Ask question

Ask question

All categories

3 years ago

7

The density of sodium is 968 g/cm3. Suppose you need 250 gm of sodium for an experiment. Which volume of sodium do you need? a 2

,420 cm3 b 0.26 cm3 c 3.87 cm3

1 answer:

Flauer [41]3 years ago

5 0

Density = 968 g/cm³

mass = 250 g

Volume = ?

Therefore:

D = m / V

968 = 250 / V

V = 250 / 968 => 0.26 cm³

Answer C

You might be interested in

Capsaicin, the compound that gives the hot taste to chili peppers, has the formula C18H27NO3. Its molar mass is 305.42 g/mol. Ca

stellarik [79]

Molar mass = 305.42 g/mol

C = ( 12 x 18 / 305.42 ) x 100 => 70.72 % of C

H = ( 1 x 27 / 305.42 ) x 100 => 8.84 % of H

N = ( 14 x 1 / 305.42 ) x 100 => 4.58 % of N

O = ( 16 x 3 / 305.42) x 100 => 15.71% of O

hope this helps!

5 0

3 years ago

Read 2 more answers

Oxidation of Nitrogen in NO2

tiny-mole [99]

Explanation:

oxidation of Nitrogen in NO2 is +4

8 0

2 years ago

Mass and energy are conserved Question 13 options: A) only in chemical changes. B) always in physical changes and sometimes in c

Roman55 [17]

Answer:

B

Explanation:

sometimes it can change chemically but not all the time.

3 0

3 years ago

Why might it be important to follow the chain of custody when handling evidence?

Ierofanga [76]

Answer:

To keep a record of who has the evidence and when

Explanation:

Don't want anyone tampering with it, hope this helps!!!

5 0

2 years ago

Chlorine has two naturally occurring isotopes, 35C1 Gsotopic mass 34.9689 amu) and 370 (isotopic mass 36.9659 amu). If chlorine

Lilit [14]

<u>Answer:</u> The percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 75.77% and 24.23% respectively.

<u>Explanation:</u>

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)

Let the fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope be 'x'. So, fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope will be '1 - x'

<u>For $_{17}^{35}\textrm{Cl}$ isotope:</u>

Mass of $_{17}^{35}\textrm{Cl}$ isotope = 34.9689 amu

Fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope = x

<u>For $_{17}^{37}\textrm{Cl}$ isotope:</u>

Mass of $_{17}^{37}\textrm{Cl}$ isotope = 36.9659 amu

Fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope = 1 - x

Average atomic mass of chlorine = 35.4527 amu

Putting values in equation 1, we get:

$35.4527=[(34.9689\times x)+(36.9659\times (1-x))]\\\\x=0.7577$

Percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope = $0.7577\times 100=75.77\%$

Percentage abundance of $_{17}^{37}\textrm{Cl}$ isotope = $(1-0.7577)=0.2423\times 100=24.23\%$

Hence, the percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 75.77% and 24.23% respectively.

6 0

3 years ago