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3 years ago

How many years old are you if you have lived 1 billion seconds 31 years round this answer to 3 sig figs

Chemistry

1 answer:

stich3 [128]3 years ago

5 0

Answer:

age = 63.2 years

Explanation:

Given data

Number of years = 31

number of seconds = 1 billion

Age = ?

Solution

First we convert seconds to minutes

minutes = 1000000000 / 60

minutes = 16666666.667

Now convert minutes to hours

hours = 16666666.667 / 60

hours = 277777.778

Now convert hours to days

days = 277777.778 / 24

days = 11574.074

Now convert days to months

months = 11574.074 / 30

months = 385.802

Now convert months to years

years = 385.803 / 12

years = 32.15

Now add these years with given years

age = 31 + 32.15

age = 63.2 years

You might be interested in

How many moles of MgCl2 are there in 319 g of the compound?

Leokris [45]

Hey there!:

Molar mass MgCl2 = 95.2110 g/mol

So:

1 mole MgCl2 -------------- 95.2110 g

moles MgCl2 ---------------- 319 g

moles MgCl2 = 319 * 1 / 95.2110

moles MgCl2 = 319 / 95.2110

=> 3.350 moles of MgCl2

Hope that helps!

3 0

3 years ago

Naturally occurring element X exists in three isotopic forms: X-28 (27.979 amu, 92.21% abundance), X-29 (28.976 amu 4.70% abunda

bezimeni [28]

Answer: The average atomic mass of X is 28.09 amu

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

For isotope 1:

Mass of isotope 1 = 27.979 amu

Percentage abundance of isotope 1 = 92.21 %

Fractional abundance of isotope 1 = 0.9212

For isotope 2:

Mass of isotope 2 = 28.976 amu

Percentage abundance of isotope 2 = 4.70 %

Fractional abundance of isotope 2 = 0.0470

For isotope 3:

Mass of isotope 3 = 29.974 amu

Percentage abundance of isotope 3 = 3.09 %

Fractional abundance of isotope 3 = 0.0309

Putting values in equation 1, we get:

$\text{Average atomic mass of X}=[(27.979\times 0.9212)+(28.976\times 0.0470)+(29.974\times 0.0309)]$

$\text{Average atomic mass of X}=28.09amu$

Hence, the average atomic mass of X is 28.09 amu

4 0

3 years ago

A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.

irga5000 [103]

Answer:

1. pH = 1.23.

2. $H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

$H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+$

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the pKa is:

$pKa=-log(Ka)=-log(5.90x10^{-2})=1.23$

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

$pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23$

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

$n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol$

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

$H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Which is also shown in net ionic notation.

Best regards!

4 0

3 years ago

What is the name of the molecule shown below?

DiKsa [7]

Answer:

the correct awnser is b 1-heptene

8 0

3 years ago

Ammonia reacts with sulfuric acid to produce the important fertilizer, ammonium hydrogen sulfate.

Liula [17]

Answer:

404.8g of (NH4)HSO4 is produced.

Explanation:

Step 1:

Data obtained from the question. This include the following:

Temperature (T) = 10°C = 10°C + 273 = 283K

Pressure (P) = 110KPa = 110/101.325 = 1.09atm

Volume (V) = 75L

Step 2:

Determination of the number of mole of ammonia, NH3.

The number of mole (n) of ammonia, NH3 can be obtained by using the ideal gas equation. This is illustrated below:

Note:

Gas constant (R) = 0.0821atm.L/Kmol

Number of mole (n) =?

PV = nRT

1.09 x 75 = n x 0.0821 x 283

Divide both side by 0.0821 x 283

n = (1.09 x 75) /(0.0821 x 283)

n = 3.52 moles

Step 3:

Determination of the number of mole ammonium hydrogen sulfate produced from the reaction.

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

NH3 + H2SO4 —> (NH4)HSO4

From the balanced equation above,

1 mole of NH3 produced 1 mole of (NH4)HSO4

Therefore, 3.52 moles of NH3 will also produce 3.52 moles of (NH4)HSO4.

Therefore, 3.52 moles of ammonium hydrogen sulfate, (NH4)HSO4 is produced.

Step 4:

Conversion of 3.52 moles of ammonium hydrogen sulfate, (NH4)HSO4 to grams. This is illustrated below:

Molar Mass of (NH4)HSO4 = 14 + (4x1) + 1 + 32 + (16x4) = 115g/mol

Number of mole of (NH4)HSO4 = 3.52 moles

Mass of (NH4)HSO4 =..?

Mass = mole x molar Mass

Mass of (NH4)HSO4 = 3.52 x 115 = 404.8g

Therefore, 404.8g of (NH4)HSO4 is produced.

5 0

3 years ago