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eimsori [14]
3 years ago
9

This formula equation is unbalanced.

Chemistry
2 answers:
Afina-wow [57]3 years ago
6 0
The balanced equation will read:

Pb(NO3)2 (aq) + LiSO4 (aq) ➡️ PbSO4 (s) + 2️⃣LiNO3 (aq)

have a great day!!
Alex_Xolod [135]3 years ago
3 0

<u>Answer:</u> The coefficient '2' must be added in front of LiNO_3 to form a balanced equation.

<u>Explanation:</u>

Every balanced chemical equation follows law of conservation of mass.

This law states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form. The total number of individual atoms on the reactant side must be equal to the total number of individual atoms on the product side.

For the given chemical equation:

Pb(NO_3)_2(aq.)+Li_2SO_4(aq.)\rightarrow PbSO_4(s)+2LiNO_3(aq.)

The coefficient '2' must be added in front of LiNO_3 to balance the equation.

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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
The natural abundances of elements in the human body, expressed as percent by mass, are: oxygen (O),65 percent; carbon (C), 18 p
almond37 [142]

Answer:

The mass in grams of each element in the body of 62 Kg are

Oxygen               40300 g

Carbon                11160 g

Hydrogen            6200 g

Nitrogen              1860 g

Calcium                992 g

Phosphorus         744 g

Other elements   744 g

It is possible to check it adding all the components, the total is 62000 g.  

Explanation:

Firstly, we have a total mass of 62 Kg and the composition as a percentage of each component.

                            Percentage ( %)

Oxygen               65.0

Carbon               18.0

Hydrogen            10.0

Nitrogen              3.0

Calcium               1.6

Phosphorus         1.2

Other elements   1.2

Taking the definition of a percentage as part of hundred total, each percentage can be expressed as a decimal number. We should divide the percentage with 100 of total. For example: Oxygen 65%, it means 65/100 = 0.65. So, Oxygen is 0.65 in decimal numbers. We do the same for each component.

                           Percentage ( %)      Decimal number

Oxygen               65.0                         65/100 = 0.65

Carbon               18.0                           18/100 = 0.18

Hydrogen            10.0                         10/100 = 0.10

Nitrogen              3.0                           3/100 = 0.03

Calcium               1.6                           1.6/100 = 0.016

Phosphorus         1.2                           1.2/100 = 0.012

Other elements   1.2                           1.2/100 = 0.012

Finally. We can multiply decimal number of each component taking the total. 1 Kg is equivalent to 1000 g.

So, 62 Kg = 62 * 1000 g = 62000 g

After that we get the mass in grams of each element, by multiplying 62000 g and the decimal number of each component as follows:

                           Percentage ( %)      Decimal number      Mass in grams (g)        

Oxygen               65.0                         65/100 = 0.65         0.65*62000=40300

Carbon               18.0                           18/100 = 0.18         0.18*62000=  11160

Hydrogen            10.0                         10/100 = 0.10         0.10*62000=   6200

Nitrogen              3.0                           3/100 = 0.03          0.03*62000= 1860

Calcium               1.6                           1.6/100 = 0.016        0.016*62000= 992

Phosphorus         1.2                           1.2/100 = 0.012        0.012*62000=744

Other elements   1.2                           1.2/100 = 0.012        0.012*62000=744  

5 0
3 years ago
How do balanced chemical equations show the conservation of mass
antiseptic1488 [7]

Answer: Matter cannot be created or destroyed

Explanation: Balanced equations are set equations we cannot change one element or compound in the equation without changing the entire equation. So balanced equation show the conservation of mass because while other substances may be formed from the synthesis or decomposition of compounds new elements are never introduced and are not created out of thin air    :)

4 0
3 years ago
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Each of the following are descriptions of physical properties except C. Flammability
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What is a geochemist
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