Choose all the right answers.
Choose the two places where continental glaciers can be seen.
Iceland
Norway
Antarctica
Greenland

Answers

Answer 1
Antarctica and Greenland
Answer 2

Answer:

the real answer is Iceland, Norway, Antarctica.

Explanation:

It is not Green land proof Glaciers are not an unusual sight in Greenland due to the gigantic ice sheet that covers most of the country. However, there are only a few places where the glaciers occur immediately adjacent to a town, and therefore can be appreciated on an excursion.  

proof for Antarctica These large-scale features of glacial erosion can be seen in Antarctica where rock protrudes above the surface of the ice sheet There are many glaciers in the Antarctic. ... The lists include outlet glaciers, valley glaciers, cirque glaciers, tidewater glaciers and ice streams. Ice streams are a type of glacier and many of them have "glacier" in their name, e.g. Pine Island Glacier

proof for Norway are the largest glaciers on mainland Norway. ... In total, Norway has around 1,600 glaciers - 900 of these are in North Norway, but 60% of the total glacier area is south of Trøndelag. 1% of mainland Norway is covered by glaciers.

proof for Iceland Glaciers in Iceland are among the best natural wonders in the world. In fact, the country's glaciers are so great that you can see them from space! With 11% of the total land area covered by ice caps, Iceland is a glacier paradise. Here are the five most amazing glaciers that you must see when visiting Iceland!


Related Questions

The radius of Pt atom in an fcc structure is 0.1386 nm. Pt has atomic mass of 195.09 g/mol. Calculate the density of this fcc structure. The Avogadro’s number NAis 6.022 x 1023 per mole

Answers

Answer:

21.51g/cm³

Explanation:

To answer this question we need to know that, in 1 unit FCC cell:

Edge length = √8 * R  

Volume = 8√8 * R³

And there are 4 atoms per unit cell

The mass of the 4 atoms of the cell, in grams, is:

4 atom * (1mol / 6.022x10²³atom) * (195.09g / mol) = 1.2958x10⁻²¹g

Volume in cm³:

0.1386nm * (1x10⁻⁷cm / 1nm) = 1.386x10⁻⁸cm

Volume = 8√8 * (1.386x10⁻⁸cm)³

Volume = 6.02455x10⁻²³cm³

And density, the ratio of mass and volume, is:

1.2958x10⁻²¹g  / 6.02455x10⁻²³cm³  =

21.51g/cm³

Determine the molar mass of CaO

Answers

56.0774 g/mol just took the test

A student asks why the ashes from a fire have a much lower mass than the wood that was burned.

Which is the correct answer to the student’s question?

Gases are released into the air.

Atoms in the wood are destroyed.

Heat causes the molecules to change color.

Water inside the wood solidifies.


(I am not in college)

Answers

Gases are released into the air.

calculate the equilibrium concentration for the nonionized bases and all ions in a solution that is 0.25M

Answers

Answer:

The equilibrium concentration of [CH₃NH₂] = 0.23965 M.

The equilibrium concentration of CH₃NH₃⁺ and  OH⁻ =  0.01035 M respectively.

Explanation:

The first step is to write out the dissociation reaction. Therefore, the equation showing the dissociation is given as below;

CH₃NH₂ + H₂0   <--------------------------------------------------------> CH₃NH₃⁺ + OH⁻.

Kindly note that ''<----------->'' arrow shows that the reaction is an equilibrium reaction.

Therefore, at the start of the reaction [that is time, t =0], we have that the concentration of CH₃NH₂ = 0.25M, thus, the concentration of CH₃NH₃⁺  and  OH⁻ is zero respectively at this time, t =0.

At equilibrium, the concentration of CH₃NH₂ = 0.25M - x, thus, the concentration of CH₃NH₃⁺  and  OH⁻ is x respectively.

Therefore, kb =  4.47 × 10-4 = [CH₃NH₃⁺ ][OH⁻]/[CH₃NH₂].  Hence, slotting in the values into this equilibrium equation showing the relationship between kb and concentration of the species involved, we have that;

kb = 4.47 × 10⁻⁴ = x² /0.25 - x.

x² +  4.47 × 10⁻⁴x - 1.1175  × 10⁻⁴ = 0.

Solving this quadratic equation gives us the value of x as 0.01035 M.

Thus, the concentration of [CH₃NH₂] = 0.25 M - 0.01035 M = 0.23965 M

The equilibrium concentration of [CH₃NH₂] = 0.23965 M.

The equilibrium concentration of CH₃NH₃⁺ and  OH⁻ =  0.01035 M respectively.

PLEASE PLEAS HELP Which of the following compounds is insoluble in water?
a) ZnSO4
b) K2SO4
c) Na2CO3
d) Ag2CO3

Answers

Answer:

Your answer is d

Explanation: silver carbonate Ag2CO3 is insoluble in water

Select the term that matches each definition:
a) A decrease in the solubility of an ionic compound as a result of the addition of a common ion.
b) The mass of a salt in grams that will dissolve in 100 mL of water.
c) A solution that has dissolved the maximum amount of a compound at a given temperature. Any further addition of salt will remain undissolved.
d) The product of the molarities of the dissolved ions, raised to a power equal to the ion's coefficient in the balanced chemical equation.
e) The maximum number of moles of a salt that will dissolve in 1 L of solution.
*** Answer options for all questions: ***
- Solubility
- Molar Solubility
- Solubility product constant
- Common ion effect
- Saturated Solution

Answers

Answer:

a) common ion effect

b) solubility

c) saturated solution

d) solubility product constant

e) molar solubility

Explanation:

When a substance, say BA2 is dissolved in a solution and another substance CA2 is dissolved in the same solution. The solubility of BA2 is decreased due to the addition of CA2. This is known as common ion effect.

The mass of a substance that will dissolve in a given Volume of solvent is known as it's solubility.

The molar solubility is the amount of moles of solvent that dissolves in 1 dm^3 of solvent.

A solution that contains just as much solute as it can normally hold at a given temperature is known as a saturated solution.

Lastly, the product of molar solubilites raised to the power of the molar coefficient is know as the solubility product constant.

The correct matches and their explanation are:

a) A decrease in the solubility of an ionic compound as a result of the addition of a common ion: Option 4. common ion effect

It relates to the equilibrium effect that occurs due to the addition of common ions.

b) The mass of salt in grams that will dissolve in 100 mL of water: Option 1. solubility

Solubility is the property of solute to form a solution with the solvent of another substance.

c) A solution that has dissolved the maximum amount of a compound at a given temperature. Any further addition of salt will remain undissolved: Option 5. saturated solution

When the solution cannot accommodate any more addition of solute of a substance is called a saturated solution.

d)  The product of the molarities of the dissolved ions, raised to a power equal to the ion's coefficient in the balanced chemical equation: Option 3. solubility product constant

It is an equilibrium constant for the solid solute dissolved in the solution.

e) The maximum number of moles of a salt that will dissolve in 1 L of solution: Option 2. molar solubility

Before the saturation of a solution, the amount of solute it can dissolve is called molar solubility.

To learn more about molar solubility and common ion effect follow the link:

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Which item has the most thermal energy?
A. 1 kg boiling water
B. 1 kg ice
C. 1 kg hot water just below the boiling point
D. 1 kg cold water

Answers

Answer:

1L of hot water just below the Boling point

Explanation:

asking questions is best to learn please ask more questions

Answer:

its 1kg of boiling water

Explanation:

Look at the image HOH. What does this image represent?

Answers

Answer:

Formation of covalent bond structures. The image is essentially a Lewis dot structure.

The image of HOH has depicted the covalent bond formation in water. Thus, option B is correct.

The image has been the Lewis structure of the water molecule. It has been consisted of two hydrogen atoms bonded to oxygen molecules.

The oxygen has a presence of 4 dots that have been the representation of the valence electrons.  Thus, oxygen has been consisted of 2 lone pairs.

The water molecule has the presence of shared electrons between hydrogen and oxygen. The bond formed by the sharing of electrons has been covalent.

Thus, the image of HOH has been the representation of the formation of covalent bonds in water. Thus, option B is correct.

For more information about Lewis structure, refer to the link:

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help me please i’ll give u a good rating

Answers

Answer:

d

Explanation:

I think possibly the answer would be “d”

Y-Chart
Explain how an invasive species can influence each aspect of an ecosystem.

Answers

Answer:

Invasive species are an organism that causes ecological or economic harm in a new environment where it's not native.

Explanation:

An invasive species can harm both the natural resources in the ecosystem as well it threaten the human use of these resources and invasive species can be introduced to a new area via the ballast water of oceangoing ships, intentional and accidental releases of aquaculture species, aquarium specimens or bait, and etc.

Invasive species is capable of causing extinctions to native plants and animals, reducing biodiversity, competing with native organisms for limited resources, and altering habitats. This can also result a huge economic impacts and fundamental disruptions of coastal and the great lakes of the ecosystems.

I hope it helps you.

The reaction between HCl and KOH results in an increase in temperature in the solution. Select the correct statement from the list below.

a) this is an endothermic reaction
b) this is a phase change reaction
c) this is a vaporization reaction
d) this is an exothermic reaction

Answers

Answer:

d) this is an exothermic reaction.

Explanation:

The reaction between HCl and KOH results in an increase in temperature in the solution. Select the correct statement from the list below.

a) this is an endothermic reaction . NO. This would cause a decrease in the temperature of the solution.

b) this is a phase change reaction . NO. All the species remain in the aqueous phase.

c) this is a vaporization reaction . NO. All the species remain in the aqueous phase.

d) this is an exothermic reaction. YES. The reaction releases heat, so it is exothermic.

13. (6C) An unknown chemical has the
following properties: it is a white crystal,
reacts with water, and has a high boiling
point. Which of these properties is
physical?
A White color only
B Reacts with water, high
boiling point
C Reacts with water only
D White color, high boiling
point

Answers

Answer: it is B

Explanation:

The empirical formula of CBr2 has a molar mass of 515.46 g/mol. What is the molecular formula of this
compound

Answers

Answer:

C3Br6

Explanation:

C= (1 X 12.011) = 12.011

Br= (2 X 79.904)= 159.808

159.808+12.011 = 171.819

515.46 divided by 171.819 = 3.00

so you mulitpy CBr2 by 3 which gives you C3Br6

A stream of oxygen enters a compressor at 298 K and 1.00 atm at a rate of 127m3/h and is compressed to 358 K and 1000 atm. Estimate the volumetric flow rate of compressed O2 using the compressibility-factor equation of state.

Answers

Answer:

The value is  [tex]V_2  =   0.246 \  m^3/h[/tex]  

Explanation:

From the question we are told that

  The temperature at which the gas enters the compressor is  

         [tex]T_i = 298 \  K[/tex]

  The pressure  at which the gas enters the compressor is  

       [tex]P_I =  1.0 \ atm[/tex]

  The volumetric rate at which the gas enters the compressor is

      [tex]V  =  127 m^3/h[/tex]

   The temperature to which the gas is compressed to is  

      [tex]T_f  =  358 \ K[/tex]

  The pressure  to which the gas is compressed to is  

     [tex]P_f=  1000 \  atm[/tex]

Generally the volumetric flow rate of compressed oxygen is evaluated from the compressibility-factor equation of state as

    [tex]V_2  =  V_1 *\frac{z_2}{z_1} * \frac{T_2}{T_1} * \frac{P_1}{P_2}[/tex]

Here [tex]z_1[/tex] is the inflow compressibility factor with value [tex]z_1 = 1[/tex]

Here [tex]z_1[/tex] is the outflow compressibility factor with value [tex]z_2 = 1.61[/tex]

So

    [tex]V_2  =  127*\frac{1.61}{1} * \frac{358}{298} * \frac{1}{1000}[/tex]  

    [tex]V_2  =   0.246 \  m^3/h[/tex]  

   

Determine the number of moles of oxygen atoms in each of the following.
1) 4.93 mol H2O2
2) 2.01 mol N2O

Answers

Answer :

Part 1: 4.93 moles of [tex]H_2O_2[/tex] contains 9.86 moles of oxygen atoms.

Part 2: 2.01 moles of [tex]N_2O[/tex] contains 2.01 moles of oxygen atoms.

Explanation :

Part 1: 4.93 mol [tex]H_2O_2[/tex]

In 1 mole of [tex]H_2O_2[/tex], there are 2 atoms of hydrogen and 2 atoms of oxygen.

As, 1 mole of [tex]H_2O_2[/tex] contains 2 moles of oxygen atoms.

So, 4.93 moles of [tex]H_2O_2[/tex] contains [tex]4.93\times 2=9.86[/tex] moles of oxygen atoms.

Thus, 4.93 moles of [tex]H_2O_2[/tex] contains 9.86 moles of oxygen atoms.

Part 2: 2.01 mol [tex]N_2O[/tex]

In 1 mole of [tex]N_2O[/tex], there are 2 atoms of nitrogen and 1 atom of oxygen.

As, 1 mole of [tex]N_2O[/tex] contains 1 mole of oxygen atoms.

So, 2.01 moles of [tex]N_2O[/tex] contains [tex]2.01\times 1=2.01[/tex] moles of oxygen atoms.

Thus, 2.01 moles of [tex]N_2O[/tex] contains 2.01 moles of oxygen atoms.

Compute the values of the diffusion coefficients for the interdiffusion of carbon in both α-iron (BCC) and γ-iron (FCC) at 900°C. Which is larger? Explain why this is the case.

Answers

Answer:

α-iron (BCC) has faster diffusion rate because of lower values in activation energy and pre-exponential value.

Explanation:

Taking each parameters or data at a time, we can determine the values/a constant for each parameters in the diffusion coefficient equation.

For α-iron (BCC), the diffusion coefficient = pre-exponential value,Ao × e^( -Activation energy,AE)/gas constant,R × Temperature.

Converting the given Temperature, that is 900°C to Kelvin which is equals to 1173.15K.

For α-iron (BCC), the pre-exponential value, Ao = 1.1 × 10^-6, and the activation energy, AE = 87400.

Thus, we have that the diffusion coefficient = 1.1 × 10^-6 × e(-87400)/1173.15 × 8.31.

Diffusion coefficient for α-iron (BCC) = 1.41 × 10^-10 m^2/s.

Also, For the γ-iron (FCC), the pre-exponential value, Ao = 2.3 × 10^-5 and the activation energy, AE = 148,00.

From these values we can see that both the exponential value, Ao and the activation energy for γ-iron (FCC) are higher than that of α-iron (BCC).

Thus, the diffusion coefficient for the γ-iron (FCC) = 2.3 × 10^-5 × e ^-(14800)/8.31 × 1173.15.

Then, the diffusion coefficient for the γ-iron (FCC) = 5.87 × 10^-12 m2/s.

Therefore, there will be faster diffusion in α-iron (BCC) because of lower activation energy and vice versa.

A 57.07 g sample of a substance is initially at 24.3°C. After absorbing of 2911 J of heat, the temperature of the substance is 116.9 CWhat is the specific heat (SH) of the substance?

Answers

Answer:

Approximately [tex]0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}[/tex].

Explanation:

The specific heat of a material is the amount of energy required to increase unit mass (one gram) of this material by unit temperature (one degree Celsius.)

Calculate the increase in the temperature of this sample:

[tex]\Delta T = (116.9 - 24.3)\; \rm ^\circ\! C= 92.6\; \rm ^\circ\! C[/tex].

The energy that this sample absorbed should be proportional the increase in its temperature (assuming that no phase change is involved.)

It took [tex]2911\; \rm J[/tex] of energy to raise the temperature of this sample by [tex]\Delta T = 92.6\; \rm ^\circ\! C[/tex]. Therefore, raising the temperature of this sample by [tex]1\; \rm ^\circ\! C[/tex] (unit temperature) would take only [tex]\displaystyle \frac{1}{92.6}[/tex] as much energy. That corresponds to approximately [tex]31.436\; \rm J[/tex] of energy.

On the other hand, the energy required to raise the temperature of this material by [tex]1\; \rm ^\circ\! C[/tex] is proportional to the mass of the sample (also assuming no phase change.)

It took approximately [tex]31.436\; \rm J[/tex] of energy to raise the temperature of [tex]57.07\; \rm g[/tex] of this material by [tex]1\; \rm ^\circ C[/tex]. Therefore, it would take only [tex]\displaystyle \frac{1}{57.07}[/tex] as much energy to raise the temperature of [tex]1\; \rm g[/tex] (unit mass) of this material by [tex]1\; \rm ^\circ \! C\![/tex]. That corresponds to approximately [tex]0.551\; \rm J[/tex] of energy.

In other words, it takes approximately [tex]0.551\; \rm J[/tex] to raise [tex]1\; \rm g[/tex] (unit mass) of this material by [tex]1\; \rm ^\circ \! C[/tex]. Therefore, by definition, the specific heat of this material would be approximately [tex]0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}[/tex].

Give Me An Atom With The Following Characteristics Lanthanide series

Answers

Atoms in the Lanthanide series are Lanthanum, Cerium, Neodymium, Europium, terbium, to name a few.

To what volume would you need to dilute 200 mL of a 5.85M solution of Ca(OH)2 to make it a 1.95M solution?

Answers

Answer: 600 mL

Explanation:

Given that;

M₁ = 5.85 m

M₂ = 1.95 m

V₁ = 200 mL

V₂ = ?

Now from the dilution law;

M₁V₁ = M₂V₂

so we substitute

5.85 × 200 = 1.95 × V₂

1170 = 1.95V₂

V₂ = 1170 / 1.95

V₂ = 600 mL

Therefore final volume is 600 mL

When 28.0 g of acetylene reacts with hydrogen, 24.5 g of ethane is produced. What is the percent yield of C2H6 for the reaction?
C2H2(g) + 2H2(g) → C2H6(g)

Answers

Answer:

[tex]Y=75.6\%[/tex]

Explanation:

Hello.

In this case, since no information about the reacting hydrogen is given, we can assume that it completely react with the 28.0 g of acetylene to yield ethane. In such a way, via the 1:1 mole ratio between acetylene (molar mass = 26 g/mol) and ethane (molar mass = 30 g/mol), we compute the yielded grams, or the theoretical yield of ethane as shown below:

[tex]m_{C_2H_6}^{theoretical}=28.0gC_2H_2*\frac{1molC_2H_2}{26gC_2H_2}*\frac{1molC_2H_6}{1molC_2H_2} *\frac{30gC_2H_6}{1molC_2H_6}\\ \\m_{C_2H_6}^{theoretical}=32.3gC_2H_6[/tex]

Hence, by knowing that the percent yield is computed via the actual yield (24.5 g) over the theoretical yield, we obtain:

[tex]Y=\frac{24.5g}{32.3g}*100\%\\ \\Y=75.6\%[/tex]

Best regards.

which element Shows very similar chemical properties to barium?​

Answers

Tanong mo sa pagong lol

How many grams of H2SO4 are needed to prepare 5.0 L of a 2 M H2SO4 solution? You must show your work in order to receive credit.

Answers

Molar mass H2SO4 = 98g/mol

To prepare 5L of a 2.0M solution you require: 98g/mol * 5mol * 2mol /L = 980g H2SO4

2.
(3x – 4y = -10
(6x + 3y = –42
SOLUTION:

Answers

Answer:

Is like for solving for the solution for both equations??

Which of the following represents a species with 16 protons and 18 electrons? A) Ar B) S C) S²⁻ D) Si⁴⁻ E) S²⁺ Question 5 of 40

Answers

Answer:S²⁻

Explanation:

Sulphur is in group 16. The atomic number of sulphur is 16.

However, sulphur can accept two electrons to form the sulphide ion S²⁻ which contains 18 electrons, hence the answer provided above.

calculate the mass of N2 gas which has a volume 0.227 at STP​

Answers

Hi I just wanted you know I have to work at the gym I will call when you I have a good night love I will

What is the density of a block of gold that occupies 1000 ml and has a mass of 3.5 kg? Show your work

Answers

Answer:

We are given:

mass of the block = 3500 grams

volume of the block = 1000 mL

Finding the density:

Density = mass of the object (in grams) / volume of the object (in mL)

Density = 3500 / 1000

Density = 3.5 g / mL

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Answers

Answer:

I would say it is true

Explanation:

I believe this is true (please give me thanks)

the half life of I -137 is 8.07 days. if 24 grams are left after 40.35 days, how many grams were in the original sample?

Answers

Answer:

768g

Explanation:

We can use to formula [tex]N(A) = N_0(\frac{1}{2})^\frac{t}{t_{1/2}}[/tex] . Here, N(A) is the final amount. N0 is the initial amount. t is the time elapsed, and [tex]t_{1/2}[/tex] is the half life. Plugging in, we get the answer above.

The half life of I -137 is 8.07 days. if 24 grams are left after 40.35 days, 800 gram  were in the original sample.

What is half life?

The half-life (symbol t12) is the amount of time it takes for a volume (of material) to be reduced to half of its original value. In nuclear physics, the phrase is typically used to indicate how rapidly unstable atoms experience radioactive decay or even how long stable nuclei survive.

The phrase is sometimes used more broadly to describe any form of exponential (or, in rare cases, non-exponential) decay. The biological ½ of medications and other compounds in the human body, for example, is referred to in the medical sciences. In exponential growth, the inverse of half-life is doubling time.

ln P = kt + C

P = amount of I-137 at time t

C = constant

k = 1/time

t = time

1st condition:

P = Po, t = 0 days

2nd condition: (half-life)

P = 0.5Po, t = 8.07 days

3rd condition:

P = 25 grams, t = 40.35 days

Po = 800 grams

mass of I-137 = 800 gram

Therefore, 800 gram were in the original sample.

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Determine the mass (in grams) of NaCl in 294 grams of a 24.1% (m/m) NaCl solution. Be sure to report to the correct number of significant figures with no units.
I NEED HELP WITH THIS URGENTLY!!!!!

Answers

Answer:

70.9 grams 3 sig figs

Explanation:

24.1% of 294 grams = 0.241(294 gram) = 70.854 grams ≅ 70.9 grams 3 sig figs

The natural abundance for boron isotopes is 19.9% 10B and 80.1%
11B Calculate boron's atomic mass.

Answers

Answer:

10.801 amu

Explanation:

From the question given above, the following data were obtained:

Isotope A (¹⁰B):

Mass of A = 10

Abundance (A%) = 19.9%

Isotope B (¹¹B):

Mass of B = 11

Abundance (B%) = 80.1%

Atomic mass of Boron =?

The atomic mass of boron can be obtained as illustrated below:

Atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100]

= [(10 × 19.9)/100] + [(11 × 80.1)/100]

= 1.99 + 8.811

= 10.801 amu

Thus, the atomic mass of boron is 10.801 amu

The atomic mass of boron with natural abundance of 19.9% of 10 B and 80.1%  of 11 B is 10.801 amu

Boron has 2 isotopes.

First isotopes

mass = 10

% abundance = 19.9%

Second Isotopes

mass = 11

% abundance = 80.1%

Therefore,

Atomic mass = (19.9% of 10) + (80.1% of 11)

Atomic mass = (19.9 / 100 × 10) + (80.1 / 100 × 11)

Atomic mass = 199 / 100 + 881.1 / 100

Atomic mass = 1.99 + 8.811

Atomic mass = 10.801

Atomic mass = 10.801 amu

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