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2 years ago

Find the opposite words from the text a.Fold b. noisy c. uncertainly

Chemistry

2 answers:

sveticcg [70]2 years ago

7 0

Unfold, peacefully, certainly

vlabodo [156]2 years ago

5 0

Answer:

a.Fold × unfold

b. noisy × peaceful

c. uncertainly × certainly

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why does ocean water found near the poles often have a higher leaves of salinity than ocean water found near the equator

zaharov [31]

Answer:

Water near the poles often have higher salinity because Cold polar air cools the water and lowers its temperature, increasing its salinity. Fresh water freezes out of seawater to become sea ice, which also increases the salinity of the remaining water.

Understands! ♥

4 0

3 years ago

How many formula units are there in 50.3 moles potassium chloride?

Alona [7]

<h3>Answer:</h3>

3.03 × 10²⁵ formula units KCl

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

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Unit 0

Reading a Periodic Table
Writing Compounds

Atomic Structure

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

<h3>Explanation:</h3>

Step 1: Define

50.3 mol KCl (Potassium chloride)

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

$\displaystyle 50.3 \ mol \ KCl(\frac{6.022 \cdot 10^{23} \ formula \ units \ KCl}{1 \ mol \ KCl} )$ = 3.02907 × 10²⁵ formula units KCl

Step 4: Check

We are given 3 sig figs. Follow sig fig rules and round.

3.02907 × 10²⁵ formula units KCl ≈ 3.03 × 10²⁵ formula units KCl

6 0

3 years ago

What is the most massive element?

Vsevolod [243]

I believe the most massive element that is naturally occurring on Earth is uranium (U), which has a nucleus of 92 protons

8 0

3 years ago

In sexual reproduction, two individuals produce offspring that have genetic characteristics from both parents. This type of repr

agasfer [191]

Answer:

genetic variation

Explanation:

7 0

3 years ago

Read 2 more answers

You mix 285.0 mL of 1.20 M lead(II) nitrate with 300.0 mL of 1.60 M potassium iodide. The lead(II) iodide is insoluble. Which of

SIZIF [17.4K]

Answer:

D. The final concentration of NO3– is 0.821 M.

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For potassium iodide :

Molarity = 1.60 M

Volume = 300.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 300.0×10⁻³ L

Thus, moles of potassium iodide :

$Moles=1.60 \times {300.0\times 10^{-3}}\ moles$

Moles of potassium iodide = 0.48 moles

For lead(II) nitrate :

Molarity = 1.20 M

Volume = 285 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 285×10⁻³ L

Thus, moles of lead(II) nitrate :

$Moles=1.20\times {285\times 10^{-3}}\ moles$

Moles of lead(II) nitrate = 0.342 moles

According to the given reaction:

$2KI_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbI_2_{(s)}+2KNO_3_{(aq)}$

2 moles of potassium iodide react with 1 mole of lead(II) nitrate

1 mole of potassium iodide react with 1/2 mole of lead(II) nitrate

0.48 moles potassium iodide react with 0.48/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.24 moles

Available moles of lead(II) nitrate = 0.342 moles

Limiting reagent is the one which is present in small amount. Thus, potassium iodide is limiting reagent.

Also, consumed lead(II) nitrate = 0.24 moles (lead ions precipitate with iodide ions)

Left over moles = 0.342 - 0.24 moles = 0.102 moles

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.102/0.585 M = 1.174 M

Statement A is correct.

The formation of the product is governed by the limiting reagent. So,

2 moles of potassium iodide gives 1 mole of lead(II) iodide

1 mole of potassium iodide gives 1/2 mole of lead(II) iodide

0.48 mole of potassium iodide gives 0.48/2 mole of lead(II) iodide

Mole of lead(II) iodide = 0.24 moles

Molar mass of lead(II) iodide = 461.01 g/mol

Mass of lead(II) chloride = Moles × Molar mass = 0.24 × 461.01 g = 111 g

Statement B is correct.

Potassium iodide is the limiting reagent. So all the potassium ion is with potassium nitrate . Thus,

2 moles of Potassium iodide on reaction forms 2 moles of potassium ion

0.48 moles of Potassium iodide on reaction forms 0.48 moles of potassium ion

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.48/0.585 M = 0.821 M

Statement C is correct.

Nitrate ions are furnished by lead(II) nitrate . So,

1 mole of lead(II) nitrate produces 2 moles of nitrate ions

0.342 mole of lead(II) nitrate produces 2*0.342 moles of nitrate ions

Moles of nitrate ions = 0.684 moles

So, Concentration = 0.684/0.585 M = 1.169 M

Statement D is incorrect.

4 0

3 years ago

Find the opposite words from the text a.Fold b. noisy c. uncertainly​

Find the opposite words from the text a.Fold b. noisy c. uncertainly