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Marta_Voda [28]

3 years ago

7

Un estudiante debe preparar una disolución 2M de NaCl (58,44 g/mol) en un recipiente de 7L. ?Cuántos gramos de NaCl debe agregar

?

1 answer:

Gnesinka [82]3 years ago

3 0

Answer:

818.2 g.

Explanation:

Molarity is the no. of moles of solute per 1.0 L of the solution.

<em>M = (no. of moles of NaCl)/(Volume of the solution (L))</em>

<em></em>

M = 2.0 M.

no. of moles of NaCl = ??? mol,

Volume of the solution = 7.0 L.

∴ (2.0 M) = (no. of moles of NaCl)/(7.0 L)

∴ (no. of moles of NaCl) = (2.0 M)*(7.0 L) = 14.0 mol.

To find the mass of NaCl, we can use the relation:

<em>no. of moles of NaCl = mass/molar mass</em>

<em></em>

<em>∴ mass of NaCl = (no. of moles of NaCl)*(molar mass) =</em> (14.0 mol)*(58.44 g/mol) = <em>818.2 g.</em>

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inessss [21]

Answer:

it is possible to remove 99.99% Cu2 by converting it to Cu(s)

Explanation:

So, from the question/problem above we are given the following ionic or REDOX equations of reactions;

Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V

Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V

In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:

Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.

Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.

Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.

Thus, ΔG° = - 92640.

This is less than zero[0]. Therefore, it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.

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3 years ago

Which of the following breaks and forms bonds?

svp [43]

Your answer would be "Chemical Change"

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3 years ago

Given the following: [G3P] = 1.5x10-5M; [BPG] = 3.0x10-3M ; [NAD+] = 1.2x10-5M; [NADH]=1.0x10-4 ; [HPO42-]= 1.2x10-5 M; pH = 7.5

mel-nik [20]

<u>Answer:</u> The given reaction is non-spontaneous in nature.

<u>Explanation:</u>

To calculate the $H^+$ concentration, we use the equation:

$pH=-\log[H^+]$

We are given:

pH of the solution = 7.5

$7.5=-\log [H^+]$

$[H^+]=10^{-7.5)=3.1\times 10^{-8}M$

For the given chemical equation:

$\text{Glyceraldehyde3-phosphate }+NAD^++HPO_4^{2-}\rightarrow \text{1,3-Biphosphoglycerate }+NADH+H^+$

The equation used to Gibbs free energy of the reaction follows:

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

where,

$\Delta G$ = free energy of the reaction

$\Delta G^o$ = standard Gibbs free energy = 6.3 kJ/mol = 6300 J/mol (Conversion factor: 1kJ = 1000J)

R = Gas constant = 8.314J/K mol

T = Temperature = $25^oC=[273+25]K=298K$

$K_{eq}$ = Ratio of concentration of products and reactants = $\frac{[BPG][NaDH][H^+]}{[G_3P][NAD^+][HPO_4^{2-}]}$

$[BPG]=3.0\times 10^{-3}M$

$[NADH]=1.0\times 10^{-4}M$

$[H^+]=3.1\times 10^{-8}M$

$[G_3P]=1.5\times 10^{-5}M$

$[NAD^+]=1.2\times 10^{-5}M$

$[HPO_4^{2-}]=1.2\times 10^{-5}M$

Putting values in above equation, we get:

$\Delta G=6300J/mol+(8.314J/K.mol\times 298K\times \ln (\frac{(3.0\times 10^{-3})\times (1.0\times 10^{-4})\times (3.1\times 10^{-8})}{(1.5\times 10^{-5})\times (1.2\times 10^{-5})\times (1.2\times 10^{-5})}))\\\\\Delta G=9917.02J/mol=9.92kJ/mol$

For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.

As, the Gibbs free energy of the reaction is positive. The reaction is said to be non-spontaneous.

Hence, the given reaction is non-spontaneous in nature.

4 0

3 years ago

A sample of water (88 grams) had a temperature change of 6.0 ℃. What was the heat change of this sample? (q = m c ΔT, c = 4.18 J

jeka94

Answer:

$Q=2209J$

Explanation:

Hello!

In this case, since the heat involved during a heating process is computed in terms of mass, specific heat and temperature change as shown below:

$Q=mC\Delta T$

Thus, since the heated mass of water was 88 g, the specific heat of water is 4.184 J/g°C and the temperature change is 6.0 °C, we can compute the heat as shown below:

$Q=88g*4.184\frac{J}{g\°C}*6.0\°C \\\\Q=2209J$

Best regards!

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