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2 years ago

Compute 4.659×104−2.14×104. Round the answer appropriately.

Chemistry

2 answers:

Flauer [41]2 years ago

7 0

The result is 25200

4.659 minus 2.14 will produce 3 significant number referring to a number that has the smallest significant number value of 2.14

<h3>Further Explanation </h3>

Significant numbers are numbers obtained from the measurement results of exact numbers and the last number estimated

In scientific notation, all numbers are displayed before the big order is called significant numbers

the scientific notation can be described as:

a, ... x 10ⁿ

a, ... called an significant numbers or mantissa

10ⁿ is called a big order

Rules for significant numbers in general:

1. all non-zero numbers are significant numbers

2. a zero which is located between two non-zero numbers including a significant number

3. all zeros are located in the final row written behind the decimal point include significant number

4. zero decimal point is not significant number
5. in scientific notation, significant numbers are coefficient

In the question it is known that

(4,659 × 10⁴) and (2.14 × 10⁴) both have a significant number in a row 4,659 and 2.14 so that it meets the rule of significant numbers 5 which is 4 and 3 significant number

So 4.659 minus 2.14 will produce 3 significant number (because we are referring to a number that has the smallest significant number value of 2.14)

So that we do the significant number/coefficient then proceed a big order

Because it has the same exponent number, then the coefficients can be directly deducted, except if the exponent number is different then it must be converted first

4.659 - 2.14 = 2.519

So the result = 2.519 . 10⁴

Because the result must have 3 significant number , the result is 25200 (2 zeros behind is not a significant number)

<h3>Learn more </h3>

significant figures

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significant figures in the following number: 5.67 x 106

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the number of significant figures is 0.025

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0.080 significant figures

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the fewest number of significant figures

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Keywords : significant number, scientific notation, the exponent number

pentagon [3]2 years ago

5 0

Answer: 25,200.

Explanation:

1) Given: 4.659 × 10⁴ - 2.14 × 10⁴

2) You have to deal with significant figures.

Since, the powers are the same (10⁴) you can directly subtract the decimal numbers. But first analyze the significant figures and the number of decimal digits.

3) The number 4.659 × 10⁴ has four significant figures (4, 6, 5, and 9), while the number 2.14 × 10⁴ has three significan figures (2, 1, and 4).

4) When you add or subtract numbers with diferent amount of decimal digits, the result must show the same number of decimal digits as the term with the least number of decimal digits.

5) Before subtracting, you must round all the terms to the least number of decimal digits. So, since 2.14 has two decimal digits and 4.659 has three decimal digits, you shall round 4.659 to 4.66.

6) Now you subtract 4.66 - 2.14 = 2.52

7) Multiply by the power of 10: 2.52 × 10⁴ = 25,200. And that is the answer.

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It takes 839./kJmol to break a carbon-carbon triple bond. Calculate the maximum wavelength of light for which a carbon-carbon tr

tresset_1 [31]

Answer:

The maximum wavelength of light for which a carbon-carbon triple bond could be broken by absorbing a single photon is 143 nm.

Explanation:

It takes 839 kJ/mol to break a carbon-carbon triple bond.

Energy required to break 1 mole of carbon-carbon triple bond = E = 839 kJ

E = 839 kJ/mol = 839,000 J/mol

Energy required to break 1 carbon-carbon triple bond = E'

$E'=\frac{ 839,000 J/mol}{N_A}=\frac{839,000 J}{6.022\times 10^{23} mol^{-1}}=1.393\times 10^{-18} J$

The energy require to single carbon-carbon triple bond will corresponds to wavelength which is required to break the bond.

$E'=\frac{hc}{\lambda }$ (Using planks equation)

$\lambda =\frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{1.393\times 10^{-18} J}$

$\lambda =1.427\times 10^{-7} m =142.7 nm = 143 nm$

$(1 m = 10^9 nm)$

The maximum wavelength of light for which a carbon-carbon triple bond could be broken by absorbing a single photon is 143 nm.

6 0

2 years ago

Copper(II) sulfide, CuS, is used in the development of aniline black dye in textile printing. What is the maximum mass of CuS wh

Naya [18.7K]

Answer:

1.82 g is the maximum mass of CuS.

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $CuCl_2$ :

Molarity = 0.500 M

Volume = 38.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 38.0×10⁻³ L

Thus, moles of $CuCl_2$ :

$Moles=0.500 \times {38.0\times 10^{-3}}\ moles$

Moles of $CuCl_2$ = 0.019 moles

For $(NH_4)_2S$ :

Molarity = 0.600 M

Volume = 42.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 42.0×10⁻³ L

Thus, moles of $(NH_4)_2S$ :

$Moles=0.600 \times {42.0\times 10^{-3}}\ moles$

Moles of $(NH_4)_2S$ = 0.0252 moles

According to the given reaction:

$CuCl_2_{(aq)}+(NH_4)_2S_{(aq)}\rightarrow CuS_{(s)}+2NH_4Cl_{(aq)}$

1 mole of $CuCl_2$ reacts with 1 mole of $(NH_4)_2S$

So,

0.019 mole of $CuCl_2$ reacts with 0.019 mole of $(NH_4)_2S$

Moles of $(NH_4)_2S$ = 0.019 mole

Available moles of $(NH_4)_2S$ = 0.0252 mole

Limiting reagent is the one which is present in small amount. Thus, $CuCl_2$ is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of $CuCl_2$ gives 1 mole of $CuS$

0.019 mole of $CuCl_2$ gives 0.019 mole of $CuS$

Moles of $CuS$ formed = 0.019 moles

Molar mass of $CuS$ = 95.611 g/mol

Mass of $CuS$ = Moles × Molar mass = 0.019 × 95.611 g = 1.82 g

1.82 g is the maximum mass of CuS.

5 0

2 years ago

When 5.00 g of FeCl3 . xH2O are heated, 2.00 g of H20 are driven off. Find the chemical formula and the name of the hydrate

cricket20 [7]

Answer:

x551x6x255

Explanation:

if its wrong sorry

3 0

2 years ago

Read 2 more answers

Phosphorous acid, h3po3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. calculate the ph f

Varvara68 [4.7K]

Answer:

Explanation:

(a)

Before the addition of KOH :-

Given pKa1 of H3PO3 = 1.30

we know , pKa1 = - log10Ka1

Ka1 = 10-pKa1

Ka1 = 10-1.30

Ka1 = 0.0501

similarly pKa2 = 6.70 ,therefore Ka2 = 1.99 x 10-7

because Ka1 >> Ka2 , therefore pH of diprotic acid i.e H3PO3 can be calculated from first dissociation only .

ICE table is :-

H3PO3 (aq) <-------------> H+ (aq) + H2PO3-(aq)

I 2.4 M 0 M 0 M

C - x + x + x

E (2.4 - x )M x M x M

x = degree of dissociation

Now expression of Ka1 is :

Ka1 = [ H+ ] [ H2PO3-] / [ H3PO3]

0.0501 = x2 / 2.4 - x

on solving for x by using quadratic formula , we have

x = 0.32

Now [ H+ ] = [ H2PO3-] = 0.32 M

pH = - log [H+]

pH = - log 0.32

pH = - ( - 0.495)

pH = 0.495

Hence pH before the addition of KOH = 0.495

(b)

After the addition of 25.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.025 L = 0.06 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

Now 0.06 moles of KOH is equal to the half of the moles required for the first equivalent point . therefore pH at this point is equal to pKa1 .

Hence pH = 1.30 M

(c)

After the addition of 50.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.050 L = 0.12 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

because Number of moles of H3PO4 = Number of moles of KOH

therefore , this point is the first equivalence point

and pH = pKa1 + pKa2 / 2

pH = 1.30 + 6.70 / 2

pH = 4.00

Hence pH = 4.00

(d)

After the addition of 75.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.075 L = 0.18 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

This is the half way of the second equivalence point , therefore pH is equal to pKa2 .

Hence pH = 6.70

5 0

2 years ago

A student reacts 20.0 mL of 0.248 M H2SO4 with 15.0 mL of 0.195 M NaOH. Write a balanced chemical equation to show this reaction

Molodets [167]

Answer: The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For NaOH:

Initial molarity of NaOH solution = 0.195 M

Volume of solution = 15.0 mL = 0.015 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol$

For sulfuric acid:

Initial molarity of sulfuric acid solution = 0.248 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, $2.925\times 10^{-3}$ moles of KOH will react with = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sulfuric acid

As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, NaOH is considered as a limiting reagent because it limits the formation of product.

Excess moles of sulfuric acid = $(4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol$

By Stoichiometry of the reaction:

2 moles of KOH produces 1 mole of sodium sulfate

So, $2.925\times 10^{-3}$ moles of KOH will produce = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sodium sulfate

For sodium sulfate:

Moles of sodium sulfate = $1.462\times 10^{-3}moles$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sodium sulfate}=\frac{1.462\times 10^{-3}}{0.035}=0.0418M$

For sulfuric acid:

Moles of excess sulfuric acid = $3.498\times 10^{-3}mol$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sulfuric acid}=\frac{3.498\times 10^{-3}}{0.035}=0.0999M$

For NaOH:

Moles of NaOH remained = 0 moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of NaOH}=\frac{0}{0.050}=0M$

Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

8 0

2 years ago