Mass of a substance divided by unit volume is called

The article is on newsela called the water cycle

garri49 [273]

Answer:

Explanation:Canguru comeu algumas folhas de 3 galhos de eucalipto. Cada galho tinha inicialmente20 folhas. Canguru comeu algumas folhas do primeiro galho e depois comeu tantas folhas do segundo galho quantas tinham sido deixadas no primeiro galho. Depois ele comeu 2 folhas do terceiro galho. No total, quantas folhas foram dei-xadas nos 3 galhos?Canguru comeu algumas folhas de 3 galhos de eucalipto. Cada galho tinha inicialmente20 folhas. Canguru comeu algumas folhas do primeiro galho e depois comeu tantas folhas do segundo galho quantas tinham sido deixadas no primeiro galho. Depois ele comeu 2 folhas do terceiro galho. No total, quantas folhas foram dei-xadas nos 3 galhos?Canguru comeu algumas folhas de 3 galhos de eucalipto. Cada galho tinha inicialmente20 folhas. Canguru comeu algumas folhas do primeiro galho e depois comeu tantas folhas do segundo galho quantas tinham sido deixadas no primeiro galho. Depois ele comeu 2 folhas do terceiro galho. No total, quantas folhas foram dei-xadas nos 3 galhos?Canguru comeu algumas folhas de 3 galhos de eucalipto. Cada galho tinha inicialmente20 folhas. Canguru comeu algumas folhas do primeiro galho e depois comeu tantas folhas do segundo galho quantas tinham sido deixadas no primeiro galho. Depois ele comeu 2 folhas do terceiro galho. No total, quantas folhas foram dei-xadas nos 3 galhos?Canguru comeu algumas folhas de 3 galhos de eucalipto. Cada galho tinha inicialmente20 folhas. Canguru comeu algumas folhas do primeiro galho e depois comeu tantas folhas do segundo galho quantas tinham sido deixadas no primeiro galho. Depois ele comeu 2 folhas do terceiro galho. No total, quantas folhas foram dei-xadas nos 3 galhos?

7 0

3 years ago

Use the following scenario to answer the question: A cell has an antiport protein on its apical surface. The cell is placed in a

balu736 [363]

Answer:

The correct answer is "Secondary active transport".

Explanation:

Secondary active transport is a form of across the membrane transport that involves a transporter protein catalyzing the movement of an ion down its electrochemical gradient to allow the movement of another molecule or ion uphill to its concentration/electrochemical gradient. In this example, the transporter protein (antiporter), move 3 Na⁺ into the cell in exchange for one Ca⁺⁺ leaving the cell. The 3 Na⁺ are the ions moved down its electrochemical gradient and the one Ca⁺⁺ is the ion moved uphill its electrochemical gradient, because Na+ and Ca⁺⁺are more concentrated in the solution than inside the cell. Therefore, this scenario is an example of secondary active transport.

6 0

3 years ago

Type of Reaction _Mg₃N2 →

nevsk [136]

Ionic reaction when a metal and non metal are together a molecule is when a metal abs metal react hope this helps

8 0

3 years ago

Which of the following would release the most heat? Assume the same mass of in each case. Specific heats of ice, liquid water, a

lesya692 [45]

Answer:

The process which releases most heat is E)

Explanation:

As we know that water freezes at 0ºC and vaporizes at 100ºC, we calculate the heat as follows:

Processes with temperatures < 0ºC : by using specific heat of ice (Sh ice) multiplied by the change in temperature (ΔT= Final Temperature - Initial Temperature)⇒ Sh ice x ΔT

Processes of ice melting (at 0ºC): by using heat of fusion of ice (ΔH fus) multiplied by a conversor factor (1 mol H20= 18 g)⇒ ΔHfus x 1mol/18g

Processes between 0ºC and 100ºC: by using specific heat of liquid water (Sh liq) multiplied by change in temperature ⇒ Sh liq x ΔT

Processes of water evaporation (at 100ºC): by using heat of vaporization (ΔH vap) multiplied by the conversor factor ⇒ ΔH vap x 1mol/18 g

Processes at a temperature >100ºC: by using specific heat of water vapor (Sh vap) multiplied by the change in temperature ⇒ Sh vap x ΔT

A) Water at -25ºC is ice. Ice is heated from -25ºC to 0ºC, then it melts at 0ºC (ice became liquid water) and then liquid water is heated from 0ºC to 70ºC. T

This is the only process in with the heat is absorbed (not releases), so it cannot be the right answer, but we calculate the heat involved to practice:

Heat= (Sh ice x ΔT) + (ΔH fus x 1/18 g) + Sh liq x ΔT

Heat= (2.05 J/g ºC x (0ºC -(-25ºC) ) + (6.01 x 10³ J/mol x 1 mol/18 g) + (4.18 J/g ºC x (70ºC-0ºC)

Heat= 51.25 J + 333,8 J +292.6 J

Heat= 677.65 J (heat is absorbed)

B) Water is cooled from 13ºC to 0ºC, then it is freezed at 0ºC and then the ice is cooled from 0ºC to -2.6 ºC

Heat= (Sh liq x ΔT) + (-ΔH melt x 1/18 g) + (Sh ice x ΔT)

Heat= 4.18 J/ºC x (0ºC- 13ºC) + (-6.01 x 10³ J/mol x 1mol/18 g) + (2.05 J/ºC x (-2.5ºc - 0ºC)

Heat= -54.34 J - 333.8 J + 5.33 J

Heat= -393.47 J (heat is released)

C) Liquid water is cooled from 74ºC to 95ºC

Heat= Sh liq x ΔT

Heat= 4.18 J/ºC x (74ºC - 95ºC)

Heat = -87.78 J (heat is released)

D) Water at 140ºC is in vapor state. Vapor at 140ºC is cooled to 110ºC (still vapor).

Heat = Sh vap x ΔT

Heat= 2.01 J/ºC x (110ºC - 140ºC)

Heat= -60.3 J (heat is released)

E) Vapor at 106ºC is cooled to 100ºC, then it condenses at 100ºC (convertion from gas to liquid), and liquid water is cooled from 100ºC to 88ºC.

Heat= (Sh vap x ΔT) + (-ΔHvap x 1mol/18g) + (Sh liq x ΔT)

Heat= (2.01 J/ºC x (100ºC-106ºC)) - (40.7 x 10³ J/mol x 1mol/18 g) + (4.18 J/ºC x (88ºC -100ºC)

Heat= -2323.32 J (heat is released) THIS IS THE RIGHT ANSWER (the more negative= the more released)

7 0

3 years ago

If a student mixes 75 mL of 1.30 M HNO3 and 150 mL of 6.5 M NaOH. is the final solution acidic, basic, or neutral

raketka [301]

Answer:

The solution is basic.

Explanation:

We can determine the nature of the solution via determining which has the large no. of millimoles (acid or base):

If no. of millimoles of acid > that of base; the solution is acidic.

If no. of millimoles of acid = that of base; the solution is neutral.

If no. of millimoles of acid < that of base; the solution is basic.

We need to calculate the no. of millimoles of acid and base:

no. of millimoles of acid (HNO₃) = MV = (1.3 M)(75.0 mL) = 97.5 mmol.

no. of millimoles of base (NaOH) = MV = (6.5 M)(150.0 mL) = 975.0 mmol.

∴ The no. of millimoles of base (NaOH) is larger by 10 times than the acid (HNO₃).

So, the solution is: basic.

4 0

3 years ago