Predict the chemical formula of the compound that will form when Li and o interact.

Answer:

If coal and petroleum will get exhausted it will be very difficult for us to transport because most vehicles depends on petroleum, Transport on Earth will became complicated, and if coal will get exhausted we will lose an unique fossil fuel. Coal is used in various domestic and industrial purposes.

Answer:

6.576 g NaCl

Explanation:

2Na + Cl2 --> 2 NaCl

Step 1: Find out moles of Na and Cl to determine the limiting reactant:

moles Na = 12/23 = 0.52 moles Na

moles Cl2 = 16/71=0.225 moles Cl2

ratio Na:Cl is 2.3:1 so Cl is the limiting reactant.

Step 2: How many moles of NaCl are formed:

moles NaCl = moles Cl2 x (1 mol NaCl/2 moles Cl2) = 0.225/2 = 0.1125 moles NaCl

Step 3:

mass NaCl = moles NaCl x MM NaCl = 0.1125 x 58.45 = 6.576 g NaCl

Answer:

the answer is D . our galaxy

Answer:

our galaxy of course

Answer:

3

Explanation:

nitrogen atom in NH3 has 3 shared electrons.

When hydrogen is passed over hot ferric oxide (FeO) hydrogen reacts with oxygen present in the compound and forms water (H2O) and pure Iron

The question is incomplete, the complete question is;

Aluminum metal and bromine liquid (red) react violently to make aluminum bromide (white powder). One way to represent this equilibrium is:

Al(s) + 3/2 Br2(l)AlBr3(s)

We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above.

1) 2 AlBr3(s) 2 Al(s) + 3 Br2(l)

2) 2 Al(s) + 3 Br2(l) 2 AlBr3(s)

3) AlBr3(s) Al(s) + 3/2 Br2(l)

Answer:

See explanation

Explanation:  

We have that; Al(s) + 3/2 Br2(l)AlBr3(s)

So;

Al(s) + 3/2 Br₂(l) = AlBr₃(s)

K = [  AlBr₃] / [ Al] [  Br₂]³/²

K² =  [  AlBr₃]² / [  Al ] ² [ Br₂]³

Now;

1) 2 AlBr₃ = 2 Al(s) + 3 Br₂(l) =

K₁ =  [  Al ] ² [ Br₂]³ /  [  AlBr₃]²

K₁ =  ( 1 / K² ) = K⁻²

For the second reaction;

2 ) 2 Al(s) + 3 Br₂(l) = 2 AlBr₃(s)

K₂ = [ AlBr₃ ]² / [  Al ]² [  Br₂ ]³

K₂ = K²

For the third reaction;

3 )

AlBr₃(s) =   Al(s) + 3/2 Br₂(l)

K₃  = [ Al ] [ Br₂ ] ³/² / [ AlBr₃ ]

=  ( 1 / K ) = K⁻¹

Answer:

Explanation:

A peptide is any class of organic compounds composed of different numbers of amino acids in which the amine of one is reacted with the carboxylic acid of the next to form an amine bond. The peptide of DTLH is composed of the following amino acids:

Asp-Thr-Leu-His

Their structures are first drawn out in the image attached below. This is followed by the isoelectric structure of the peptide DTLH

Divide the number of molecules N by avogadro's constant.

Let n be the number of moles then,

[tex]n=\frac{N}{N_A}=\frac{3.3\cdot10^{24}}{6.022\cdot10^{23}}=5.48\mathrm{mol}[/tex]

Hope this helps.

Given

Atomic symbol

Required

Atomic composition

Solution

Atomic number = number of protons = number of electrons

Mass Number (A) is the sum of protons and neutrons

Number of protons and Number of Neutrons  in nucleus

Number of electrons in the shell

From the picture:

protons = 3

neutrons = 4

electrons = 3

atomic number = protons = electrons = 3

mass number = protons + neutrons = 3+4 = 7

Answer:

ccc

Explanation:

Answer:

A

Explanation:

trust

Answer:

m = 79.1 grams

Explanation:

Given that,

Density, d = 2.26 g/mL

Volume occupies, V = 35 mL

We need to find the mass of the substance. We know that the density of an object is equal to the mass per unit volume. So,

d = m/V

[tex]m=d\times V\\\\m=2.26\ g/mL\times 35\ mL\\\\m=79.1\ g[/tex]

So, the mass of the substance is 79.1 grams.

Answer:

Follows are the solution to the given question:

Explanation:

Dry Soil weight = solid soil weight = [tex]284 \ grams[/tex]

solid soil volume =[tex]205 \ cc[/tex]

saturated mass soil = [tex]361 \ g[/tex]

The weight of the soil after drainage is =[tex]295 \ g[/tex]

Water weight for soil saturation = [tex](361-284) = 77 \ g[/tex]

Water volume required for soil saturation =[tex]\frac{77}{1} = 77 \ cc[/tex]

Sample volume of water: [tex]= \frac{\text{water density}}{\text{water density input}}[/tex]

[tex]= 361- 295 \\\\ = 66 \ cc[/tex]

Soil water retained volume = (draining field weight - dry soil weight)

                                             [tex]= 295 - 284 \\\\ = 11 \ cc.[/tex]

[tex]\text{POROSITY}= \frac{\text{Vehicle volume}}{\text{total volume Soil}}[/tex]

                    [tex]= \frac{77}{(205 + 77)} \\\\= \frac{77}{(282)} \\\\ = 27.30 \%[/tex]

(Its saturated water volume is equal to the volume of voids)

[tex]\text{YIELD SPECIFIC} = \frac{\text{Soil water volume}}{\text{Soil volume total}}[/tex]

                              [tex]= \frac{66}{(205+77)}\\\\= \frac{66}{(282)}\\\\=0.2340\\\\ = 0.23[/tex]

[tex]\text{Specific Retention}= \frac{\text{Volume of soil water}}{\text{Total soil volume}}[/tex]

                            [tex]= \frac{11}{282} \\\\= 0.0390 \\\\ = 0.04[/tex]

Answer:

combustion reaction. hope this helps!

Answer: We start by doing a 1/10 serial dilution, using 100 µL of 1M solution into 900 µL of water, until we get a 1.0x10^-5M (0.00001M) solution. Then use 1 mL of this 1.0x10^-5M solution into 9 mL of water.

Explanation:

To answer this question, we must use the law of conservation of mass, which states: In every chemical reaction mass is conserved, this means the total mass of the reactants is equal to the total mass of the products. The law implies that mass can neither be created nor destroyed, but it can be transformed. For example, in chemical reactions, the mass of the chemical components before the reaction is equal to the mass of the components after the reaction. Therefore, during any chemical reaction and low-energy thermodynamic processes in an isolated system, the total mass of the reactants or starting materials must be equal to the mass of the products. This law is quite accurate for low-energy processes, such as chemical reactions.

So, if the solution to be used is 1M , and a 10 mL of a 1.0x10^-6M solution is needed, we use the following equation:

Initial concentration x initial volume = final concentration x final volume.

The initial concentration is 1M, the final concentration is 1.0x10^-6M and the final volume required is 10 mL.

1M x initial volume = 1.0x10^-6M x 10 mL

initial volume=  1.0x10^-5 mL= 0.01 µL of 1M solution.

Since the final volume is 10 mL, we have to add the difference in volume with water, which is 10 mL - 1.0x10^-5 mL= 9.99999 mL.

However, since 0.01 mL is a very small volume that is difficult to take, the best option in this case is to make serial dilutions.

Usually, we start from a concentrated solution and prepare a series of dilutions to the tenth (1:10) or half (1:2). In this way a series of solutions is obtained, related for example by a dilution factor of 10, i.e. 1/10; 1/100; 1/1000 and so on.

Here we can prepare a series of dilutions to the tenth, from 1M to 1.0x10^-6.

We start by doing a 1/10 dilution, using 100 µL of 1M solution into 900 µL of water. This is a 0.1M solution. Then we take 100 µL of it into 900 µL of water to get a 0.01M solution. We continue doing that until we get a 1.0x10^-6M (0.000001M) solution. This final solution is the desired concentration, however we need 10 mL of it, and actually we have 1 mL. So we can just take 1 mL of the 1.0x10^-5M solution into 9 mL of water:

1.0x10^-5M x 1 mL = 1.0x10^-6M x 10 mL.

Answer:

The smaller numbers that result from measuring with larger units are easier to use .

That’s the one, good luck!

The smaller numbers that result from measuring with larger units are easier to use.

Light year is a unit of astronomical distance that measures the distance traveled by light in one year. The distance between stars can also be measured using parsecs.

The equivalent of this distance in various units is written as follows;

Using light year to represent the unit of distance is more concise because big measurement can be represented easily with small numbers using this large unit.

Thus, we can conclude that the smaller numbers that result from measuring with larger units are easier to use.

Learn more here:https://brainly.com/question/803764

Answer:

atoms

Explanation:

i hope that help you

Answer:

atoms

Explanation:

Answer:

26.982 g

i think so..........

Answer:

The number of particles in the 41.8g sample of Bismuth is 12.044 × 10²³

Explanation:

In a 15.5 g sample of bismuth there are 4.47 x 10²² atoms are present.

The term molar mass is defined as the ratio between the mass and the amount of substance of any sample of said compound. The molar mass is a property of a substance.

The molar mass of a compound can be calculated by adding the standard atomic masses in g/mol of the constituent atoms.

Given:

Mass of Bismuth  = 15.5g

Number of atoms  = ?

To find the number atoms, find the number of moles in this element first.

Number of moles  = mass/ molar mass

Number of moles  =  15.5g / 209    

= 0.074mole

1 mole of a substance contains 6.023 x 10²³ atoms

0.074 mole of Bismuth will contain 0.074 x 6.023 x 10²³ atoms

= 4.47 x 10²² atoms

Thus, 4.47 x 10²² atoms are found in a 15.5 g sample of bismuth.

To learn more about the molar mass, follow the link;

https://brainly.com/question/22997914

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Answer:

150.17 g/mol

Explanation:

Answer:

Continuing on across the periodic table we see that fluorine is the next element after oxygen. ... Rather than forming seven bonds fluorine only forms a single bond for basically the same reasons that oxygen only forms two bonds. Hydrogen fluoride, HF, has one bond, but four centers of electron density around the fluorine.