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sesenic [268]
3 years ago
7

How do you write the compound Mg3N2? Are Roman numerals included as well? ​

Chemistry
1 answer:
GREYUIT [131]3 years ago
5 0

Answer:

It is written exactly like that; Mg3N2

Explanation:

Roman numerals are used sometimes to show the charge of an ion that can be multiple charges.

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Write the chemical equation for the reaction of Fe(II) ion (Fe2 ) with phenanthroline. (Use the abbreviation phen for the phenan
miskamm [114]

Answer:

[Fe(phen)3]2+    Reactants  Fe+2 clear phen white solid   product orange red

mol ratios 1;1 OF fE

Explanation:

8 0
3 years ago
Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti
gayaneshka [121]

Answer : The value of K for this reaction is, 2.6\times 10^{15}

Explanation :

The given chemical reaction is:

CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)

Now we have to calculate value of (\Delta G^o).

\Delta G^o=G_f_{product}-G_f_{reactant}

\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]

where,

\Delta G^o = Gibbs free energy of reaction = ?

n = number of moles

\Delta G^0_{(HCH_3CO_2(g))} = -389.8 kJ/mol

\Delta G^0_{(CH_3OH(g))} = -161.96 kJ/mol

\Delta G^0_{(CO(g))} = -137.2 kJ/mol

Now put all the given values in this expression, we get:

\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]

\Delta G^o=-89.4kJ/mol

The relation between the equilibrium constant and standard Gibbs, free energy is:

\Delta G^o=-RT\times \ln K

where,

\Delta G^o = standard Gibbs, free energy  = -89.4 kJ/mol = -89400 J/mol

R = gas constant  = 8.314 J/L.atm

T = temperature  = 30.0^oC=273+30.0=303K

K = equilibrium constant = ?

Now put all the given values in this expression, we get:

-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K

K=2.6\times 10^{15}

Thus, the value of K for this reaction is, 2.6\times 10^{15}

4 0
3 years ago
Which is an electronic tool that can be used to improve science?
Vanyuwa [196]
The answer is probeware
6 0
3 years ago
Read 2 more answers
A 5.00-mL sample of blood was treated with trichloroacetic acid to precipitate proteins. After centrifugation, the resulting sol
Anarel [89]

Explanation:

According to the Beer-lambert' s law,

        Absrobance = absrobptitvity × lenght × concentration

So, if we will plot a graph between absrobance and concentration then we will obtain a straight line

Hence, formula to calculate slope of the graph is as follows.

   The slope of graph = \frac{\text{difference in absrobance at two points}}{\text{difference in concentrations at those points}}

The given data is as follows.

     A_{1} = 0.412 (first point)

     A_{2} = 0.642 (second point)

     C_{1} = 0.240 ppm (first point)

     C_{2} = 0.475 ppm (second point)

Hence, putting these values into the above formula we will calculate the value of slope as follows.

           Slope = \frac{\text{difference in absrobance at two points}}{\text{difference in concentrations at those points}}

                     = \frac{0.642 - 0.412}{0.475 - 0.240}

                      = \frac{0.23}{0.235}

                     = 0.978

It is known that for the graph, line equation can be written as follows.

                       y = mx + c  ........... (1)

where,      C = intercept

                 m = slope

Hence, calculate the value of intercept as follows.

           0.412 = 0.24 \times 0.978 + c

               c = 0.178

As it is given that the absorbance values 0.454 (the y-axis value). Therefore, putting this value into equation (1) we get the following.

                     y = mx + c

                  0.454 = x \times 0.978 + 0.178

                   0.276 = 0.978x

                         x = 0.282 ppm

Thus, we can conclude that the concentration of Pb in the given sample is 0.282 ppm.

3 0
3 years ago
How many ml of a 0.50m solution of hno3 solution are needed to make 500 ml of 0.15m hno3
Anna71 [15]

Answer:

150.0 mL.

Explanation:

  • It is known that the no. of millimoles of HNO₃ before dilution = the no. of millimoles of HNO₃ after dilution.

∵ (MV) before dilution = (MV) after dilution.

<em>∴ V before dilution = (MV) after dilution / M before dilution</em> = (0.15 M)(500.0 mL)/(0.50 M) = <em>150.0 mL.</em>

7 0
3 years ago
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