1. Fill in the blanks.<br> 1. Anything which occupies space and has muss is called

Will give brainliest, like, 5 stars and 5 points to best answer

Oliga [24]

evaporation systems allow for an endless source of water. you can grab cups of water straight from the sea or even a lake. the use of evaporation allows for you to drink water thats even healthier than getting it from a cloud and it will leave all of the bad parts that used to be in the water in the first container you pour into. this system is most useful in hot climates such as places near the equator.

6 0

3 years ago

How many particles would be found in a 1.224 g sample of K2O

lbvjy [14]

Answer:

9.96*10^21

Explanation:

Molar mass of K2O=29*2+16

= 74g per mol

number of moles in the sample= 1.224/ 74

=0.1654

Number of particles in 1 mole=6.0221409*10^23

Number of particles= 0.01654*6.0221409*10^23

=9.96*10^21

4 0

4 years ago

Read 2 more answers

Which pairing below does not correctly match the scientist with the contribution to the understanding of the atom? A. Proust and

Lyrx [107]

The answer would be C. In fact, letter A already defines the Law of Definite Proportions. On the other hand, J.J. Thomson was responsible for the discovery of electrons through cathode ray tube experiments. The rest of the choices are true.

5 0

4 years ago

Read 2 more answers

A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution is added to precipitate out

scZoUnD [109]

<u>Answer:</u> The mass percent of potassium bromide in the mixture is 9.996%

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For lead (II) bromide:</u>

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

$\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol$

The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

$2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3$

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = $\frac{2}{1}\times 0.0021=0.0042mol$ of potassium bromide

Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

$0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g$

To calculate the percentage composition of KBr in the mixture, we use the equation:

$\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100$

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

$\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%$

Hence, the percent by mass of KBr in the mixture is 9.996 %

5 0

3 years ago

How many moles of copper are in 1.51 x 1024 Cu atoms?

kkurt [141]

<h3>Answer:</h3>

2.51 mol Cu

<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>Chemistry</u>

<u>Atomic Structure</u>

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

<h3>Explanation:</h3>

<u>Step 1: Define</u>

1.51 × 10²⁴ atoms Cu

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

Avogadro's Number

<u>Step 3: Convert</u>

Set up: $\displaystyle 1.51 \cdot 10^{24} \ atoms \ Cu(\frac{1 \ mol \ Cu}{6.022 \cdot 10^{23} \ atoms \ Cu})$
Multiply/Divide: $\displaystyle 2.50747 \ mol \ Cu$

<u>Step 4: Check</u>

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

2.50747 mol Cu ≈ 2.51 mol Cu

8 0

3 years ago

1. Fill in the blanks. 1. Anything which occupies space and has muss is called