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anzhelika [568]
3 years ago
5

How many moles are in 4.3 x 10 ^22 molecules of Na3PO4

Chemistry
1 answer:
Sever21 [200]3 years ago
3 0

Answer:

0.071 moles of Na₃PO₄ .

Explanation:

Given data:

Number of molecules of Na₃PO₄ = 4.3× 10²² molecules

Number of moles = ?

Solution:

1 mole contain 6.022 × 10²³ molecules

4.3× 10²² molecules × 1 mol / 6.022 × 10²³ molecules

0.71× 10⁻¹ mol

0.071 mol

The number 6.022 × 10²³ is called Avogadro number.

"It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance"

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When two different substances are mixed together, does a chemical reaction always take place?
vladimir2022 [97]
No it depends on the substances
3 0
3 years ago
Read 2 more answers
An unknown metal has a mass of 86.8 g. When 5040 J of heat are added to the sample, the sample temperature changes by 64.7 ∘ C .
grandymaker [24]

Answer: The specific heat of the unknown metal is 0.897J/g^0C

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q=m\times c\times \Delta T

Q = Heat absorbed=5040 Joules

m= mass of substance = 86.8 g

c = specific heat capacity = ?

Initial temperature of the water = T_i

Final temperature of the water = T_f

Change in temperature ,\Delta T=T_f-T_i=(64.7)^0C

Putting in the values, we get:

5040=86.8\times c\times 64.7^0C

c=0.897J/g^0C

The specific heat of the unknown metal is 0.897J/g^0C

4 0
3 years ago
Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp
Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}

For calculating edge length,

a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}

For calculating M_{ave}, we use the formula

M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}

Similarly for calculating (\rho)_{ave}, we use the formula

\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}

From the periodic table, masses of the two elements can be written

M_{Fe}= 55.85g/mol

M_{V}=50.941g/mol

Specific density of both the elements are

(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3

Putting  M_{ave} and \rho_{ave} formula's in edge length formula, we get

a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}}  \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}}  \right )}  \right ]^{1/3}

a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}}  \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}}  \right )}  \right ]^{1/3}

By calculating, we get

a=2.89\times10^{-8}cm=0.289nm

7 0
3 years ago
Webb has calculated the percent composition of a compound. How can he check his result?
fenix001 [56]
Webb has calculated the percent composition of a compound. He can check his result by adding them to see if they equal up to 100. Why? Well, percent composition tells the quantity of elements with 100 as a base of total amount. This means that it will have to add to 100 to check the result. You would add up all of the values of percent composition of elements to see if they equal 100, and if they do, the results are accurate.

Your final answer: Webb can check his result by seeing if they add up to 100, considering that is the base total quantity.
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2 years ago
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Answer:

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