All categories

2 years ago

How many moles are in 1.93x1023 particles of Na3PO4?

Chemistry

2 answers:

Tresset [83]2 years ago

6 0

Answer:

<h3>The answer is 0.32 moles</h3>

Explanation:

To find the number of moles given the number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question

N = 1.93 × 10²³ particles

We have

$n = \frac{1.93 \times {10}^{23} }{6.02 \times {10}^{23 } } \\ = 0.320598006...$

We have the final answer as

<h3>0.32 moles</h3>

Hope this helps you

wariber [46]2 years ago

5 0

<h3><u>Answer</u> :</h3>

No. of particles of Na₃PO₄ = 1.93×10²³

We have to find number of moles of the given compound $.$

Number of moles cam be found by using this formula :

$\bigstar\:\boxed{\bf{\purple{n=\dfrac{N}{N_A}}}}$

n denotes number of moles
N denotes number of particles
$\sf{N_A}$ denotes avogadro's constant

We know that,

$\bf{N_A}$ = 6.022×10²³

★ <u>Calculation</u> :

➠ n = N/ $\sf{N_A}$

➠ n = (1.93×10²³)/(6.022×10²³)

➠ <u>n = 0.32 moles</u>

★ <u>Explore More</u> :

n = weight / Molar mass
n = volume / 22.4

<h3>Hope It Helps!</h3>

You might be interested in

When scientists looked at the polarity of bands of rock on either side of the mid-ocean ridges, what did they find? A. that ther

alisha [4.7K]

I am pretty sure the answer is D. That the patterns of a polarity matched up on both sides.

3 0

2 years ago

Read 2 more answers

Using the equations 2 Sr(s) + O₂ (g) → 2 SrO (s) ∆H° = -1184 kJ/mol SrO (s) + CO₂ (g) → SrCO₃ (s) ∆H° = -234 kJ/mol CO₂ (g) → C(

kkurt [141]

<u>Answer:</u> The $\Delta H^o_{rxn}$ for the reaction is 72 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

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

The given chemical reaction follows:

$2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $2Sr(s)+O_2(g)\rightarrow 2SrO(s)$ $\Delta H_1=-1184kJ$

(2) $SrO(s)+CO_2(g)\rightarrow SrCO_3(s)$ $\Delta H_2=-234kJ$ ( × 2)

(3) $CO_2(g)\rightarrow C(s)+O_2(g)$ $\Delta H_3=394kJ$ ( × 2)

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[2\times (-\Delta H_2)]+[2\times (\Delta H_3)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-1184))+(2\times -(-234))+(2\times (394))]=72kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is 72 kJ.

4 0

2 years ago

Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang

Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

$\Delta (H_{f})_{MnO_{2}}$ = -520.0 KJ/mole

$\Delta (H_{f})_{Al_{2}O_{3}}$ = -1699.8 KJ/mole

The balanced chemical reaction is,

$3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)$

Formula used :

$\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}$

$\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))$

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

$\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]$