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3 years ago

In order to control pollution, you should reduce the need for _____________ goods and transportation.

Chemistry

2 answers:

Tems11 [23]3 years ago

5 0

Answer:

nonrenewable

Explanation:

It's just an educated guess but you know plastic bottles and stuff like that contribute to pollution

sergey [27]3 years ago

3 0

Answer:manufacture

Explanation:

You might be interested in

What are the rules for naming hydrocarbons?

GrogVix [38]

Answer:

IUPAC Rules for Alkane Nomenclature

Find and name the longest continuous carbon chain.

Identify and name groups attached to this chain.

Number the chain consecutively, starting at the end nearest a substituent group.

Designate the location of each substituent group by an appropriate number and name.

Explanation:

3 0

2 years ago

Mrs.Jacobs dropped an object from 10 meters she knows it did 50 joules of work how much did it weigh?

Alenkasestr [34]

It is 30 hope it help's.

5 0

3 years ago

Iridium has two naturally occurring isotopes, Iridium-191 and Iridium-193. If the average atomic mass of orodum s 192.217, what

Lina20 [59]

Answer:

The percent isotopic abundance of Ir-193 is 60.85 %

The percent isotopic abundance of Ir-191 is 39.15 %

Explanation:

we know there are two naturally occurring isotopes of iridium, Ir-191 and Ir-193

First of all we will set the fraction for both isotopes

X for the isotopes having mass 193

1-x for isotopes having mass 191

The average atomic mass is 192.217

we will use the following equation,

193x + 191(1-x) = 192.217

193x + 191 - 191x = 192.217

193x- 191x = 192.217 - 191

2x = 1.217

x= 1.217/2

x= 0.6085

0.6085 × 100 = 60.85 %

60.85% is abundance of Ir-193 because we solve the fraction x.

now we will calculate the abundance of Ir-191.

(1-x)

1-0.6085 =0.3915

0.3915× 100= 39.15 %

6 0

3 years ago

Write the balanced nuclear equation for the following. (Use the lowest possible coefficients. Omit states-of-matter in your answ

olga55 [171]

Answer:

$_{93}^{232}\text{Np} + _{-1}^{0}\text{e} \longrightarrow _{92}^{232}\text{U}$

Explanation:

The unbalanced nuclear equation is

$_{93}^{232}\text{Np} + _{1}^{0}\text{e} \longrightarrow ?$

It is convenient to replace the question by an atomic symbol, $_{x}^{y}\text{Z}$ , where x = the atomic number, y = the mass number, and Z = the symbol of the element.

$_{93}^{232}\text{Np} + _{1}^{0}\text{e} \longrightarrow _{x}^{y}\text{Z}$

Then your equation becomes

$_{93}^{232}\text{Np} + _{1}^{0}\text{e} \longrightarrow _{x}^{y}\text{Z}$

The main point to remember in balancing nuclear equations is that the sums of the superscripts and of the subscripts must be the same on each side of the equation.

Then

93 – 1 = x, so x = 92

232 + 0 = y, so y = 232

Element 92 is uranium, so the nuclear equation becomes

$_{93}^{232}\text{Np} + _{-1}^{0}\text{e} \longrightarrow _{92}^{232}\text{U}$

3 0

3 years ago

Calculate the percent ionization of nitrous acid in a solution that is 0.230 M in nitrous acid. The acid dissociation constant o

djyliett [7]

Ok first, we have to create a balanced equation for the dissolution of nitrous acid.

HNO2 <-> H(+) + NO2(-)

Next, create an ICE table

HNO2 <--> H+ NO2-
[]i 0.230M 0M 0M
Δ[] -x +x +x
[]f 0.230-x x x

Then, using the concentration equation, you get

4.5x10^-4 = [H+][NO2-]/[HNO2]

4.5x10^-4 = x*x / .230 - x

However, because the Ka value for nitrous acid is lower than 10^-3, we can assume the amount it dissociates is negligable,

assume 0.230-x ≈ 0.230

4.5x10^-4 = x^2/0.230

Then, we solve for x by first multiplying both sides by 0.230 and then taking the square root of both sides.

We get the final concentrations of [H+] and [NO2-] to be x, which equals 0.01M.

Then to find percent dissociation, you do final concentration/initial concentration.

0.01M/0.230M = .0434 or

≈4.34% dissociation.

8 0

3 years ago