All categories

1 year ago

Which of the following is not found in the tropical rainforest?

Chemistry

2 answers:

natita [175]1 year ago

4 0

Answer:

B. Wild Pigs!!!!!!!!

ICE Princess25 [194]1 year ago

3 0

ILang are not found in the tropical rainforests. All the others can surprisingly be found in rainforests

You might be interested in

Calculate the theoretical density of iron, and then determine the number of vacancies per cm3 needed for a BCC iron crystal to h

Ann [662]

Answer:

Therefore the theoretical density of iron is 7.877 g/cm³ .

Therefore the number of vacancy per cm³ is $3.27 \times 10^{20}$

Explanation:

BCC structure contains 2 atoms per cell.

BCC : Body centered cubic.

Atomic mass of iron = 55.845 gram/mole.

Atomic mass of a chemical element is the mass of 1 mole of the chemical element.

Density: Density of a matter is the ratio of mass of the matter and volume of the matter.

Avogrdro Number is number of atoms per 1 mole.

Avogrdro Number= 6.023×10²³

Lattice parameter of iron = 2.866×10⁻⁸ cm

The theoretical density:

$\rho=\frac{\textrm{(x atom/cell)} \times \textrm{(atomic mass)}}{\textrm{avogadro's number}\times \textrm{(lattice parameter)}^3}$

$=\frac{2\times 55.845 }{(6.023\times 10^{23})\times (2.866\times 10^{-8})^3}$ g/cm³ [x= 2,Since it is BCC structure]

=7.877 g/cm³

Therefore the theoretical density of iron is 7.877 g/cm³ .

Now we have to find out the number of unit cell of iron crystal having density 7.874 g/cm³.

$\rho=\frac{\textrm{(x atom/cell)} \times \textrm{(atomic mass)}}{\textrm{avogadro's number}\times \textrm{(lattice parameter)}^3}$

$\Rightarrow 7.874 =\frac{x\times 55.845}{6.023\times 10^{23}\times (2.866\times 10^{-8})^3}$

$\Rightarrow x= \frac{7.874\times 6.023\times 10^{23}\times (2.866\times 10^{-8})^3}{55.845}$

⇒ x =1.9923 atom/cell

Therefore the vacancy of atom per cell = (2- 1.9923)=0.0077

$Vacancy\ per\ cm^3=\frac{\textrm{Vacancy per cm} ^3 }{\textrm{( lattice parameter)} ^3}$

$=\frac{0.0077}{(2.866\times 10^-8)^3}$

$=3.27 \times 10^{20}$

Therefore the number of vacancy per cm³ is $3.27 \times 10^{20}$

5 0

1 year ago

Combustion of hydrogen releases 142 j/g of hydrogen reacted. How many kj of energy are released by the combustion of 16.0 oz of

Andrej [43]

Given the mass of hydrogen = 16.0 oz

Converting 16.0 oz hydrogen to pounds (lb) using the conversion factor 1 lb = 16 oz:

$16.0 oz * \frac{1 lb}{16 oz} =1 lb$

Converting 16.0 lb to g using the conversion factors 1 kg = 2.2 lb, 1 kg = 1000 g:

$1lb * \frac{1kg}{2.2lb}*\frac{1000g}{1kg}= 454.5 g$

Heat of combustion of hydrogen = 142 J/g

Calculating the heat released when 16.0 oz is combusted:

$454.5g H_{2} * \frac{142 J}{g} *\frac{1 kJ}{1000J}=64.5kJ$

5 0

1 year ago

Read 2 more answers

The volume of a gas is decreased from 100 liters at 173.0°C to 50 liters at a constant pressure. After the decrease in volume, w

Minchanka [31]

Answer:

223.08 K

Explanation:

First we convert 173.0 °C to K:

173.0 °C + 273.16 = 446.16 K

With the absolute temperature we can use Charles' law to solve this problem:

T₁V₂=T₂V₁

Where in this case:

T₁ = 446.16 K

V₂ = 50 L

T₂ = ?

V₁ = 100 L

We input the data:

446.16 K * 50 L = T₂ * 100 L

And solve for T₂:

T₂ = 223.08 K

5 0

1 year ago

Suppose you wanted to make a buffer of exactly ph 7.00 using kh2po4 and na2hpo4. if the final solution was 0.10 m in kh2po4, wha

OleMash [197]

Answer:- 0.138 M

Solution:- The buffer pH is calculated using Handerson equation:

$pH=pKa+log(\frac{base}{acid})$

$KH_2PO_4$ acts as a weak acid and $Na_2HPO_4$ as a base which is pretty conjugate base of the weak acid we have.

The acid hase two protons(hydrogen) where as the base has only one proton. So, we could write the equation as:

$H_2PO_4^-\rightleftharpoons H^++HPO_4^-^2$

Phosphoric acid gives protons in three steps. So, the above equation is the second step as the acid has only two protons and the base has one proton.

So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.

Let's plug in the values in the equation:

$7.00=6.86+log(\frac{base}{0.10})$

$7.00-6.86=log(\frac{base}{0.10})$

$0.14=log(\frac{base}{0.10})$

Taking antilog:

$10^0^.^1^4=\frac{base}{0.10}$

$1.38=\frac{base}{0.10}$

On cross multiply:

[base] = 1.38(0.10)

[base] = 0.138

So, the concentration of the base that is $Na_2HPO_4$ required to make the buffer is 0.138M.

5 0

1 year ago

Americium-242 has a half-life of 6 hours. If you started with 24 g and you now have 3 g, how much time

kogti [31]

How many times has it halved?

24/2 = 12
12/2 = 6
6/2 = 3

It halved three times.
It halves once every 6 hours.

18 hours have passed.

7 0

2 years ago