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3 years ago

Match each term with the correct location on a trophic pyramid.

Chemistry

1 answer:

makkiz [27]3 years ago

3 0

Answer:

Consumers B) Second level from the bottom

Secondary Consumers C) Top of the pyramid

Tertiary Consumers D) Bottom of the pyramid

Ape x Predator E) Middle of the pyramid

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Will give lots of points if answered correctly. Determine the kb for chloroform when 0.793 moles of solute in 0.758 kg changes t

Liono4ka [1.6K]

Answer: The value of $K_{b}$ for chloroform is $3.62^{o}C/m$ when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

Explanation:

Given: Moles of solute = 0.793 mol

Mass of solvent = 0.758

$\Delta T_{b} = 3.80^{o}C$

As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.

$Molality = \frac{no. of moles}{mass of solvent (in kg)}\\= \frac{0.793 mol}{0.758 kg}\\= 1.05 m$

Now, the values of $K_b$ is calculated as follows.

$\Delta T_{b} = i\times K_{b} \times m$

where,

i = Van't Hoff factor = 1 (for chloroform)

m = molality

$K_{b}$ = molal boiling point elevation constant

Substitute the values into above formula as follows.

$\Delta T_{b} = i\times K_{b} \times m\\3.80^{o}C = 1 \times K_{b} \times 1.05 m\\K_{b} = 3.62^{o}C/m$

Thus, we can conclude that the value of $K_{b}$ for chloroform is $3.62^{o}C/m$ when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

7 0

3 years ago

I need help with my homework.Calculate the number of formula units in: 41.5 grams CaBr2

lana66690 [7]

Answer:

$\text{ }1.25\times10^{23}\text{ formula units}$

Explanation:

Here, we want to calculate the number of formula units in the given molecule

We start by getting the number of moles

To get the number of moles, we have to divide the mass given by the molar mass

The molar mass is the mass per mole

The molar mass of calcium bromide is 200 g/mol

Thus, we have the number of moles as follows:

$\frac{41.5}{200}\text{ = 0.2075 mol}$

The number of formula units in a mole is:

$1\text{ mole = 6.02 }\times10^{22}\text{ formula units}$

The number of formula units in 0.2075 mole will be:

$0.2075\text{ }\times6.02\times10^{23}\text{ = }1.25\times10^{23}\text{ formula units}$

5 0

1 year ago

Convert 15.6 g Na2Cr2O7 to moles of Na2Cr2O7.

Leviafan [203]

Answer:

261.96754

Explanation:

3 0

3 years ago

If 42.7 of 0.208 M hydrochloric acid are needed to completely neutralize a solution of calcium hydroxide, how many grams of calc

Montano1993 [528]

Answer:

0.329 g

Explanation:

In the context of this problem, we have a chemical reaction between hydrochloric acid and calcium hydroxide. HCl is the acid here and calcium hydroxide is the base. Hence, we have an acid-base reaction, also known as neutralization reaction.

In a neutralization reaction, water is produced as a product, as well as a salt that we obtain after we exchange the cations: calcium bonds to chloride and hydrogen bonds to hydroxide (the latter is the formation of water). This means we also produce calcium chloride as a product. The overall reaction represents this as:

$Ca(OH)_2(aq)+2 HCl (aq)\rightarrow CaCl_2 (aq)+2 H_2O (l)$

Firslt of all, we wish to find the number of moles of HCl present. Having molarity and volume, this is done by applying the molarity formula. It states that molarity is equal to the rate between moles and volume:

$c_{HCl}=\frac{n_{HCl}}{V_{HCl}}$

Rearranging for moles of HCl, n:

$n_{HCl}=c_{HCl}V_{HCl}$

Based on stoichiometry of the balanced chemical equation, notice that 1 mole of calcium hydroxide reacts with 2 moles of HCl, meaning:

$n_{Ca(OH)_2}=\frac{1}{2} n_{HCl}=\frac{1}{2}c_{HCl}V_{HCl}$

Now that we have the expression for moles, we may also express moles of calcium hydroxide as the ratio between its mass and molar mass:

$n_{Ca(OH)_2}=\frac{m_{Ca(OH)_2}}{M_{Ca(OH)_2}}$

Using the last two equations, we see that:

$\frac{1}{2}c_{HCl}V_{HCl}=\frac{m_{Ca(OH)_2}}{M_{Ca(OH)_2}}\\\therefore m_{Ca(OH)_2}=\frac{1}{2}c_{HCl}V_{HCl}M_{Ca(OH)_2}$

Substitute the given data, as well as the molar mass of calcium hydroxide:

$m_{Ca(OH)_2}=\frac{1}{2}\cdot0.208 M\cdot0.0427 L\cdot74.093 g/mol=0.329 g$

8 0

3 years ago

(a) Compute the radius r of an impurity atom that will just fit into an FCC octahedral site in terms of the atomic radius R of t

11Alexandr11 [23.1K]

Answer:

The radius of an impurity atom occupying FCC octahedral site is $0.414{\rm{R}}$

The radius of an impurity atom occupying FCC tetrahedral site is $0.225{\rm{R}}$ .

Explanation:

In order to get a better understanding of the solution we need to understand that the concept used to solve this question is based on the voids present in a unit cell. Looking at the fundamentals

An impurity atom in a unit cell occupies the void spaces. In FCC type of structure, there are two types of voids present. First, an octahedral void is a hole created when six spheres touch each other usually placed at the body center. On the other hand, a tetrahedral void is generated when four spheres touch each other and is placed along the body diagonal.

Step 1 of 2

(1)

The position of an atom that fits in the octahedral site with radius $\left( r \right)$ is as shown in the first uploaded image.

In the above diagram, R is the radius of atom and a is the edge length of the unit cell.

The radius of the impurity is as follows:

$2r=a-2R------(A)$

The relation between radius of atom and edge length is calculated using Pythagoras Theorem is shown as follows:

Consider $\Delta {\rm{XYZ}}$ as follows:

$(XY)^ 2 =(YZ) ^2 +(XZ)^2$

Substitute $XY$ as ${\rm{R}} + 2{\rm{R + R}}$ and ${\rm{YZ}}$ as a and ${\rm{ZX}}$ as a in above equation as follows:

$(R+2R+R) ^2 =a ^2 +a^ 2\\16R ^2 =2a^ 2\\ a =2\sqrt{2R}$

Substitute value of aa in equation (A) as follows:

$r= \frac{2\sqrt{2}R -2R }{2} \\ =\sqrt{2} -1R\\ = 0.414R$

The radius of an impurity atom occupying FCC octahedral site is $0.414{\rm{R}}$

Note

An impure atom occupies the octahedral site, the relation between the radius of atom, edge length of unit cell and impure atom is calculated. The relation between the edge length and radius of atom is calculated using Pythagoras Theorem. This further enables in finding the radius of an impure atom.

Step 2 of 2

(2)

The impure atom in FCC tetrahedral site is present at the body diagonal.

The position of an atom that fits in the octahedral site with radius rr is shown on the second uploaded image :

In the above diagram, R is the radius of atom and a is the edge length of the unit cell.

The body diagonal is represented by AD.

The relation between the radius of impurity, radius of atom and body diagonal is shown as follows:

$AD=2R+2r----(B)$

In $\Delta {\rm{ABC}},$

$(AB) ^2 =(AC) ^2 +(BC) ^2$

For calculation of AD, AB is determined using Pythagoras theorem.

Substitute ${\rm{AC}}$ as a and ${\rm{BC}}$ as a in above equation as follows:

$(AB) ^2 =a ^2 +a ^2$

$AB= \sqrt{2a} ----(1)$

Also,

$AB=2R$

Substitute value of $2{\rm{R}}$ for ${\rm{AB}}$ in equation (1) as follows:

$2R= \sqrt{2} aa = \sqrt{2} R$

Therefore, the length of body diagonal is calculated using Pythagoras Theorem in $\Delta {\rm{ABD}}$ as follows:

$(AD) ^2 =(AB) ^2 +(BD)^2$

Substitute ${\rm{AB}}$ as $\sqrt 2$ a and ${\rm{BD}}$ as a in above equation as follows:

$(AD) ^2 =( \sqrt 2a) ^2 +(a) ^2 AD= \sqrt3a$

For calculation of radius of an impure atom in FCC tetrahedral site,

Substitute value of AD in equation (B) as follows:

$\sqrt 3a=2R+2r$

Substitute a as $\sqrt 2$ ${\rm{R}}$ in above equation as follows:

$( \sqrt3 )( \sqrt2 )R=2R+2r\\\\$

$r = \frac{2.4494R-2R}{2}\\$

$=0.2247R$

$\approx 0.225R$

The radius of an impurity atom occupying FCC tetrahedral site is $0.225{\rm{R}}$ .

Note

An impure atom occupies the tetrahedral site, the relation between the radius of atom, edge length of unit cell and impure atom is calculated. The length of body diagonal is calculated using Pythagoras Theorem. The body diagonal is equal to the sum of the radii of two atoms. This helps in determining the relation between the radius of impure atom and radius of atom present in the unit cell.

7 0

3 years ago