

Therefore, 7 moles of (NH4)2CO3 will also decompose to produce 7 moles of CO2.

Therefore, 7 moles of CO2 were produced.

3. Determination of the number of mole of H2O produced.

From the balanced equation above,

1 mole of (NH4)2CO3 decompose to produce 1 mole of H2O.

Therefore, 7 moles of (NH4)2CO3 will also decompose to produce 7 moles of H2O.

Therefore, 7 moles of H2O were produced.

Answer:

a) 0.0379 M

b) 24.89 mL

c)  0.09 mL

Explanation:

The equation of the reaction is:

2HBr(aq) + Ca(OH)2(aq) ---> CaBr2(aq) + 2H20(l)

a)

Concentration of acid CA = 0.117 M

Concentration of base CB= ??

Volume of acid VA = 16.0 ml

Volume of base VB = 24.7 ml

number of moles of acid NA = 2

number of moles of base NB= 1

From;

CA VA/CB VB = NA/NB

CAVANB =CBVBNA

CB = CAVANB/VBNA

CB = 0.117 *16 * 1/ 24.7 *2

CB  = 1.872/49.4

CB = 0.0379 M

b)

CA = 0.270 M

CB = 0.113 M

VA = ??

VB =22.8mL

NA =2

NB =1

From;

CA VA/CB VB = NA/NB

CAVANB =CBVBNA

VA = CBVBNA/CANB

VA = 0.113 * 22.8 * 2/0.207 *1

VA = 24.89 mL

c)

using the dilution formula

C1V1 =C2V2

Where:

C1= initial concentration =10.8 M

V1 = initial volume = ??

C2= final concentration =0.400 M

V2= final volume = 2.5 L

V1 = C2V2/C1

V1 = 0.400 * 2.5/10.8

V1 = 0.09 mL

Answer:

0.0836 mol

Explanation:

A chemist measured the amount of calcium sulfide produced during an experiment. She finds that 6.03 g of calcium sulfide is produced. Calculate the number of moles of calcium sulfide produced.

Step 1: Given data

Mass of calcium sulfide: 6.03 g

Step 2: Determine the chemical formula of calcium sulfide

Calcium cation has a charge +2 and sulfide anion has a charge -2. Then, the chemical formula of calcium sulfide is CaS

Step 3: Calculate the moles corresponding to 6.03 g of CaS

The molar mass of CaS is 72.14 g/mol. The moles corresponding to 6.03 g of CaS are:

6.03 g × (1 mol/72.14 g) = 0.0836 mol

Answer:

(A) A ball sitting on the ground

Explanation:

The option with the least gravitational potential energy in the question is : A: A ball sitting on the ground.

[tex]U= mgh[/tex]

h= height of the object

m= mass of the object.

Therefore, option A is right.

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Answer:

its a physical change

Explanation:

its b

Answer:

2 C 2 H 2 ( g ) + 5 O 2 ( g ) ⟶ 4 C O 2 ( g ) + 2 H 2 O ( l )- combustion reaction

N H 4 N O 3 ( s ) ⟶ N 2 O ( g ) + 2 H 2 O ( l )- decomposition reaction

C O ( g ) + 2 H 2 ( g ) ⟶ C H 3 O H ( l ) - combination reaction

2 F e ( s ) + 6 H C l ( a q ) ⟶ 2 F e C l 3 ( a q ) + 3 H 2 ( g )- Redox reaction

C a C l 2 ( a q ) + N a 2 C O 3 ( a q ) ⟶ 2 N a C l ( a q ) + C a C O 3 ( s )- double displacement reaction

Explanation:

We can determine the type of reaction by considering the reactants and products.

Combustion is a reaction between a substance and oxygen which produces heat and light. The first reaction is the equation for the combustion of ethyne.

A decomposition reaction is one in which a single reactant breaks down to form products. The second reaction is the decomposition of ammonium nitrate.

A  combination reaction is said to occur when two elements or compounds react to form a single product. The third reaction is the combination of carbon dioxide and methane to form  methanol.

An oxidation-reduction reaction is a reaction in which there is a change in oxidation number of species from left to right of the chemical reaction equation. The fourth reaction is the oxidation of iron (0 to +3 state) and reduction of hydrogen (+1 to 0 state).

A double displacement reaction is a reaction in which ions exchange partners from left to right in the reaction equation. The fifth reaction is a double displacement reaction. Both Na^+ and Ca^2+ exchanged partners from left to right of the reaction equation.

Reactions are the formation of the products from the reactant. The types of reactions are combustion, decomposition, combination, Redox and double displacement.

The reaction is a chemical change in the properties of the reactant that forms the products. It can be of various types based on the formation of the product.

The first reaction is combustion as the reactants react and use oxygen to form heat, carbon dioxide and water. The combustion reaction of ethyne can be shown as,

[tex]\rm 2 C _{2} H _{2} ( g ) + 5 O _{2} ( g ) \rightarrow 4 C O _{2} ( g ) + 2 H _{2} O ( l )[/tex]

The second reaction is decomposition in which a single reactant decomposes to form two or more products. The decomposition of ammonium nitrate can be shown as,

[tex]\rm N H _{4} N O _{3} ( s ) \rightarrow N _{2} O ( g ) + 2 H _{2} O ( l )[/tex]

The third reaction is a combination reaction in which two compound or elements combines to form one product. The combination reaction between carbon monoxide and hydrogen to form methanol can be shown as,

[tex]\rm C O ( g ) + 2 H _{2} ( g ) \rightarrow C H _{3} O H ( l )[/tex]

The fourth reaction is redox and includes the oxidation and the reduction of the species of the reaction. In the reaction, iron undergoes oxidation and hydrogen reduction. The redox reaction can be shown as,

[tex]\rm 2 F e ( s ) + 6 H C l ( a q ) \rightarrow 2 F e C l _{3} ( a q ) + 3 H _{2} ( g )[/tex]

The fifth reaction is a double displacement reaction in which the calcium and sodium interchange their position in the product formation. The reaction can be shown as,

[tex]\rm C a C l _{2} ( a q ) + N a _{2} C O _{3} ( a q ) \rightarrow 2 N a C l ( a q ) + C a C O _{3} ( s )[/tex]

Therefore, the type of reactions is 1. combustion, 2. decomposition, 3. combination, 4. redox and 5. double displacement.

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Answer:

Solution

Explanation:

In science, the solution is a homogeneous mixture of two or more pure substances in proportional concentrations that can be continually varied up to what has been called the solubilization limit. The word solution is usually applicable to the fluid subatomic particle, but gaseous formulations are conceivable.