What was Albert Einstein's contribution to our understanding of the

How many CaF are in a 1.7x10^25 please I need help fast!!

Gemiola [76]

<h3>Answer:</h3>

28 mol CaF

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

<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>

[Given] 1.7 × 10²⁵ molecules CaF

[Solve] moles CaF

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

Avogadro's Number

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 1.7 \cdot 10^{25} \ molecules \ CaF(\frac{1 \ mol \ CaF}{6.022 \cdot 10^{23} \ molecules \ CaF})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 28.2298 \ moles \ CaF$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

28.2298 mol CaF ≈ 28 mol CaF

7 0

2 years ago

When N2(g) reacts with H2(g) to form NH3(g), 92.2 kJ of energy are evolved for each mole of N2(g) that reacts. Write a balanced

Naddika [18.5K]

Answer:

N₂ + 3H₂ → 2NH₃ ΔH = - 92.2KJ

Explanation:

Let's write out the chemical equation between Nitrogen and Hydrogen to Form Ammonia.

Nitrogen + Hydrogen = Ammonia

N₂ + H₂ → NH₃

A Thermochemical Equation is a balanced stoichiometric chemical equation that includes the enthalpy change, ΔH.

The balanced stoichiometric chemical equation is given as;

N₂ + 3H₂ → 2NH₃

92.2 kJ of energy are evolved for each mole of N2(g) that reacts. And from the equation, 1 mole of N2 reacts.

The enthalpy change, ΔH = - 92.2KJ. The negative sign is because heat is being evolved.

The balanced thermochemical equation;

N₂ + 3H₂ → 2NH₃ ΔH = - 92.2KJ

6 0

2 years ago

What physical property makes metal pots good for cooking?

jeka94

Metal pots are good for cooking because they have heat conductivity.

4 0

3 years ago

Read 2 more answers

In Part A, you found the amount of product (1.80 mol P2O5 ) formed from the given amount of phosphorus and excess oxygen. In Par

Finger [1]

First step is to balance the reaction equation. Hence we get P4 + 5 O2 => 2 P2O5

Second, we calculate the amounts we start with

P4: 112 g = 112 g/ 124 g/mol – 0.903 mol

O2: 112 g = 112 g / 32 g/mol = 3.5 mol

Lastly, we calculate the amount of P2O5 produced.

2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4 mol of P2O5.

This is 1.4 * (31*2 + 16*5) = 198.8 g

3 0

3 years ago

Balance these equations!<br> 1)<br> AIBr; +<br> K-<br> KBr +<br> AI

Tema [17]

Answer:

Instructions on balancing chemical equations:

Enter an equation of a chemical reaction and click 'Balance'. The answer will appear below

Always use the upper case for the first character in the element name and the lower case for the second character. Examples: Fe, Au, Co, Br, C, O, N, F. Compare: Co - cobalt and CO - carbon monoxide

To enter an electron into a chemical equation use {-} or e

To enter an ion specify charge after the compound in curly brackets: {+3} or {3+} or {3}.

Example: Fe{3+} + I{-} = Fe{2+} + I2

Substitute immutable groups in chemical compounds to avoid ambiguity.

For instance equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced,

but PhC2H5 + O2 = PhOH + CO2 + H2O will

Compound states [like (s) (aq) or (g)] are not required.

If you do not know what products are enter reagents only and click 'Balance'. In many cases a complete equation will be suggested.

Reaction stoichiometry could be computed for a balanced equation. Enter either the number of moles or weight for one of the compounds to compute the rest.

Limiting reagent can be computed for a balanced equation by entering the number of moles or weight for all reage

Explanation:

3 0

2 years ago