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2 years ago

Will give brainliest and extra points please help ASAP

Chemistry

2 answers:

Keith_Richards [23]2 years ago

8 0

The answer is translation, the figure moves to another side, does not rotate or reflect
-Hope it helps!

Viefleur [7K]2 years ago

3 0

The Correct answer to this question is translation

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In composting, gardeners allow a pile of plant material such as dead leaves to decay. Composting is a form of combustion.

maks197457 [2]

Answer:

This is true

Explanation:

Because it mixes in with the earth making the earth more suitable for gardening and/or farming.

8 0

3 years ago

How many moles in 4.93 x 10E23 atoms of silver?

leva [86]

<h3>Answer:</h3>

0.819 mol Ag

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

4.93 × 10²³ atoms Ag

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

Set up: $\displaystyle 4.93 \cdot 10^{23} \ atoms \ Ag(\frac{1 \ mol \ Ag}{6.022 \cdot 10^{23} \ atoms \ Ag})$
Divide: $\displaystyle 0.818665 \ mol \ Ag$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

0.818665 mol Ag ≈ 0.819 mol Ag

8 0

2 years ago

Which statement is an opinion?

Zepler [3.9K]

C. or A. Is your best bet

8 0

2 years ago

Read 2 more answers

32 gm of O2 to mole of O2

lana [24]

Molar mass of O2:-

$\\ \rm\longmapsto 2(16u)=32g/mol$

Now

$\boxed{\sf No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}}$

$\\ \rm\longmapsto No\:of\;moles=\dfrac{32}{32}$

$\\ \rm\longmapsto No\:of\;moles=1mol$

5 0

2 years ago

A city continuously disposes of effluent from a wastewater treatment plant into a river. The minimum flow in the river is 130 m3

Vera_Pavlovna [14]

Answer:

$2.54\ \text{mg/L}$

Explanation:

C = Allowable concentration = 1.1 mg/L

$Q_1$ = Flow rate of river = $130\ \text{m}^/\text{s}$

$Q_2$ = Discharge from plant = $37\ \text{m}^3/\text{s}$

$C_1$ = Background concentration = 0.69 mg/L

$C_2$ = Maximum concentration that of the pollutant

The concentration of the mixture will be

$C=\dfrac{Q_1C_1+Q_2C_2}{Q_1+Q_2}\\\Rightarrow C_2=\dfrac{C(Q_1+Q_2)-Q_1C_1}{Q_2}\\\Rightarrow C_2=\dfrac{1.1(130+37)-130\times 0.69}{37}\\\Rightarrow C_2=2.54\ \text{mg/L}$

The maximum concentration that of the pollutant (in mg/L) that can be safely discharged from the wastewater treatment plant is $2.54\ \text{mg/L}$ .

6 0

2 years ago