Help idk how to do this

Reil [10]

Answer:

There are 5 significant digits in 0.23100.

Explanation:

This is because all non-zero digits are considered significant and zeros after decimal points are considered significant.

4 0

3 years ago

A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

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:

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

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0

3 years ago

Read 2 more answers

Calculate the solubility of ( = ) in moles per liter. Ignore any acid–base properties. s = mol/L Calculate the solubility of ( =

BaLLatris [955]

This is an incomplete question, here is a complete question.

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties.

CaCO₃, Ksp = 8.7 × 10⁻⁹

Answer : The solubility of CaCO₃ is, $9.33\times 10^{-5}mol/L$

Explanation :

As we know that CaCO₃ dissociates to give $Ca^{2+}$ ion and $CO_3^{2-}$ ion.

The solubility equilibrium reaction will be:

$CaCO_3\rightleftharpoons Ca^{2+}+CO_3^{2-}$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ca^{2+}][CO_3^{2-}]$

Let solubility of CaCO₃ be, 's'

$K_{sp}=(s)\times (s)$

$K_{sp}=s^2$

$8.7\times 10^{-9}=s^2$

$s=9.33\times 10^{-5}mol/L$

Therefore, the solubility of CaCO₃ is, $9.33\times 10^{-5}mol/L$

4 0

3 years ago

What are the effects on the bodies of water? use terms algal bloom, dead zone and hypotoxic

Anna35 [415]

Water pollution is defined as the mixing of water with unwanted substances and that makes water unsafe.

Water pollution includes runoff of excess fertilizers,

herbicides, and insecticides from agricultural lands

and residential areas; oil, grease, and toxic chemicals

from urban runoff and energy production; and

sediment from improperly managed construction sites,

crop and forest lands, and eroding stream banks.

Polluted waters have high BOD which affects aquatic biodiversity.

Hpotoxic substances such as lead, arsenic, and fluoride cause many problems to aquatic animals and humans. Water contaminated with Arsenic results in diseases such as arsenicosis. Fluoride had been reported to cause depression in DNA and RNA synthesis in cultured cells, significant reductions in DNA and RNA levels, and conditions including aging, cancer, and arteriosclerosis are associated with DNA damage and its disrepair. Lead causes problems related to the central nervous system. Children and pregnant women are most at risk. Routine applications of fertilizers and pesticides for agriculture and uncontrolled runoff in water bodies. Adds nitrogen and phosphorus to water. It adds nitrogen and phosphorus to water causing eutrophication and algal blooms.

Therefore, all these activities make our water bodies a death zone for the marine ecosystem as well as humans.

To know more about water pollution, refer to the below link:

brainly.com/question/2976496

#SPJ4

8 0

2 years ago

How many grams of N2 gas are present in 1.13 L of gas at 2.09 atm and 291 K?

solong [7]

Answer:

Mass= 2.77g

Explanation:

Applying

P=2.09atm, V= 1.13L, R= 0.082, T= 291K, Mm of N2= 28

PV=nRT

NB

Moles(n) = m/M

PV=m/M×RT

m= PVM/RT

Substitute and Simplify

m= (2.09×1.13×28)/(0.082×291)

m= 2.77g

6 0

3 years ago