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3 years ago

(they don't have science so I guess cemastry will do.)

Chemistry

1 answer:

DedPeter [7]3 years ago

7 0

warm air rises because of convection.

A student holds the back of his hand near an iron to see if it's hot. The heat transfer is conduction.

Warm water moves throughout a mattress because of ... why is there warm water in a mattress again? Convection, I guess.

In a swimming pool, the water near the surface is warmer because of convection and solar radiation; it's closer to the Sun than the bottom of the pool, so it receives more thermal radiation, making it warmer.

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I don't get it Is a Flubber solid, liquid or gas?

Marianna [84]

Flubber would be considered an amorphous solid.

7 0

4 years ago

Read 2 more answers

In the Mond process for the purification of nickel, carbon monoxide is reacted with heated nickel to produce Ni(CO)4, which is a

igor_vitrenko [27]

Answer: The equilibrium constant for this reaction is $1.068\times 10^{6}$

Explanation:

The equation used to calculate standard Gibbs free change is of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_{(product)}]-\sum [n\times \Delta G^o_{(reactant)}]$

For the given chemical reaction:

$Ni(s)+4CO(g)\rightleftharpoons Ni(CO)_4(g)$

The equation for the standard Gibbs free change of the above reaction is:

$\Delta G^o_{rxn}=[(1\times \Delta G^o_{(Ni(CO)_4(g))})]-[(1\times \Delta G^o_{(Ni(s))})+(4\times \Delta G^o_{(CO(g))})]$

We are given:

$\Delta G^o_{(Ni(CO)_4(g))}=-587.4kJ/mol\\\Delta G^o_{(Ni(s))}=0kJ/mol\\\Delta G^o_{(CO(g))}=-137.3kJ/mol$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(1\times (-587.4))]-[(1\times (0))+(4\times (-137.3))]\\\\\Delta G^o_{rxn}=-38.2kJ/mol$

To calculate the equilibrium constant (at 58°C) for given value of Gibbs free energy, we use the relation:

$\Delta G^o=-RT\ln K_{eq}$

where,

$\Delta G^o$ = Standard Gibbs free energy = -38.2 kJ/mol = -38200 J/mol (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = 8.314 J/K mol

T = temperature = $58^oC=[273+58]K=331K$

$K_{eq}$ = equilibrium constant at 58°C = ?

Putting values in above equation, we get:

$-38200J/mol=-(8.314J/Kmol)\times 331K\times \ln K_{eq}\\\\K_{eq}=e^{13.881}=1.068\times 10^{6}$

Hence, the equilibrium constant for this reaction is $1.068\times 10^{6}$

4 0

3 years ago

For this heterogeneous system 2 A ( aq ) + 3 B ( g ) + C ( l ) − ⇀ ↽ − 2 D ( s ) + 3 E ( g ) the concentrations and pressures at

maw [93]

Answer:

$2.55*10^{11$

Explanation:

Equation for the heterogeneous system is given as:

$2A_{(aq)} + 3 B_{(g)} + C_{(l)}$ ⇄ $2D_{(s)}$ $+$ $3E_{(g)}$

The concentrations and pressures at equilibrium are:

$[A] = 9.68*10^{-2}M$

$P_B = 9.54*10^3Pa$

$[C]=14.64M$

$[D]=10.11M$

$P_E=9.56*10^4torr$

If we convert both pressure into bar; we have:

$P_B = 9.54*10^3Pa$

$P_B = (9.54*10^3)*\frac{1}{10^5} bar$

$P_B=9.54*10^{-2}bar$

$P_E=9.56*10^4torr$

1 torr = 0.001333 bar

$9.54*10^4 *0.001333 = 127.5 bar$

$K=\frac{[P_E]^3}{[A]^2[P_B]^3}$

$K=\frac{(127.5)^3}{(9.68*10^{-2})^2(9.54*10^{-2})^3}$

$K=2.55*10^{11$

3 0

3 years ago

How many moles of H₂O are produced from 3 moles of oxygen

BabaBlast [244]

Answer:

6 mols

Explanation:

Notice that on the left, there is one "O2" molecule, and one O2 results in two "H2O" molecules. So the ratio of O2 to H2O is 1 to 2
If we have 3 mols of oxygen, then per the ratio, we would have 6 mols of H2O, since 3*2=6

4 0

2 years ago

Someone answer this please...

boyakko [2]

Answer:

120 mol Mg

General Formulas and Concepts:

Chemistry - Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

120 moles H₂

Step 2: Identify Conversions

RxN: 3 mol H₂ = 3 mol Mg

Step 3: Stoichiometry

$120 \ mol \ H_2(\frac{3 \ mol \ Mg}{3 \ mol \ H_2} )$ = 120 mol Mg

7 0

3 years ago

Read 2 more answers