Which of the following sentences describe how latitude affects a biome? (There may be more than one answer).

Question 4 options:

In the far north, the summers are too short and cold for trees.


As you travel east, the climate gets colder.


The closer the area is near to the equator, the hotter the weather.


The higher in elevation, the colder the environment.

Answer:

B-100

Explanation:

ppm is an unit of concentration that could be defined as the mass in mg of solute (In this case, NaCl) per kg of solution.

Now, a solution of NaCl that is 0.01% by mass, contains 0.01g of NaCl in 100g of solution.

To solve this question, we must convert the mass of NaCl to mg and the mass of solution to kg:

Mass NaCl:

0.01g * (1000mg / 1g) = 10mg

Mass Solution:

100g * (1kg / 1000g) = 0.10kg

The ppm are:

10mg / 0.10kg =

100ppm

Right answer is:

Answer:

78.75L

Explanation:

According to Charle's Law:

V1/T1=V2/T2

35/200=V2/450

V2=(35*450)200

V2=78.75L

Answer:

c and d

Explanation:

mitochondria and vacoule

Answer:

It is actually D and A

Explanation:

I took this assignment before on edge and the guy who said C and D is wrong

Answer:

due to the pollution the light months tended to stand out against the the wood that was dark due to the smoke the darker months population grew due to the fact that they were more harder to see for the predators

Answer:

The dark colored months became more common

Explanation:

Thats the answer !

Answer:

175.5

Explanation:

Answer: The answer is a. limiting, excess. I took the test.

Explanation:

Answer:

Dominant

Explanation:

hope this help

Answer:

They refer to energy that moves. Kinetic energy is energy that is currently moving. Potential refers to energy that has yet to move, or simply energy in wait

Explanation:

Answer:

I'm not 100% but a possible answer could be: No energy could either be created or destroyed

LMK if not but good luck!

Answer:

The correct answer is - 1 and 3 are true.

Explanation:

Hydrogen bonds require a lot of energy to be broken as these bonds are relatively strong intermolecular forces which makes the boiling point of liquid like water very high, as it requires a lot of heat to break the intermolecular bonds holding the water molecules together.

An increase in the surface area of the liquid also increases the rate of evaporation as it takes place among the molecules of the liquid on the surface.

Answer:

(C) the physical state of each reactant and product

Explanation:

Hope this helps

Answer:

3.7 x 10²¹ O₂ molecules

Explanation:

2KClO₃ => 2KCl + 3O₂

given 0.500g KClO₃ => 0.500g/122.55g/mol = 0.0041 mole KClO₃

0.0041 mole KClO₃ => 3/2(0.0041) mole O₂ = 0.0061 mole O₂

0.0061 mole O₂ = 0.0061 mole O₂ x 6.02 x 10²³ molecules O₂/mole O₂

= 3.7 x 10²¹ molecules O₂

Answer:

D

Explanation:

Answer:

Explanation:

There are 7 protons in nitrogen

7 TYPE OF NITRONGEN

. . .

Answer: The percentage of potassium iodide in the sample is 2.63 %.

Explanation:

The chemical equation for the reaction of iodide ions with bromine gas follows:

[tex]I^-+3Br_2+3H_2O\rightarrow 6Br^-+IO_3^-+6H^+[/tex]                (i)

The chemical equation for the reaction of iodate ions with barium ions follows:

[tex]Ba^{2+}+2IO_3^-\rightarrow Ba(IO_3)_2[/tex]                 ......(ii)

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Given mass of barium iodate = 0.0596 g

Molar mass of barium iodate = 487.13 g/mol

Using equation 1:

[tex]\text{Moles of barium iodate}=\frac{0.0596 g}{487.13 g/mol}\\\\\text{Moles of barium iodate}=1.22\times 10^{-4} moles[/tex]

By Stoichiometry of the reaction (ii):

1 mole of barium iodate is produced by 2 moles of iodate ions

So, [tex]1.22\times 10^{-4} moles[/tex] of barium iodate will be produced by [tex]\frac{2}{1}\times 1.22\times 10^{-4} =2.44\times 10^{-4}moles[/tex] of iodate ions

By the stoichiometry of the reaction (i):

1 mole of iodate ions are produced by 1 moles of iodine ions

So, [tex]2.44\times 10^{-4} moles[/tex] of iodate ions will be produced by [tex]\frac{1}{1}\times 2.44\times 10^{-4} =2.44\times 10^{-4}moles[/tex] of iodine ions

Moles of potassium iodide = Moles of iodide ions = [tex]2.44\times 10^{-4}[/tex]

Since, the molar mass of potassium iodide = 166 g/mol

Using equation 1:

[tex]\text{Mass of potassium iodide}=2.44\times 10^{-4}mol\times 166 g/mol\\\\\text{Mass of potassium iodide}=0.0405 g[/tex]

To calculate the percentage by mass of a substance, the equation used is:

[tex]\text{Percent by mass}=\frac{\text{Mass of a substance}}{\text{Mass of solution}}\times 100[/tex]

Mass of a solution = 1.54 g

Mass of potassium iodide = 0.0405 g

Using above equation:

[tex]\text{Percent potassium iodide}=\frac{0.0405 g}{1.54g}\times 100\\\\\text{Percent potassium iodide}=2.63\%[/tex]

Hence, the percentage of potassium iodide in the sample is 2.63 %.

Expressing the results of potassium iodide in percentage = 2.63%

The chemical reaction of iodine ions with Bromine gas can be expressed as :

I⁻  + 3Br₂  + 3H₂O  -- > 6Br⁻ + IO₃ + 6H⁺ ----- ( 1 )

Chemical reaction between iodate ions with barium ions  can be expressed as :  Ba²⁺  + 2IO⁻₃ ------> Ba ( IO₃ )₂   --------- ( 2 )

Step 1 : Calculate the number of Barium iodate moles

mass of Barium iodate = 0.0596 g

molar mass of Barium iodate = 487.13 g/mol

from equation ( 1 )

moles of Barium iodate = ( 0.0596 ) / ( 487.13 ) = 1.22 * 10⁻⁴ moles

also from equation ( 1 ) the moles of potassium iodide = moles of iodide ions

= 2.44 * 10⁻⁴

molar mass of potassium iodide = 166 g/mol

Next step : Determine the mass of potassium iodide

moles of potassium * molar mass

= 2.44 * 10⁻⁴  * 166 g/mol  = 0.0405 g

Final step : Determine the percentage of potassium iodide in the solution

Percentage = ( mass of potassium iodide / mass of solution ) * 100

                   = ( 0.0405 / 1.54 ) * 100

                   = 2.63%

Hence we can conclude that potassium iodide in percentage = 2.63%

Learn more : https://brainly.com/question/3388761

Answer:

A. mushroom

Explanation:

Answer:

Dont completly trust me here but i think its a but id try and wait for someone else just be careful

Explanation:

Answer:

The weak acid is - E. [tex]NH_3[/tex]

Explanation:

Ammonia is a strong base, which in turn make it to be a weak acid.

As, the rest of the options are acid and therefore furnish hydrogen ions when dissolved in water, making them strong acid.

And ammonia dissolves in water to form its conjugate ammonium ion and hydroxide ion, therefore making the solution basic.

Hence, from the given options, the correct option is E. [tex]NH_3[/tex]

The compound which when dissolved in water, forms a weak acid is; Choice E: NH3

According to the question;

Unlike other compounds (HNO3, HClO, HBr, HClO) which dissolve in water to yield H+ ion, Ammonia dissolves in water to form ammonium ion and hydroxide ion, OH-.

In essence, NH3 is a strong base when dissolved in water and consequently, is a weak acid when dissolved in water.

Read more:

https://brainly.com/question/15192126

Answer: [tex]2NH_3(g)+2O_2(g)\rightarrow N_2O(g)+3H_2O(l)+684kJ[/tex]

Explanation:

The skeletal thermochemical equation for the reaction is:

[tex]NH_3(g)+O_2(g)\rightarrow N_2O(g)+H_2O(l)+342kJ[/tex]

The balanced  thermochemical equation for the reaction is:

[tex]2NH_3(g)+2O_2(g)\rightarrow N_2O(g)+3H_2O(l)+684kJ[/tex]

When 1 mole of [tex]NH_3[/tex] reacts with oxygen , heat released = 342 kJ

Thus when 2 moles of [tex]NH_3[/tex] reacts with oxygen , heat released = [tex]\frac{2}{1}\times 342 kJ=684kJ[/tex]

Answer:

C

Explanation:

It is run by Elon Musk, who is determined to get to Mars.

Answer:

The answer is 3 valance electrons.

Explanation:

Hope this helped Mark BRAINLIEST!!!

Answer:

3 valence electrons

Explanation:

The atoms in group 13 have three valence electrons.

Answer:

This happened because,the comb was charged with static electricity when it was rubbed on your hair.

(¬_¬)ﾉ( ˘ ³˘)♥

The data set is missing in the question. The data set is given in the attachment.

Solution :

a). In the table, there are four positive examples and give number of negative examples.

Therefore,

[tex]$P(+) = \frac{4}{9}$[/tex]   and

[tex]$P(-) = \frac{5}{9}$[/tex]

The entropy of the training examples is given by :

[tex]$ -\frac{4}{9}\log_2\left(\frac{4}{9}\right)-\frac{5}{9}\log_2\left(\frac{5}{9}\right)$[/tex]

= 0.9911

b). For the attribute all the associating increments and the probability are :

  [tex]$a_1$[/tex]   +   -

  T   3    1

  F    1    4

Th entropy for   [tex]$a_1$[/tex]  is given by :

[tex]$\frac{4}{9}[ -\frac{3}{4}\log\left(\frac{3}{4}\right)-\frac{1}{4}\log\left(\frac{1}{4}\right)]+\frac{5}{9}[ -\frac{1}{5}\log\left(\frac{1}{5}\right)-\frac{4}{5}\log\left(\frac{4}{5}\right)]$[/tex]

= 0.7616

Therefore, the information gain for [tex]$a_1$[/tex]  is

  0.9911 - 0.7616 = 0.2294

Similarly for the attribute [tex]$a_2$[/tex]  the associating counts and the probabilities are :

  [tex]$a_2$[/tex]  +   -

  T   2    3

  F   2    2

Th entropy for   [tex]$a_2$[/tex] is given by :

[tex]$\frac{5}{9}[ -\frac{2}{5}\log\left(\frac{2}{5}\right)-\frac{3}{5}\log\left(\frac{3}{5}\right)]+\frac{4}{9}[ -\frac{2}{4}\log\left(\frac{2}{4}\right)-\frac{2}{4}\log\left(\frac{2}{4}\right)]$[/tex]

= 0.9839

Therefore, the information gain for [tex]$a_2$[/tex] is

  0.9911 - 0.9839 = 0.0072

[tex]$a_3$[/tex]     Class label      split point       entropy        Info gain

1.0         +                        2.0            0.8484        0.1427

3.0        -                         3.5            0.9885        0.0026

4.0        +                        4.5            0.9183         0.0728

5.0        -

5.0        -                        5.5            0.9839        0.0072

6.0        +                       6.5             0.9728       0.0183

7.0        +

7.0        -                        7.5             0.8889       0.1022

The best split for [tex]$a_3$[/tex]  observed at split point which is equal to 2.

c). From the table mention in part (b) of the information gain, we can say that [tex]$a_1$[/tex] produces the best split.

Answer:

9.24

Explanation:

-logofH

Answer:

The answer is "Option b and Option c".

Explanation:

This buffer is a buffer of ammonia and ammonium ion. Thus it requires the solution [tex]NH_3 \ \ and \ \ NH_4^{+}[/tex].

In point 1:

The solution containing [tex]NH_3 \ \ and \ \ NH_4^{+}[/tex] at 1M concentration would be given by mixing the two solutions. Thus, this buffer is a legitimate route.

In point 2:

It gives the ions you want but they are not the same.

In point 3:

[tex]1 M[/tex] [tex]NH_3[/tex] and [tex]1 M[/tex][tex]HCl[/tex] volume would not produce the same [tex]NH_3 \ \ and \ \ NH_4^{+}[/tex] concentrations. Therefore, this buffer isn't a valid route.

In point 4:

Some [tex]1 M[/tex] [tex]NH_3[/tex] volume and half [tex]HCl[/tex] . This offers the same rate as half.

Answer:

The answer is C.

Explanation:

A P E X approved :)