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Agata [3.3K]
3 years ago
5

Balance the following equations:

Chemistry
1 answer:
GuDViN [60]3 years ago
8 0

<em><u>ANSWERS</u></em><em><u> </u></em><em><u>:</u></em><em><u>-</u></em>

<em>1</em><em>)</em>

\red{2}Na + Br2 → \red{2}NaBr

<em>2</em><em>)</em>

\blue{2}C4H6 + \blue{11}O2 → \blue{8}CO2 + \blue{6}H2O

<em>3</em><em>)</em>

<em>H3PO4 + \green{3}LiOH → \green{3}H2O + Li3PO4</em>

<em>4</em><em>)</em>

<em>Ca(NO3)2 + K2SO4 → CaSO4 + \pink{2}KNO3</em>

<em>5</em><em>)</em>

<em>\orange{2}H2O2→\orange{2}H2O+O2</em>

<em>(</em><em>mark</em><em> </em><em>me</em><em> </em><em>brainliest</em><em> </em><em>please</em><em> </em><em>I</em><em> </em><em>need</em><em> </em><em>3</em><em> </em><em>more</em><em> </em><em>:</em><em>)</em><em>)</em>

<em>\large\sf\red{thank \: you}</em>

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The number of oxygen atoms in 48.0g of oxygen gas
Doss [256]

Answer:

Now u have 48 g of O2. There. Fore mole=weight/M. W. Of oxygen. Therefor 3mole.

After that if we to multiply the avogadro number with it. So 3 *NA

Now u want only atom calculation then we have 2 molecule of oxygen then multiply it with 2 too.

So final claculation is =3*2*NA.

Explanation:

your welcome

brainlest PLEASEEEEEEE!

3 0
3 years ago
How many molecules of Ca are found in a sample with 0.2 mols?
DochEvi [55]

Answer:

\boxed{1.2 \times 10^{23}\text{ atoms}}

Explanation:

6.023 × 10²³ atoms of Ca are in 1 mol of Ca

\text{No. of atoms} = \text{0.2 mol} \times \dfrac{6.023 \times 10^{23}\text{atoms }}{\text{1 mol }} = \mathbf{1.2 \times 10^{23}} \textbf{ atoms}}\\\\\text{There are }\boxed{\mathbf{1.2 \times 10^{23}} \textbf{ atoms}} \text{ atoms in 0.20 mol of Ca}

6 0
3 years ago
A liquid-fueled rocket engine has two liquids, usually called the fuel and the oxidizer. Which two liquids are most commonly use
hammer [34]

Answer:

A liquid-fueled rocket has two liquids (liquids are good because of the density, they need less space than a gas to be stored), such that these liquids are called the fuel and the oxidizer.

These liquids are injected into a system that leads to a combustion chamber, where the liquids are mixed (we need to mix the fuel with the oxidizer to enable the combustion of the fuel) and burned to produce thrust.

Some common examples of oxidizers are liquid oxygen, which may be combined with fuels like liquid hydrogen, liquid methane, kerosene and hydrazine.

Other oxidizers are liquid fluorine (which also can be combined with the fuels liquid hydrogen and hydrazine), nitrogen tetroxide (which can be combined whit kerosene, hydrazine and other fuels) and FLOX-70, which can only be combined with kerosene.

The "most commonly used" may depend on the country and the type of liquid propellant ( petroleum, cryogens, and hypergols)

Such that the most common oxidizer may be liquid oxygen, and the most common fuel the kerosene.

8 0
3 years ago
Of the following substances, only __________ has London dispersion forces as its only intermolecular force. CH3OH NH3 H2S CH4 HC
QveST [7]

Answer:

CH₄

Explanation:

CH₃OH has hydrogen bonding due to the OH group present

NH₃ also has hydrogen bonding due to the NH bonds

H₂S has dipole-dipole forces present due to the polar SH bonds

HCl also has dipole-dipole forces due to the polar HCl bond

7 0
3 years ago
How many moles are in 8.63 x 103 atoms of Li?
Ira Lisetskai [31]
<h3>Answer:</h3>

1.43 × 10⁻²⁰ mol Li

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. 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>

8.63 × 10³ atoms Li

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 8.63 \cdot 10^3 \ atoms \ Li(\frac{1 \ mol \ Li}{6.022 \cdot 10^{23} \ atoms \ Li})
  2. Multiply/Divide:                \displaystyle 1.43355 \cdot 10^{-20} \ moles \ Li

<u>Step 4: Check</u>

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

1.43355 × 10⁻²⁰ mol Li ≈ 1.43 × 10⁻²⁰ mol Li

4 0
3 years ago
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