categorize the given phrases as descriptions of α‑keratin, collagen, or fibroin.

Answer:the answer is a c and e

Explanation:

Answer:carbon dioxide is released during combustion

Answer:

55.0125 grams NO2

Explanation:

So we have 80 grams of O2, first convert to moles

80 g O2 * 1 mol/32 g O2 = 2.5 mol O2

Next use the mole ratio of O2 to NO2

2.5 mol O2* 1 mol NO2/2 mol NO2=1.25 mol NO2

Since the question is asking how many grams, convert NO2 to grams

1.25 mol NO2 * 44.01 NO2/1 mol NO2= 55.0125 grams

Answer: a. +2, cation and magnesium ion .

b. -1,  anion, chloride

c. -2, anion, oxide

d. +1. cation , potassium ion

Explanation:

When an atom accepts an electron negative charge is created on atom and is called as anion.

When atom loses an electron positive charge is created on atom and is called as cation.

Magnesium (Mg) with atomic number of 12 has electronic configuration of 2,8,2 and thus it can lose 2 electrons to form [tex]Mg^{2+}[/tex] cation and becomes magnesium ion.

Chlorine (Cl) with atomic number of 17 has electronic configuration of 2,8,7 and thus it can gain 1 electron to form [tex]Cl^{-}[/tex] anion and becomes chloride.

Oxygen (O) with atomic number of 8 has electronic configuration of 2,6 and thus it can gain 2 electrons to form [tex]O^{2-}[/tex] anion and becomes oxide.

Potassium (K) with atomic number of 19 has electronic configuration of 2,8,8,1 and thus it can lose 1 electron to form [tex]K^{+}[/tex] cation and becomes potassium ion.

Answer:

To run a spectrophotometry experiment, begin by warming up the spectrophotometer and preparing the samples. Be sure to select the correct wavelength, then run a measurement on the blank solution. Follow up by running measurements on sample solutions. Once data is collected, turn off the instrument, clean the area, and discard the samples.

Explanation:

The spectrophotometer is a device used in laboratories to carry out analysis in experiments that need to measure and compare the amount of light absorbed or transmitted between samples, regardless of whether the samples are transparent or opaque solid. The spectrophotometer can be single-beam and double-beam. Regardless of the nature of the beam, to perform a spectrophotometry experiment, you must start by heating the spectrophotometer and preparing the samples. Make sure to select the correct wavelength and perform a measurement on the blank solution. Follow up by performing measurements on sample solutions. Once the data is collected, turn off the instrument, clean the area and discard the samples.

Answer:

I promise it is the 2nd one

Explanation:

The following diagram shows a stage of a cell during mitosis. Two identical circles are joined in the middle and they have the nucleolus inside the nucleus in the center of the circles.

Answer:

The Second One

Explanation:I did the test and got this question correct

Pls comment :D

690 g AgCl

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Stoichiometry

Step 1: Define

[RxN - Unbalanced] AgNO₃ + ZnCl₂ → AgCl + Zn(NO₃)₂

↓

[RxN - Balanced] 2AgNO₃ + ZnCl₂ → 2AgCl + Zn(NO₃)₂

[Given] 2.4 mol ZnCl₂

[Solve] x g AgCl

Step 2: Identify Conversions

[RxN] 1 mol ZnCl₂ → 2 mol AgCl

[PT] Molar Mass of Ag - 107.87 g/mol

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of AgCl - 107.87 + 35.45 = 143.32 g/mol

Step 3: Stoich

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

687.936 g AgCl ≈ 690 g AgCl

Answer:

16.5g of CO₂ could be produced

Explanation:

The combustion of hexane occurs as follows:

C₆H₁₄(l) + 19/2O₂ → 6CO₂ + 7H₂O

Where 1 mole of hexane reacts with 19/2 moles of O₂.

To solve this question we need to find the moles of each reactant in order to find limiting reactant. The moles of the limiting reactant will determine the moles of CO₂ produced:

Moles C₆H₁₄ -Molar mass: 86.18g/mol-:

17.2g hexane * (1mol / 86.18g) = 0.200 moles hexane

Moles O₂ -Molar mass: 32g/mol-:

19g O₂ * (1mol / 32g) = 0.594 moles oxygen.

For a complete reaction of 0.594 moles of oxygen are required:

0.594 moles O₂ * (1mol C₆H₁₄ / 19/2 moles O₂) = 0.0625 moles C₆H₁₄.

As there are 0.200 moles of hexane, hexane is the excess reactant and oxygen the limiting reactant.

The moles of CO₂ produced assuming a yield of 100% -All moles of oxygen react producing carbon dioxide.:

0.594 moles O₂ * (6mol CO₂ / 19/2 moles O₂) = 0.375 moles of CO₂ could be produced. The mass is:

0.375 moles of CO₂ * (44.01g / mol) =  

Answer:

Option D and Option B

When comparing the two elements K and Ge , the more metallic element is_____K_____ based on periodic trends alone.

When comparing the two elements Sb and Pb , the more metallic element is_____Pb______ based on periodic

Explanation:

The metallic characteristic increases when we move down a column in a periodic table or when we move left in the row.

Potassium and Germanium are located on the same row, but germanium lies on the right side of potassium. Thus, potassium (K) is more metallic than Germanium (Ge)

While Lead (Pb) lies to left of Sb in the adjacent column and is also lies below Sb. Hence Pb is more metallic than Sb

Answer:

Dividend Divisor Quotient/ Remainder SAGOT 75 21 16.) 54 17.) 108 18.) 380 19.) 633 20.) 648 27 76 57 32. Solo nihanda ni: NNA MARIE S.

Explanation:

Answer:

(i) Change in colour (ii) Change in temperature  (iii) Formation of precipitate

Explanation:

(i) Change in colour: Reaction between lead nitrate solution and potassium iodide solution. Pb(NO3)2(aq)+2KI → PbI2(s)+2KNO3(aq) In this reaction, colour changes from colourless to yellow. (ii)Change in temperature: Action of dilute sulphuric acid on zinc. Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2 In this reaction, heat is evolved (iii) Formation of precipitate: Action of barium chloride on sodium sulphate. BaCl2(aq) +Na2SO4(aq) →  BaSO4(s) +2NaCl(aq) BaSO4(s)

Answer:

Cellular respiration

Explanation:

Answer:

Percent Composition of 41K = 6.7302%

Explanation:

The explination is in the image.

The percentage composition of 41K isotope of potassium according to the given data is 6.7302 percent.

Given that the atomic masses of the potassium isotopes are:

mass of 39K = 38.9637amu

mass of 40K = 39.9639amu

mass of 41K = 40.9618amu

and the percentages are:

percentage of 39K = 93.2581%

percentage of 40K =  0.0117%

percentage of 41K = x

Given average mass of potassium = 39.0983amu

sum (mass of isotopes × percentage abundance) =  average mass

38.9637 × 93.2581% +  39.9639 ×  0.0117%  + 40.9618 × x = 39.0983

x = 6.7302%

So the percentage composition of 41K is 6.7302%

Learn more about composition of isotopes:

https://brainly.com/question/24291471?referrer=searchResults

Answer:

MnSO4.5H2O(s) ---------> MnSO4(s) + 5H2O(g)

Explanation:

The dehydration of a hydrate implies that the water of crystallization is lost. Water of crystallization is included in the crystal structure as it is formed and is incorporated into the structure of the compound.

Now the equation for the dehydration of manganese(II) sulfate pentahydrate is;

MnSO4.5H2O(s) ---------> MnSO4(s) + 5H2O(g)

The  MnSO4 is now said to be anhydrous.

Carbon-14 is a radioisotope of carbon that decays following first-order kinetics. There are four values of interest in this problem: the "normal" (or original) amount of carbon-14 for a jawbone ([tex]\mathrm{N_0}[/tex]), the actual amount of carbon-14 in a jawbone ([tex]\mathrm{N}[/tex]), the half-life of carbon-14 ([tex]\mathrm{t_{1/2}}[/tex]), and the actual time elapsed ([tex]\mathrm{t}[/tex]) from the original time. There is an equation that ties all these values in together,

[tex]N= N_0 e^{-kt}[/tex]

where k is the rate constant, which, for first-order decay, is related to the half-life by

[tex]k = \dfrac{\ln 2}{ t_{1/2} }.[/tex]

What you want to find here is the time elapsed (t). So, you can substitute the latter equation for k into the k in the former equation to get

[tex]N= N_0 e^{\frac{-\ln 2 \;t}{t_{1/2}}.[/tex]

Rearranging to solve for t, the equation becomes

[tex]t = \left(\dfrac{\ln \dfrac{N_0}{N}}{\ln 2} \right) t_{1/2}.[/tex]

You are given all three of the values necessary to solve for t: The normal amount of carbon-14 is 200 mg; the actual amount of carbon-14 in the sample is 50 mg; and the half-life of carbon-14 is 5730 years. Plugging them into the above equation, we get

[tex]t = \left(\dfrac{\ln \dfrac{200 \text{ mg}}{50 \text{ mg}}}{\ln 2} \right) \left(5730 \text{ years} \right) = 11460 \text{ years}.[/tex]

So the jawbone found is 11460 years old (or 11000 if accounting for sig figs).

Answer:

Sn(s) + MgBr2(aq) --> SnBr2(aq) + Mg(s)

Explanation:

Decide whether a chemical reaction happens in either of the following situations.

- A reaction happens. The type of this reaction is the single replacement reaction.

If a reaction does happen, write the chemical equation for it.

Sn(s) + MgBr2(aq) --> SnBr2(aq) + Mg(s)

Answer:

[tex]n =1.905mol[/tex]

Explanation:

Hello!

In this case, by considering helium gas as an ideal one, we can use the following equation:

[tex]PV=nRT[/tex]

Whereas P should go in atmospheres and T in Kelvins; thus we proceed as follows:

[tex]n=\frac{PV}{RT}=\frac{132kPa*\frac{1atm}{101.325kPa}*36L}{0.08206\frac{atm*L}{mol*K}(27+273)K} =1.905mol[/tex]

Best regards!

Answer: a)  The maximum mass of sulfur trioxide that can be formed is 31.4 grams

b) The FORMULA for the limiting reagent is [tex]O_2[/tex]

c) Mass of excess reagent remains is 4.8 grams

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} SO_2=\frac{29.9g}{64g/mol}=0.467moles[/tex]

[tex]\text{Moles of} O_2=\frac{6.26g}{32g/mol}=0.196moles[/tex]

[tex]2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)[/tex]  

According to stoichiometry :

1 mole of [tex]O_2[/tex] require = 2 moles of [tex]SO_2[/tex]

Thus 0.196 moles of [tex]O_2[/tex] will require=[tex]\frac{2}{1}\times 0.196=0.392moles[/tex]  of [tex]SO_2[/tex]

Thus [tex]O_2[/tex] is the limiting reagent as it limits the formation of product and [tex]SO_2[/tex] is the excess reagent as (0.467-0.392) = 0.075 moles or [tex]0.075mol\times 64g/mol=4.8g[/tex] are left.  

As 1 mole of [tex]O_2[/tex] give = 2 moles of [tex]SO_3[/tex]

Thus 0.196 moles of [tex]O_2[/tex] give =[tex]\frac{2}{1}\times 0.196=0.392moles[/tex]  of [tex]SO_3[/tex]

Mass of [tex]SO_3=moles\times {\text {Molar mass}}=0.392moles\times 80g/mol=31.4g[/tex]

Thus 31.4 g of [tex]SO_3[/tex] will be produced from the given masses of both reactants.

Answer:

C,A,B,D

Explanation:

That's the correct order, hope this helps

Answer:

pH=11.

Explanation:

Hello!

In this case, since the data is not given, it is possible to use a similar problem like:

"An analytical chemist is titrating 185.0 mL of a 0.7500 M solution of ethylamine(C2HNH2) with a 0.4800 M solution of HNO3.ThepK,of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 114.4 mL of the HNO3 solution to it"

Thus, for the reaction:

[tex]C_2H_5NH_2+H^+\rightleftharpoons C_2H_5NH_4^+[/tex]

Tt is possible to compute the remaining moles of ethylamine via the following subtraction:

[tex]n_{ethylamine}=0.1850L*0.7500mol/L=0.1365mol\\\\n_{acid}=0.1144L*0.4800mol/L=0.0549mol\\\\n_{ethylamine}^{remaining}=0.1365mol-0.0549mol=0.0816mol[/tex]

Thus, the concentration of ethylamine in solution is:

[tex][ethylamine]=\frac{0.0816mol}{0.1850L+0.1144L}=0.2725M[/tex]

Now, we can also infer that some salt is formed, and has the following concentration:

[tex][salt]=\frac{0.0549mol}{0.1850L+0.1144L}=0.1834M[/tex]

Therefore, we can use the Henderson-Hasselbach equation to compute the resulting pOH first:

[tex]pOH=pKb+log(\frac{[salt]}{[base]} )\\\\pOH=3.19+log(\frac{0.1834M}{0.2725M})\\\\pOH=3.0[/tex]

Finally, the pH turns out to be:

[tex]pH=14-pOH=14-3\\\\pH=11[/tex]

NOTE: keep in mind that if you have different values, you can just change them and follow the very same process here.

Best regards!

Answer:

[tex]NO_3[/tex]

Explanation:

Hello!

In this case, since percent compositions are used to set up empirical formulas when assuming those percentages are masses, we can fist compute the moles of nitrogen and oxygen in the compound as shown below:

[tex]n_N=22.5g/14.01g/mol=1.61mol\\\\n_O=77.5g/16.00g/mol=4.84mol[/tex]

Now, we divide by the fewest moles to compute the subscripts:

[tex]N=\frac{1.61}{1.61} =1\\\\O=\frac{4.84}{1.61} =3[/tex]

Thus, the empirical formula turns out:

[tex]NO_3[/tex]

Best regards!

Answer:

Convection

Explanation:

Convection is the circular motion that happens when warmer air or liquid — which has faster moving molecules, making it less dense — rises, while the cooler air or liquid drops down.

Same as in the case of sea breeze where the the lighter wind in surface of land rises up and the cooler wind from the sea comes to fill up that place. This is an example of convection.

Explanation:

1.undetectable to electromagnetic waves

2.hiding an object from an illumination containing diffre t wave lengths become difficult as the object sizes grow.

3. reduce the scattering by two orders.

Electromagnetic waves are unable to detect it. With increasing item sizes, it becomes more challenging to conceal an object from an illumination with various wave lengths. two orders down on the scattering.

Current clocking technology is defined as a hypothetical or made-up stealth technology that makes things like people or spaceships completely or partially invisible to certain electromagnetic (EM) spectrum wavelengths. In a specific frequency band, this gadget renders an object "invisible" to electromagnetic radiation. Obviously, cloaks that operate in the visible spectrum have the most fascinating potential uses.

These metamaterials, according to the researchers, can shield a medium-sized antenna from radio waves over a wide range of bandwidths, resulting in clearer communications. However, as visible light wavelengths are far shorter than radio waves, it is practically difficult to conceal huge things from them, such as the human body. The issue with present designs is that they can only handle a certain amount of bandwidth. Even this "perfect" 3D cloak that was previously exhibited could only conceal items from microwaves.

Thus, electromagnetic waves are unable to detect it. With increasing item sizes, it becomes more challenging to conceal an object from an illumination with various wave lengths. two orders down on the scattering.

To learn more about current cloaking technology, refer to the link below:

https://brainly.com/question/10964418

#SPJ5

Answer:

okay that is cool

Explanation:

Answer:

I hope it helps u! :)

Explanation:

Answer:

i think its Yes. The bubbles contain hydrogen and oxygen that separated from the water but i could be wrong so i would just wait to see if anybody else says anything

Explanation: