An unknown element is a mixture of isotopes 120X and 122X. The
average atomic mass of X is 120.37 amu. What is the percent
abundance of 122X?

Answer:

C)Grassland biomes contain mostly grasses.

Explanation:

The main differences between a savanna biome and a grassland is that grasslands biomes contain mostly grasses.

Savanna are made up of both grasses and shrubs.

Answer:

[tex]PtCl_4[/tex]

Explanation:

Hello!

In this case, since HCl and Pt react according to the following chemical equation:

[tex]HCl+Pt\rightarrow PtCl_x+H_2[/tex]

Whereas PtClx is the compound containing Pt and Cl; thus, since 1.018 g out of 1.778 g correspond to Pt and therefore 0.760 g to chlorine, so we determine the empirical formula of this compound by firstly computing the moles of each element:

[tex]n_{Pt}=1.018gPt*\frac{1molPt}{195.084gPt}=0.00522molPt\\\\\\n_{Cl}=0.760gCl*\frac{1molCl}{35.45gCl} =0.0214molCl[/tex]

Now, we divide the each moles by those of Pt as the fewest ones in order to compute their subscripts in the empirical formula:

[tex]Pt=\frac{0.00522}{0.00522}=1 \\\\Cl=\frac{0.0214}{0.00522} =4[/tex]

Thus, the required formula is:

[tex]PtCl_4[/tex]

Best regards!

Answer:

Explanation:

From the given information:

The compound in question here is [tex]\mathbf{C_2H_5Br}[/tex]

The molar mass of [tex]\mathbf{C_2H_5Br}[/tex] with 79 Br  is;

= (2 × 12) + (1 × 5) + 79

= 108 g/mol

The molar mass of [tex]\mathbf{C_2H_5Br}[/tex] with 81 Br is:

=  (2 × 12) + (1 × 5) + 81

= 110 g/mol

Thus, the mass at which the molecular ions occurs are:

108 and 110.

The relative percentage of mass 108 occurred at 50.7%

The relative percentage of mass 110 occurred at 49.3%

Answer:

A) Mass = 32 g of KCl

Explanation:

Given data:

Mass of potassium chloride produced = ?

Mass of potassium chlorate = 52 g

Solution:

Chemical equation:

2KClO₃     →       2KCl + 3O₂

Number of moles of KClO₃:

Number of moles = mass/molar mass

Number of moles = 52 g/ 122.55 g/mol

Number of moles = 0.424 mol

Now we will compare the moles of KClO₃ and KCl

             KClO₃      :       KCl

               2            :       2

          0.424         :       0.424

Mass of KCl:

Mass = number of moles × molar mass

Mass = 0.424 mol × 74.55 g/mol

Mass = 32 g

Answer:

1. B 2.B

Explanation:

Answer:

ANSWER OS HERE

Explanation:

THE O2- has two more extra electrons due to presence of 2 negative charges so a total of 10 electrons and the element that has 10 electrons in the periodic table in its neutral state is Neon (Ne) ( atomic number (10) = no.of protons = no. of electrons) so neutral element with same electron configuration as O2- is Ne. Log in or register to post comments.

The electron configuration of the neutral atom is an useful place to start when trying to determine the electron configuration of an ion.

The neutral atom in your situation is sulfur (S), which is found in group 16 of period 3 of the periodic chart. A neutral sulfur atom has a total of 16 electrons surrounding its nucleus since sulfur has an atomic number of 16. Thus, a neutral sulfur atom will have the following electron configuration. Now, a neutral sulfur atom receives two more electrons to create the sulfide anion, or S2. These two electrons will be added to the 3p-orbitals, which can hold a maximum of two protons, as you can see in the configuration of the neutral atom. between them, a maximum of six electrons. Thus, the sulfide anion's electron configuration will be the electron configuration of neon, the noble gas that appears before sulfur in the periodic table, will be used in the shorthand notation for noble gases for the sulfide anion.

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Answer:

$4.49 g/cm3

Explanation:

Density=Mass/Volume

plzz put brainiest

Answer:

In chemical reactions, elements tend to combine with other elements to INCREASE_ stability and DECREASE total energy.

Explanation:

The chemical compounds when joining are more stable in external media such as solvents or solutions already formed, although this union with another chemical compound removes total energy from the compound.

Answer:

45.7g of CO₂ is the maximum amount of carbon dioxide that could be produced.

Explanation:

Hexane, C₆H₁₄, reacts with O₂ to produce H₂O and CO₂ as follows:

C₆H₁₄ + 13/2O₂ → 7H₂O + 3CO₂

Where 1 mole of hexane reacts with 13/2 moles of O₂.

To solve this question we need to convert the mass of each reactant to moles in order to find the limiting reactant. With limiting reactant we can find the maximum amount of CO₂ produced:

Moles C₆H₁₄ -Molar mass: 86.18g/mol-:

64g C₆H₁₄ * (1mol / 86.18g) = 0.7426 moles C₆H₁₄

Moles O₂ -Molar mass: 32g/mol-:

72.0g O₂ * (1mol / 32g) = 2.25 moles O₂

For a complete reaction of 2.25 moles of O₂ are required:

2.25 moles O₂ * (1mol C₆H₁₄ / 13/2 moles O₂) = 0.346 moles of C₆H₁₄

As there are 0.7426 moles of C₆H₁₄, C₆H₁₄ is the excess reactant and:

Oxygen is limiting reactant.

The maximum moles of CO₂ that could be produced are:

2.25 moles O₂ * (3mol CO₂ / 13/2 moles O₂) = 1.038 moles CO₂

In grams:

1.038 moles CO₂ * (44.01g / mol) =

Answer:

3

Explanation:

Answer:

best guess is polystyrene

Explanation:

Answer:

B. Synthesizing a complex molecule

Explanation:

Hello!

In this case, since catabolic reactions are related to the synthesis of complex biomolecules and metabolic reactions to the breaking down of complex molecules to obtain energy from (say start), we infer that the appropriate answer is:

B. Synthesizing a complex molecule

Because regardless C. also mentions the synthesis of a compound, carbonic acid is not a complex one in comparison to the biomolecules we produce during our cellular cycles.

Best regards!

The coefficient and symbol for iodine be I₂.

bromine + potassium iodide → potassium bromide + iodine

The balanced chemical equations:

Br₂ + 2KI  → 2KBr + I₂

2KBr means 2 molecules of potassium iodide.

From the above balanced equation, we know that the coefficient of bromine = 2, iodine = 2, potassium iodide = 2and potassium bromide = 2.

Hence, option D is the correct answer.

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Answer:

56 J

Explanation:

Step 1: Given data

Step 2: Calculate the work (w) done on the object

We can find the work done on the object using the following expression.

w = F × d × cosθ

where

θ is the angle between F and d

Since F and d occur in the same direction, θ = 0° and cosθ = 1. Then,

w = F × d × 1

w = F × d

w = 7 N × 8 m = 56 J

Answer:

57.26 %

Explanation:

Step 1: Write the balanced equation

Pt + 2 HCl ⇒ PtCl₂ + H₂

Step 2: Calculate the mass of Pt in 1.778 g of PtCl₂

According to the law of conservation of the mass, all the mass of Pt in the reactants must be present in the products. Since PtCl₂ is the only product that contains Pt, the mass of Pt is 1.018 g

Step 3: Calculate the percent by mass of platinum in PtCl₂

We will use the following expression.

%Pt = mPt / mPtCl₂ × 100%

%Pt = 1.018 g / 1.778 g × 100% = 57.26 %

Answer:

257 mL

Explanation:

Step 1: Given data

Step 2: Calculate the moles (n) corresponding to 32.8 g of HCl

The molar mass of HCl is 36.46 g/mol.

32.8 g × 1 mol/36.46 g = 0.900 mol

Step 3: Calculate the volume (V) of solution required

We will use the following expression.

C = n/V

V = n/C

V = 0.900 mol/(3.50 mol/L)

V = 0.257 L = 257 mL

Answer:

C. Plate tectonic collisions can create mountains.

Answer:

They can make them larger

Explanation:

They go up under a mountain and then the mountains rise up

Answer:

25 mM Tris HCl and 0.1% w/v SDS

Explanation:

A 10X solution is ten times more concentrated than a 1X solution. The stock solution is generally more concentrated (10X) and for its use, a dilution is required. Thus, to prepare a buffer 1X from a 10X buffer, you have to perform a dilution in a factor of 10 (1 volume of 10X solution is taken and mixed with 9 volumes of water). In consequence, all the concentrations of the components are diluted 10 times. To calculate the final concentration of each component in the 1X solution, we simply divide the concentration into 10:

(250 mM Tris HCl)/10 = 25 mM Tris HCl

(1.92 M glycine)/10 = 0.192 M glycine

(1% w/v SDS)/10 = 0.1% w/v SDS

Therefore the final concentrations of Tris and SDS are 25 mM and 0.1% w/v, respectively.

Answer:

$18.30

Explanation:

The change is the amount left over after the selling price has been matched by the buyer.

Let us find the total cost of what of Ms. Kuhn bought;

For the shirt;

    Cost  = 4 shirts for $15.95 each  = 4(15.95) = $63.8

                2 new belts for $8.95 each  = 2(8.95)  = $17.9

Total cost  = $63.8 + $17.9  = $81.7

The amount of change left  = $100  - $81.7  = $18.3

Answer:

d. 10 million years

Explanation:

These are the options for the question

Choose one answer.

a. 2.5 million years

b. 5 million years

c. 15 million years

d. 10 million years

✓half-life of the radioactive substance= 5 million years

✓ Then for every 5 million years, original radioactive given (100%) percentage is reduced by factor of 1/2.

✓ 1st 5 million year; After the first 5 million years, the original radioactive is reduced as;

(100%) × 1/2 = 50% remains

✓2nd 5 million year: At 10 millions year, the 50% remaining radioactive is reduced by factor of 1/2

(50%) × 1/2 = 25% remains

Hence, at At 10 millions years 25% of the original radioactive atoms remain

.

Answer:

10 million years

Answer:

Yes, 0.11g/mL is the solubility of X in water at 17.0°C.

Explanation:

As the sample of water has crystals around it, the solution is oversaturated at 17.0°C. The water contains the maximum amount of X at 17.0°C, this concentration is called solubility. Thus, yes, we can calculate solubility in water at 17.0°C

The concentration of the crystals is 3.96g in 36.0mL, that is:

3.96g / 36.0mL =

The  uncertainties  of  the  delta  measurements  and  the  uncertainty  of  the  atomic  weight  derivedfrom  the  best  measurement  of  isotopic  abundances  constrain  the  number  of  significant  figures  in  theatomic-weight values of the upper and lower bounds. For carbon, the fifth digit after the decimal pointis uncertain because of the uncertainty value of 0.000 027. Therefore, the number of significant digitsin the atomic-weight value is reduced to four figures after the decimal point. The Commission may rec-ommend  additional  conservatism  and  reduce  the  number  of  significant  figures  further.  For  the  lowerbound of carbon, 12.009 635 is truncated to 12.0096. For an upper bound, the trailing digit is increasedto ensure the atomic-weight interval encompasses the atomic-weight values of all normal materials. Inthe case of carbon, the upper bound is adjusted from 12.011 532 to 12.0116 to express four digits afterthe decimal point. The lower and upper bounds are evaluated so that the number of significant digits ineach  is  identical.  If  a  value  ends  with  a  zero,  it  may  need  to  be  included  in  the  value  to  express  therequired number of digits. The following are examples of lower and upper atomic-weight bounds foroxygen that could be published by the Commission in its various tables.

---------------------------

Extracted from" Atomic weights of the elements 2009 (IUPAC Technical Report)"

Answer:

24.9%

Explanation:

According to this question, mole fraction of NaCl in an aqueous solution is 0.0927. This means that the mole percent of NaCl in the solution is:

0.0927 × 100 = 9.27%

Let's assume that the solution contains water (solvent) + NaCl (solute), hence, the mole fraction of water will be;

100% - 9.27% = 90.73%

THEREFORE, it can be said that, NaCl contains 0.0927moles while H2O contains 9.073moles

N.B: mole = mass/molar mass

Given the Molar Mass

NaCl: 58.44 g/mol

H2O: 18.016 g/mol

For NaCl;

0.0927 = mass/58.44

mass = 0.0927 × 58.44

5.42g

For H2O;

9.073 = mass/18.016

mass = 9.073 × 18.016

= 16.35g

Total mass of solution = 16.35g + 5.42g = 21.77g

Mass percent of NaCl = mass of NaCl/total mass × 100

% mass of NaCl = 5.42g/21.77g × 100

= 0.249 × 100

= 24.9%

Answer:

24.9

Explanation:

Answer: Photosynthesizing organisms use carbon dioxide and water to produce glucose.

Explanation:

Photosynthesis is a phenomenon in which green plants containing chlorophyll use sunlight as a source of energy to convert carbon dioxide and water to form glucose and oxygen.

Photosynthesis is the process used by plants, algae and certain bacteria to convert energy from sunlight and turn it into chemical energy in the form of glucose which is used a s a source of energy by many organisms.

[tex]6CO_2+6H_2O\overset{sunlight}\rightarrow C_6H_{12}O_6+6O_2[/tex]

Answer:

Percent relative error = 0.191%

Explanation:

Relative error is a measure of accuracy (How closeness is the measure to the accepted value) for a determined data. The formula is:

Percent relative error = |Experimental - Accepted| / Accepted * 100

Where experimental is the average of the data:

(40.220g/mol + 40.654g/mol + 40.314g/mol + 40.165g/mol + 40.554g/mol) / 5 =

40.381g/mol

Replacing using accepted value = 40.304g/mol:

|40.381g/mol - 40.304g/mol| / 40.304g/mol * 100

Answer:

iconic bond is the answer

I hope it helps you ✌

Answer is given below

Explanation:

given data

weight = 6N

temp = 105° C

weight reduced = 5N

solution

weight of container is 1N

SO W = (6-1) = 5

And Wd = 5 - 1 = 4

so

moisture content is

moisture content = [tex]\frac{W-Wd}{Wd} \times 100[/tex]       .......1

moisture content = [tex]\frac{5-4}{4} \times 100[/tex]

moisture content = 25%

and

as we know density of soil soild = 2700 kg/m³

density of water = 1000 kg/m³

and sp gravity of soil =  [tex]\frac{2700}{1000}[/tex]  = 2.7

so

now we get here bulk unit weight

bulk unit wt = [tex]Yw \times [\frac{G+e}{1+e}][/tex]     ..........2

bulk unit wt = [tex]9.01 \times [\frac{2.7 + 0.675}{1+0.675}][/tex]

bulk unit wt = 19.766 KN/m³

and

so dry unit wt  will be

dry unit wt = [tex]\frac{Ysat}{1+w}[/tex]           ..............3

dry unit wt = [tex]\frac{19.766}{1+0.25}[/tex]

dry unit wt = 15.813 kN/m³

Answer:

Incorrect

Explanation:

Mole ratios are derived from the coefficients in front of the number, N2O does not have a 2 in front of it, the real mole ratio would be 4/1

Answer:

Correct

Explanation:

I had the same problem and I put correct, and it was correct.

Explanation:

Nucleic acids are the most important macromolecules for the continuity of life. They carry the genetic blueprint of a cell and carry instructions for the functioning of the cell.