Chemical buffers are compounds that will bind (click to select) when they are in abundance in a solution. This prevents an unwanted drop in pH. True False

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Newton’s first law of motion was a giant leap forward in scientific thought during Newton’s time. Even today, the idea is sometimes difficult at first for people to understand.

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The correct answer for the question that is being presented above is this one: "A rolling ball eventually slows down and comes to a stop." Newton’s first law of motion was a giant leap forward in scientific thought during Newton’s time. Even today, the idea is sometimes difficult at first for people to understand.

At standard pressure, the total amount of heat required to completely vaporize a 100. gram sample of water at its boiling point is

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Answer:

The heat of vaporization relates the amount of heat required to transform a certain phase of a certain amount of substance from liquid to gas. The heat of vaporization of substances can be expressed in terms of joules per gram or joules per mole.

ExplanaDetermine the total amount of heat, q, that is required to evaporate the given amount of water at its boiling point. For this problem, we simply apply the equation,q

=

m

Δ

H

v

a

p

where m is the mass and

Δ

H

v

a

p

is the enthalpy of vaporization of water. We use the following values:

m

=

100

g

Δ

H

v

a

p

=

2.26

k

J

/

g

We proceed with the solution.

q

=

m

Δ

H

v

a

p

=

(

100

g

)

(

2.26

k

J

/

g

)

=

226 k

j tion:

#1: Which of the following is the abbreviation for a unit of energy? A. K / B. °C/ C. W / D. cal............... #2: A 200 g block of a substance requires 1.84 kJ of heat to raise its temperature from 25°C to 45°C. Use the table attached to identify the substance. A. iron/ B. aluminum/ C. gold/ D. copper.....................#3: In a calorimeter, the temperature of 100 g of water decreased by 10°C when 10 g of ice melted. How much heat was absorbed by the ice? A. 418 kJ / B. 100 kJ / C. 10 J / D. 4.18 kJ ..................#4: The amount of heat needed to raise the temperature of 50 g of a substance by 15°C is 1.83 kJ. What is the specific heat of the substance? A. 2.44 J/g-°C / B. 2.22 J/g-°C / C. 2.13 J/g-°C / D. 2.05 J/g-°C ................#5: In a calorimeter, 3.34 kJ of heat was absorbed when 10 g of ice melted. What is the enthalpy of fusion of the ice? A. 6.68 J/g / B. 334 J/g / C. 6.68 kJ/g/ D. 334 kJ/g

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Among the choices, the unit of energy is calories. Answer in 1) is D. In 2) we are given with te mass , heat and temperature change. we just need to get the heat capacity and compare it with the following metals. The calculated heat capacity is 0.46 kJ/kg K. The answer is A. iron. In 3) we can compute the heat absorbed by the formula ΔH=mCpΔT. Cp of water is 4.18 J/g K. Answer of 3) is D. In 4) the formula used in Cp=ΔH/mΔT. Answer in 4) is A. The heat of enthalpy of fusion of ice is 80 cal/g. We convert this to J/g. Answer  in 5) is B.334 J/g.

Answer:

1. D. cal......

2.A. iron

3. D

4.2.44j/g°C   A

5,Lf=334J/g   B

Explanation:

1: Which of the following is the abbreviation for a unit of energy? A. K / B. °C/ C. W / D. cal...............

calorie is the unit of energy

#2: A 200 g block of a substance requires 1.84 kJ of heat to raise its temperature from 25°C to 45°C. Use the table attached to identify the substance. A. iron/ B. aluminum/ C. gold/ D. copper.....................

Q=mcdt

1840=0.2*C*(45-25)

C=460J/KgK

if the specific heat capacity is the above then he substance is iron

#3: In a calorimeter, the temperature of 100 g of water decreased by 10°C when 10 g of ice melted. How much heat was absorbed by the ice? A. 418 kJ / B. 100 kJ / C. 10 J / D. 4.18 kJ .................

Q=mcdT

Q=0.1*10*4180

Q=4180j. answer D

.#4: The amount of heat needed to raise the temperature of 50 g of a substance by 15°C is 1.83 kJ. What is the specific heat of the substance? A. 2.44 J/g-°C / B. 2.22 J/g-°C / C. 2.13 J/g-°C / D. 2.05 J/g-°C ................

Q=mcdT

1830=50/1000*C*15

C=2440j/kg/k

change it to j/g°C

2.44j/g°C   A

#5: In a calorimeter, 3.34 kJ of heat was absorbed when 10 g of ice melted. What is the enthalpy of fusion of the ice? A. 6.68 J/g / B. 334 J/g / C. 6.68 kJ/g/ D. 334 kJ/g

Q=mLf

Lf=enthalpy of fusion

3340/10=Lf

Lf=334J/g   B

Enthalpy of fusion quantity of heat to convert 1 unit mass of a solid to liquid without any noticeable change in temperature.

What is the name of the giant bubble of charged gas that surrounds the sun?a. heliosphere
b. galaxy
c. Kuiper Belt
d. Oort Cloud

Answers

The correct answer is A. The gin bubble of charged gas around the sun is called the heliosphere. Basically, the heliosphere is a sphere of helium. This is a region where the solar wind is said to have an influence.

C2H5OH is able to dissociate in water to conduct an electric current. This compound could be classified as

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The fact that when in solution CHOH conducts electricity means that it is an electrolyte so you can automatically cross out A and C (the definition of an electrolyte is a substance that will conduct electricity when in a solution).  I think the correct answer is B due to the -OH group (the - indicates a covalent bond) which allows the molecule to give up a proton (H⁺).
 I hope this helps.  Let me know if anything is unclear.

________ is a strong, parallel alignment of coarse mica flakes and/or of different mineral bands in a metamorphic rock.

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Answer:

Foliation is a strong, parallel alignment of coarse mica flakes and/or of different mineral bands in a metamorphic rock.

Explanation:

Foliation is most commonly prevalent in metamorphic rocks. Foliation is the parallel alignment of textural and structural features of a rock.

Differential stress plays a major role on the texture of metamorphic rocks because it forces the mineral constituent of the rocks to align parallel to each other.  Foliation can occur in different ways for example a mineral like mica which is usually platy can crystallize in rocks , due to differential stress the mineral grows in such a way it remains parallel to the movement in which the part of the rock slide relative to one another and parallel to the forces applied or the mineral might grow perpendicular to the direction of the compressed stress.  

Notice the foliation in the texture of gneiss .You can see the light and dark mineral found in separate or parallel layer.