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kvv77 [185]
4 years ago
5

Procede by wich wind removes surface materiales

Chemistry
1 answer:
grin007 [14]4 years ago
8 0
Weathering
very close to erosion, but not quite.
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How many grams are in 1.2 x 1024 atoms of sodium?
jarptica [38.1K]

Answer:

46 g

Explanation:

First we <u>convert 1.2 x 10²⁴ atoms of sodium into moles</u>, using <em>Avogadro's number</em>:

  • 1.2x10²⁴ atoms ÷ 6.023x10²³ atoms/mol = 2.0 mol

Then we <u>convert 2.0 moles of sodium into grams</u>, using <em>sodium's molar mass</em>:

  • 2.0 mol Na * 23 g/mol = 46 g

Thus, there are 46 grams in 1.2x10²⁴ atoms of sodium.

3 0
3 years ago
Which of the following would diffuse through a cell membrane the most easily?
ZanzabumX [31]
You didn’t give us the options???

anyways: Water diffusion is called osmosis. Oxygen is a small molecule and it's nonpolar, so it easily passes through a cell membrane. Carbon dioxide, the byproduct of cell respiration, is small enough to readily diffuse out of a cell. Small uncharged lipid molecules can pass through the lipid innards of the membrane.
8 0
3 years ago
Read 2 more answers
Sarah is writing a paper explaining protein formation. Choose the correct way to complete each sentence.
Amanda [17]

Answer :

Amino acids join by forming (1) peptide bonds.  Water is released.

The chains of amino acids vary according to their (2) side chains, which each have differing makeup.  

These chains determine the protein’s structure and they fold to create a unique (3) third-dimensional shape.

Explanation :

Amino acid : The amino acids are the building blocks of protein. Amino acids are the compound which consist both amine (-NH_2) and carboxylic (-COOH) groups along with the side chains. In the protein, many amino acids are linked by the peptide bonds.

Th peptide bonds are formed by the combining of amino group of one amino acid to the carboxylic group of another amino acid by releasing of water.

Some examples of amino acids are glycine, lysine, alanine, etc.

Peptide bond formation image is shown below.

In the image 'R' can be contained by different groups like hydrogen, methyl, phenyl, etc

8 0
3 years ago
0.20 g of sodium hydroxide (NaOH) pellets are dissolved in water to make 4.5 L of solution. What is the pH of this solution?
amid [387]
Thats allllllll otttt but ill help

step by step
8 0
3 years ago
An industrial synthesis of urea obtains 87.5 kg of urea upon reaction of 68.2 kg of ammonia with excess carbon dioxide. Determin
dolphi86 [110]

Answer:

The theoretical yield of urea = <u>120.35kg</u>

The percent yield for the reaction = <u>72.70%</u>

Explanation:

Lets calculate -

The given reaction is -

2NH_3(aq)+CO_2 →CH_4N_2O(aq)+H_2O (l)

Molar mass of urea CH_4N_2O= 60g/mole

Moles of NH_3 = \frac{62.8kg/mole}{17g/mole} (since Moles=\frac{mass  of  substance}{mass of one mole})

                     = 4011.76 moles

Moles of CO_2 = \frac{105kg}{44g/mole}

                = \frac{105000g}{44g/mole}

                = 2386.36 moles

Theoritically , moles of NH_3 required = double the moles of CO_2

    but , 4011.76 , the limiting reagent is NH_3

Theoritical moles of urea obtained = \frac{1 mole CH_4N_2O}{2mole NH_3}\times4011.76 mole NH_3

                                                      = 2005.88mole CH_4N_2O

Mass of 2005.88 mole of CH_4N_2O =2005.88 mole \times\frac{60g CH_4N_2O}{1mole CH_4N_2O}

                                                     = 120352.8g

                                                     120352.8g\times \frac{1kg}{1000g}

                                                     = 120.35kg

Therefore , theroritical yeild of urea = 120.35kg

Now , Percent yeild = \frac{87.5kg}{120.35kg}\times100

                                 72.70%

Thus , the percent yeild for the reaction is 72.70%

8 0
3 years ago
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