Which sample of matter can be separated into different substances by physical means

Answer:

x

Explanation:

becaue neutron equal to mass num -atomic num

Fe is the pure element (Iron) while Fe2+ is an isotope of Fe.

Answer:

a plus sign: +

Explanation:

a plus sign is used to separate one reactant or product from another

Answer:

(Chemical Formulas test)

1.Which formula contains a metal and a nonmetal?

answer:(MGO)

2.Which formula represents a covalent compound?

answer:(H2O2)

3.In the formula for water (H2O), what does the lack of a subscript after the O indicate? answer:(There is one oxygen atom in the compound.)

4.What symbol is used to separate the products from each other or the reactants from each other in a chemical reaction?answer: (A PLUS SIGHN)

5.Which option is a covalent compound?

answer:(NO2)

Explanation:

Took the Test and got 100% thank me later

Answer:

I think its b

Explanation:

Answer:

See explanation

Explanation:

A coin is money made from metal stuff. Common coinage metals are copper, silver, nickel etc.

One consideration in coin production is the density of the coin. A coin of less density will have lesser mass and occupy a lesser volume. This makes it easier to carry the coin around.

Coins should therefore be made of low density metals so that they can easily be carried about.

Explanation:

Number of molecules: 4.30×10²⁵

We need to find out the total number of moles.

Now, as per Avogadro's law, 6.02×10²³ of anything (Avogadro's number) constitutes a unit called one mole. Just like 12 is a dozen, 6.02×10²³ is a mole.

If you were given 24 bananas and you had to find out the number of dozens, you would divide the total number of bananas (24) by the number of bananas in one dozen (12).

We do exactly the same here: divide the total number of molecules (4.30×10²⁵) by the number of molecules in one mole (6.02×10²³) to get the number of moles.

4.30×10²⁵ / 6.02×10²³ = number of moles

Solving, we get number of NH3 moles = 71.4 which is your final answer!

Hope this helps :D

Answer:

Tings that effect weather is basically the following:

Answer:

144 g of phosphorus contain 4.65 moles.

Explanation:

Given data:

Mass of phosphorus = 144 g

Number of moles = ?

Solution:

Formula:

Number of moles = mass/ molar mass

Molar mass of phosphorus is 31 g/mol

Now we will put the values in formula.

Number of moles = 144 g/ 31 g/mol

Number of moles = 4.65 mol

Thus, 144 g of phosphorus contain 4.65 moles.

Explanation:

Anions have a negative charge resulting from the gain of one or more electrons to become stable.

Cations have a positive charge due to the loss of two or more electrons in order to become stable or have a noble gas configuration.

Hope this helps you

Answer:

the types of minerals in the rock

Answer:

C

Explanation: I took the test and got it right (first person is right)

Answer: 0.275 moles

Explanation: Na- 22.99g (on periodic table)

O- 16g and H- 1.01

22.99 + 16 + 1.01 = 40g/mol of NaOH

11g of NaOH / 40g / mol of NaOH = 0.275mol NaOH

The hydronium ion concentration [H₃O⁺]=7.826 x 10⁻⁶

In general, the weak acid ionization reaction  

HA (aq) ---> H⁺ (aq) + A⁻ (aq)  

Ka's value  

[tex]\large {\boxed {\bold {Ka \: = \: \frac {[H ^ +] [A ^ -]} {[HA]}}}}[/tex]

Reaction

HCN(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CN⁻ (aq)

0.125

x                                   x               x

0.125-x                         x               x

[tex]\tt Ka=\dfrac{[H_3O^+][CN^-]}{[HCN^-]}[/tex]

[tex]\tt 4.9\times 10^{-10}=\dfrac{x^2}{0.125-x}\approx \dfrac{x^2}{0.125}\\\\x^2=6.125\times 10^{-11}\\\\x=7.826\times 10^{-6}[/tex]

Answer:

metal atoms lose electrons to form positivity charged ions. non metal atoms gain electrons to form negatively charged ions

Answer:

Work done = 664.69 J

Explanation:

Given data:

Work done = ?

Initial volume = 13.43 L

Final volume = 6.87 L

Pressure = 1.91 atm

Solution:

work done = W = -PΔV

ΔV = V₂- V₁

work done = w = -P(V₂- V₁)

Now we will put the values in formula.

W = - 1 atm (6.87 L - 13.43 L)

W = - 1 atm (-6.56 L)

W = 6.56 atm.L

atm.L to joule:

6.56 atm.L × 101.325 J / 1L.atm

664.69 J

Answer:

When an electron temporarily occupies an energy state greater than its ground state, it is in an excited state. An electron can become excited if it is given extra energy, such as if it absorbs a photon, or packet of light, or collides with a nearby atom or particle.

Explanation:

Hope this helps.

Have a good night ma´am/sir.

Be safe!

Answer: C.

Explanation:

This is called entropy I believe. (it wouldn't let me answer before)

Answer: median :4.44 mean :4.11

Explanation:

i did test on Edge

Answer:

As it is given that

[tex]activity=\lambda= 100\\[/tex]

[tex]Number~of ~counts=100/second\\[/tex]

Using formula

[tex]ln(\frac{N}{N_o})=-\lambda\cdot t\\\\[/tex]

or

[tex]-\frac{ln(\frac{N}{N_o})}{\lambda}= t\\\\[/tex]

where [tex]N=\frac{N_o}{3}[/tex]

[tex]-\frac{ln(\frac{1}{3})}{100}= t\\\\\-\frac{-1.0986}{100}=t\\\\[/tex]

[tex]\frac{1.0986}{100}=t\\1.0986\cdot 10^{-2}seconds=t\\[/tex]

Answer:

4.52×1000=4520

Explanation:

4.52*1000=4520

Answer:

(a)Double Replacement

(b)Double Replacement

(c)Single replacement

(d)single replacement

(e) synthesis

(D) At equilibrium, the concentration of the products will be much higher than the concentration of the reactants.

Answer:

The amino sequence of an octapeptide can be determined by :

i.Amino acids present in the peptide

ii.By the behaviour shown by the treatment of these material with different biological and chemical reactions

As shown ammonium was released from the peptide,which shows that the peptide contains either Asn or Gln.

As there is presence of Asp in the amino acid analysis which indicates that the octapeptide contains Asp.

Acid hydrolysis hydrolyzes the peptide bonds which yields a Ala-Val  which is the result of partial acid hydrolysis.

Chymotrypsin is on the C-terminal side of the aromatic amino acids Phe, Tyr, and Trp. The peptide has Tyr which indicates that  the two present tetrapeptides after chymotrypsin treatment are fourth and eighth amino acids i-e N--Tyr-C and Tyr as the fourth amino acid and Ala-Val somewhere in the octapeptide

As Arg is only present on the C terminal side  of the Trypsin cleavage therefore Arg must be the second or the sixth amino acid for the two dipeptides to form as a result of arginine treatment of a tetrapeptide

Cyanogen bromide cleaves on the C-terminal side of Met residues. For free Tyr to be produced from cyanogen bromide treatment of one of the tetrapeptides, Met must be either the third or seventh amino acid. Because Ala was present in both tetrapeptides, Ala must be either the first or fifth amino acid, present in the same tetrapeptide as Arg and Met,i-e N.-Tyr-C and Ala-Val somewhere in the octapeptide and Ala-Arg-Met as the first, second, and third or fifth, sixth, and seventh amino acids and Tyr as the fourth amino acid.

One of the tetrapeptides contains Ala-Val as its second and third amino acids and Asn as its N-terminal amino acid. Therefore two combinations are possible as below

N-Asn-Ala-Val-Tyr-Ala-Arg-Met-Tyr-C or N-Ala-Arg-Met-Tyr-Asn-Ala-Val-Tyr-C

Answer:

The amino aciid sequence of an octapeptide can be determined from the amino acids that are present in the peptide and from the results of different chemical and biological treatments.

One equivalent of ammonium was released fro the peptide, indicating that the peptide contains either Asn or Gln. The presence os Asp in the amino analysis indicates that the octapeptide contains Asn.

A) Acid hydrolysis hydrolyzes peptide bonds. The dipeptide Ala-Val is obtained by partial acid hydrolysis. The indicates that Ala-Val is present in the peptide

B) Chymotrypsin cleaves on the C-terminal side of the aromatic amino acids. Phe, Tyr, and Trp. The peptide contains only Tyr. The presence of tetrapeptides after the treatment of ohymotrypsin indicates that Tyr is the fourth and eighth amino acids.

N-Tyr-C and Tyr present as the fourth amino acid. in constrast, Ala-Val is present in the octapeptide

C) Trpsin cleaves on the C-terminal side of Arg and Lys residues. only Arg is present in the octapeptide. Arg must be the second or sixth amino acid, for two dipeptide to be formed from arginine during the treatment of one tetrapeptide.

D) Cyanogen bromide cleaves on the C-terminal side of Met residues. For free Tyr to be produced from cynogen bromide treatment of one of the tetrapeptides, Met must be either the third or seventh amino acid. Because Ala is present in both tetrapeptides, Ala must be either the first or fifth amino acid, present in the same tetrapeptide as Arg and Met.

In addition, presence of N-...-Tyr-C and Ala-Val in the octapeptide and Ala-Arg-Met as the first, second, third or fifth, sixth or seventh amino acids and Tyr as the fourth amino acid.

E) One of the tetrapides contains Asn as its N-terminal amino acid, and Ala-Val as its second and third amino acids. Two octapeptides are possible from this peptide analysis.

The amino acid sequence of the octapeptide is the following:

N-Asn-Ala-Val-Tyr-Ala-Arg-Met-Tyr-C:

N-Ala-Arg-Met-Tyr-Asn-Ala-Val-Tyr-C:

For more information visit,

https://brainly.com/question/18122741

C.

centi- is essentially 10^2 of one meter.

If you had 100m, multiplying 100 by 10^2 (or 100) would give you 10000 cm.

Answer:

ionic bonds also called electrovalent bonds areee bonds that exist two metals.It is an electrostatic attraction between positive and negative ions in a Crystal lattice.

Explanation:

examples of elements with this bond are

hydrogen chloride

copper oxide

potassium bromate

Answer:

Final volume=V₂ = 216.3 mL

Explanation:

Given data:

Initial volume = 120.0 mL

Initial temperature = -12.3 °C (-12.3 +273 = 260.7 K)

Final volume = ?

Final temperature = 197.0 °C (197+273 = 470 K)

Solution:

We will apply Charles Law to solve the problem.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 120 mL × 470 K /260.7K

V₂ = 56400 mL.K /260.7K

V₂ = 216.3 mL

Answer:

0.8M

Explanation:

Given parameters:

Number of moles of KNO₃  = 0.2mol

Volume of KNO₃ = 250mL

Unknown:

Molarity of the solution  = ?

Solution:

The molarity of a solution is the number of moles in a given volume of a solution;

         Molarity  = [tex]\frac{number of moles }{volume}[/tex]

 We need to convert the given volume to L;

                1000mL  = 1L

        250mL will give  250 x 10⁻³L = 0.25L

So;

        Molarity  = [tex]\frac{0.2}{0.25}[/tex]   = 0.8M

Answer:

0.8 M

Explanation:

The formula is mol divided by liters. Converting 250 ml to .25 liters is important. Dividing .2 by .25 gets the answer. (I swear it’s right but it’s graded wrong so take this with a grain of salt)

Answer:

i cant see what it says

Explanation:

OUTLINE

Paragraph 1: Introduction

Paragraph 2: Body Paragraph

Paragraph 3: Body Paragraph

Paragraph 4: Body Paragraph

Paragraph 5: Conclusion

Think of the introduction and conclusion as “bookends” that serve to hold the essay tightly together. The

introduction will “push” into or initiate the examination of your topic and the angle you decide to focus on, while

the conclusion will “pull” tight all the ideas that you have gathered together for a unified essay.

Remember, the five-paragraph model can be expanded to include more body paragraphs that probe more deeply

into your subject. Check with your instructor to ensure whether or not you can exceed this length for an

assignment.

The introductory paragraph should include the following elements:

Background information: Enough information necessary for your reader to understand your topic

Thesis statement: Indicates your paper’s topic, makes your paper’s purpose clear, and provides an

overview of the three main supporting points that will unify the essay. The thesis statement is typically

the last sentence.

If you are writing in response to a text, the introduction should include the title, author, and genre of that

piece.

Begins with a topic sentence that identifies one main idea that will be discussed as support or proof for the

thesis statement

Supporting sentences use specific details, demonstrated through closely related examples or evidence, to

expand and explain the main idea. Generally, a well-developed paragraph has at least five to eight

sentences.

Paragraph unity means that all ideas in a paragraph are closely related to its topic sentence and further

develop that topic sentence. That is, all sentences in a single paragraph must be unified around a central

point or idea.

This paragraph, and any subsequent body paragraph, should begin with a topic sentence that signals the

reader that a new idea or point is being introduced.

As you organize your essay, keep in mind its coherence. Coherence refers to connections among

paragraphs and ideas—the logical sequence of your thoughts.

o Use transition words or phrases at the outset of your body paragraphs and to move from one idea

to another within your paragraphs.

o Have you transitioned logically from the main idea in the previous paragraph to this one? Are you

making clear connections among the paragraphs and ideas? Be sure to think about coherence during

the revision stage of the writing process.

This paragraph begins with the final topic sentence that relates back to the remaining point mentioned in the

thesis statement. Each paragraph should contain a new main idea.

Again, flesh out this main idea with specific examples, details, and relevant support.

Be sure to maintain paragraph unity. That is, each sentence must relate to your topic sentence.

The conclusion revisits your overall purpose for writing and often invites your reader to consider the

implications of why your ideas are significant.

The conclusion may restate the thesis, summarize the paper’s major points, or leave the reader with a final

thought to ponder. Several other methods for writing conclusions are included on a separate Tutoring Center

handout. If you choose to restate the thesis or summarize the essay’s main ideas, do not repeat the same

wording from the introduction or body paragraphs. Remember not to introduce new, unrelated ideas in the

conclusion