All categories

3 years ago

How many atoms are in 1.4 moles of H3PO4?

Chemistry

1 answer:

sdas [7]3 years ago

3 0

<h3>Answer:</h3>

8.4 × 10²³ atoms H₃PO₄

<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>

Moles
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] 1.4 moles H₃PO₄

[Solve] atoms H₃PO₄

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

Avogadro's Number

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 1.4 \ mol \ H_3PO_4(\frac{6.022 \cdot 10^{23} \ atoms \ H_3PO_4}{1 \ mol \ H_3PO_4})$
[DA] Multiply [Cancel out units]: $\displaystyle 8.4308 \cdot 10^{23} \ atoms \ H_3PO_4$

<u>Step 4: Check</u>

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

8.4308 × 10²³ atoms H₃PO₄ ≈ 8.4 × 10²³ atoms H₃PO₄

You might be interested in

g + In a coffee-cup calorimeter, when 3.25 g of NaOH is dissolved in 50.00 g of water initially at 22.0 oC, the temperature of t

kicyunya [14]

Answer:

THE STANDARD HEAT OF SOLUTION OF SODIUM HYDROXIDE IN WATER IS -7.68 KJ PER MOLE.

Explanation:

Variables:

Mass of NaOH = 3.25 g

Mass of water = 50 g

Initial temperature of water = 22°C = 22 + 273 K = 295 K

Final temperature of the reaction mixture = 24.8 °C = 24.8 + 273 K = 297.8 K

Assuming that:

1. specific heat of water = 4.184 J/g °C

2. total mass of the reaction mixture = 50 g + 3.25 g = 53.25 g

3. the rise in temperature = (297.8 K - 295 K ) = 2.8 K

4. Molar mass of sodium hydroxide = ( 23 + 16 + 1) = 40 g/mol

5. number of mole of sodium hydroxide = mass / molar mass

n = 3.25 g / 40 g/mol

n = 0.08125 moles

The rise in temperature for the reaction mixture produces how much of heat:

Heat = mass * specific heat * change in temperature

Heat = 53.25 * 4.184 * 2.8

Heat = 623.8344 J of heat.

Equation of reaction:

NaOH + H2O -------> NaOH + H2O + Heat

This is not a reaction but a dissolution as sodium hydroxide is very soluble in water and this reaction is exothermic where heat is given off.

So since 3.25 g having 0.08125 moles produces 623.8344 J of heat, 1 mole of the sodium hydroxide used will produce:

0.08125 mole of sodium hydroxide = 623.8344 J of heat

1 mole of sodium hydroxide = ( 623.8344 / 0.08125 J of heat

= 7677.96 J of heat per mole of sodium hydroxide.

= 7.68 kJ of heat

So therefore, the standard heat of solution of sodium hydroxide in water is -7.68 kJ of heat since its an exothermic reaction.

5 0

3 years ago

Calculate the change in pH of a 1.00 L of a buffered solution preparing by mixing 0.50 M acetic acid (Ka = 1.8 x 10^-5) and 0.50

tankabanditka [31]

The change in pH of a 1.00 L of a buffered solution preparing by mixing 0.50 M acetic acid (Ka = 1.8 x 10^-5) and 0.50 M sodium acetate when 0.010 mole of NaOH is added is 4.75

when the same amount 0.010 mole of NaOH was added to 1.00 L of water the pH = 12

Explanation:

given that:

concentration of acetic acid = 0.50 M

Concentration of base sodium acetate = 0.50 M

ka = 1.8 x 10^-5)

pka = -log [ka]

pka = 4.74

From Henderson-Hasselbalch Equation:

pH = pKa + log $\frac{[base]}{[acid]}$

pH = 4.74 + Log $\frac{[0.5]}{[0.5]}$

pH = 4.74 + 0

pH = 4.74

Number of moles of NaOH = 0.010 moles

volume 1 litre

molarity = 0.010 M

Moles of acetic acid and sodium acetate before addition of NaOH

FORMULA USED:

molarity = $\frac{number of moles}{volume in litres}$

acetic acid,

0.5 = number of moles

0.5 is the number of moles of sodium acetate.

number of moles of NaOH 0.010 moles

NaOH reacts in 1:1 molar ratio with acetic acid so

number of moles in acetic acid = 0.5 - 0.010 = 0.49

number of moles in sodium acetate = 0.5 +0.010 = 0.51

new pH

pH = pKa + log $\frac{[base]}{[acid]}$

pH= 4.74 + log[0.51] - log[0.49]

pH= 4.75

PH of NaOH of 0.01 M (BASE)

pOH = -Log[0.01]

pOH = 2

pH can be calculated as

14= pH +pOH

pH= 14-2

pH = 12

8 0

3 years ago

Given that ΔH = −571.6 kJ/mol for the reaction 2 H2(g) + O2(g) → 2 H2O(l), calculate ΔH for these reactions. (a) 2 H2O(l) → 2 H2

pashok25 [27]

Answer : The value of $\Delta H$ for the reaction is +571.6 kJ/mole.

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction is,

$2H_2(g)+O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_1=-571.6kJ/mole$