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Naddik [55]
2 years ago
9

Investigate: Note the empty jars on the shelf that can be filled by using the slider. Set the amount to 2.000 moles of carbon (m

ol C), then press Start. Each jar holds exactly one mole of carbon. Your goal is to determine the mass in grams of two moles of carbon. Before you can find the mass, what do you need to know
Chemistry
1 answer:
balandron [24]2 years ago
3 0

Answer:

The molar mass of carbon

Explanation:

Before the mass (in grams) of two moles of carbon can be determined, <u>the molar mass of the element would be needed.</u>

<em>This is because the number of mole of an element is the ratio of its mass and the molar mass</em>. That is,

number of mole = mass/molar mass

Hence, the mass of elements can be obtained by making it the subject of the formular;

mass = number of mole x molar mass

<em>Therefore, the molar mass of carbon would be needed before the mass of 2 moles of the element can be determined.</em>

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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
If the p-side has a higher doping concentration, explain how to keep tuning to the same radio channel?
zzz [600]

Answer:

increase in temperature of the intrinsic semiconductor

Explanation:

  • If the p-side has a higher doping concentration, it implies that number of holes (positive ion) increased which is greater than number of electron (negative ion) in the n-side
  • in order to balance the intrinsic concentration, that is to balance the number of holes and electrons which depends on temperature.
  • an increase in the temperature of the intrinsic semiconductor (p-side), increases the number of electron but number of holes remains constant.

A balance in the intrinsic concentration helps in tuning to the same radio channel.

6 0
3 years ago
Calculate the density of the items below. (D = M/V)<br> 1) M = 15g V = 5.0ml *
grin007 [14]

Answer:

<h3>The answer is 3.0 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 15 g

volume = 5 mL

We have

density =  \frac{15}{5}  \\

We have the final answer as

<h3>3.0 g/mL</h3>

Hope this helps you

5 0
3 years ago
What tool do you use to measure a rectangle?
pashok25 [27]

Answer:

Ruler

Explanation:

3 0
2 years ago
Read 2 more answers
There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In
KIM [24]

Answer:

ΔH = -470.4kJ

Explanation:

It is possible to sum 2 or more reactions to obtain the ΔH of the reaction you want to study (Hess's law). Using the reactions:

1. CaC2(s) + 2H2O(l) → C2H2(g) + Ca(OH)2(s)ΔH = −414kJ

2. 6C2H2(g) + 3CO2(g) + 4H2O(g) → 5CH2CHCO2H(g)ΔH = 132kJ

6 times the reaction 1.

6CaC2(s) + 12H2O(l) → 6C2H2(g) + 6Ca(OH)2(s)ΔH = −414kJ*6 = -2484kJ

This reaction + 2:

6CaC2(s) + 3CO2(g) + 16H2O(l) →  + 6Ca(OH)2(s) + 5CH2CHCO2H(g) ΔH = -2484kJ + 132kJ = -2352kJ

As we want to calculate the net change enthalpy in the formation of just 1 mole of acrylic acid we need to divide this last reaction in 5:

6/5CaC2(s) + 3/5CO2(g) + 16/5H2O(l) →  + 6/5Ca(OH)2(s) + CH2CHCO2H(g) ΔH = -2352kJ / 5

<h3>ΔH = -470.4kJ</h3>

4 0
3 years ago
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