Helpppp!!!!! Please help me

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  =

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.

Answer:

рн

Explanation:

From the given choices, pH is the only chemical property that can be measured quantitatively.

A chemical property is one that tells us about what a substance can do as regards to whether or not the substance reacts with other substances.

Examples are flammability, rusting of iron, precipitation, decomposition of water e.t.c

Answer:

Percent error = 3.7%

Explanation:

Given data:

Density of Al cylinder = ?

Weight of cylinder = 18 g

Diameter = 1.3 cm

Height = 5.2 cm

Actual density of Al = 2.7 g/cm³

Percent error = ?

Solution:

First of all we will calculate the volume of cylinder through given formula.

V = πr²h

r = diameter /2

V = 22/7 × (0.65 cm)²× 5.2 cm

V = 22/7 × 0.4225cm²× 5.2 cm

V = 6.89 cm³

Now we will calculate the density.

d = m/v

d = 18 g/ 6.89 cm³

d = 2.6 g/cm³

Percent error:

Percent error = measured value - actual value /actual value × 100

Percent error = 2.6g/cm³ - 2.7g/cm³ /2.7g/cm³  × 100

Percent error = 3.7%

Negative sign shows that measured or experimental value is less than actual value.

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

Answer:

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

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

Answer:

Steel is a mixture because iron does not mix with the carbon and the other components.

Answer:

Steel is a mixture because it does not have a set chemical compound formula. It's a mixture of iron and carbon fused together with other metals or nonmetals to create a mixture.

Explanation:

Steel is an alloy , a solid solution

Steel is a homogenous mixture. It is not a pure substance because it is made form more than one component.

Answer:

7.5%

Explanation:

I just know it's that

Explanation:

U need to draw the graph first and make a line at 17 pennies, where the line of 17 pennies and your graph meet is the mass of it(at y axis)

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.

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.

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.

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.

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.

To learn more about  half life, here:

https://brainly.com/question/23485913

#SPJ6

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.

Answer:

6 protons 8 neutrons 5 electrons

Explanation:

protons are blue neutrons are red electrons are green

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

read more: https://brainly.com/question/14440832?referrer=searchResults

Answer:

1) Increases

2) decreases

3) increases

4) decreases

Explanation:

When the intermolecular forces in a liquid increases, the greater vapour pressure of the liquid decreases accordingly.

Since the vapour pressure is proportional to temperature, as temperature increases, the vapour pressure increases alongside.

As intermolecular forces increases, the boiling point increases accordingly since more energy is required to break intermolecular bonds.

Lastly, the greater the surface area, tell greater the vapour pressure since more liquid surface area is now available.

Answer: 2681.81

Explanation:

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:

https://brainly.com/question/4144781

Answer:

major ecological grouping of plants and animals

food chain

*biome*

bacteria and fungi that break down dead matter

*decomposers*

a state of change in which the end result is equal or

balanced

*dynamic equilibrium*

the basic relationships that show how a community of plants,

animals, and bacteria live and grow and how these living

*ecosystem*

things are dependent on each other as well as the Sun, soil,

and other nonliving parts of their environment; a cycle of

relationships

line of plants and animals that shows the order in which

*Food chain*

organisms are eaten

a condition or conditions of the nonliving surroundings, such

*environmental factor*

Explanation:

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:

Answer:

I would say it is true

Explanation:

Answer:

A graph can be very handy in collecting data and storing it in one place. A graph can be a very effective tool in presenting visual information rather swiftly. Hope this helps! I am learning about graphs in my course!

Explanation:

Answer:

graph) a diagram showing the relation between variable quantities, typically of two variables, each measured along one of a pair of axes at right angles.

Explanation:

Graphing is used daily. From stockbrokers to performance evaluation in companies. All use them to boost sales and meet deadlines. Even simple calculations can be assessed better by using a graph.

(let me know if this helps?)

Answer:

[tex]\%\ Error = 21.5\%[/tex]

Explanation:

Given

[tex]Measured = 24.5[/tex]

[tex]Actual = 31.2[/tex]

Required

Determine the percentage error

First, we need to determine the difference in the measurement

[tex]Difference = |Actual - Measured|[/tex]

[tex]Difference = |31.2 - 24.5|[/tex]

[tex]Difference = |6.7|[/tex]

[tex]Difference = 6.7[/tex]

The percentage error is calculated as thus:

[tex]\%\ Error = \frac{Difference * 100\%}{Actual}[/tex]

[tex]\%\ Error = \frac{6.7 * 100\%}{31.2}[/tex]

[tex]\%\ Error = \frac{670\%}{31.2}[/tex]

[tex]\%\ Error = 21.4743589744\%[/tex]

[tex]\%\ Error = 21.5\%[/tex] approximated

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].