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geniusboy [140]

3 years ago

12

The proprietor of a rock shop insists that a nugget is pure gold. if the nugget occupies a volume of 5.40 ml, what would its mas

s have to be if it were truly pure gold? (density gold = 19.3 g/ml)

1 answer:

statuscvo [17]3 years ago

7 0

Density of gold = 19.3 g/mL

Density can be calculated from volume and mass of a substance as follows,

$Density = \frac{Mass}{Volume}$

Given the volume of gold nugget = 5.40 mL

Calculating the mass of gold nugget from density and volume:

$Density = \frac{Mass}{Volume}$

$Mass = Density * Volume$

$5.40 mL * 19.3 \frac{g}{mL} =104.22 g$

Therefore, the mass of the nugget is 104.22 g

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Consider the following reaction where K. = 154 at 298 K: 2NO(g) + Brz(9) 2NOBr(g) A reaction mixture was found to contain 2.69x1

bekas [8.4K]

Explanation:

$2NO(g) + Br_2(g)\rightleftharpoons 2NOBr(g)$

Equilibrium constant of reaction = $K=154$

Concentration of NO = $[NO]=\frac{2.69\times 10^{-2} mol}{1 L}=2.69\times 10^{-2} M$

Concentration of bromine gas = $[Br_2]=\frac{3.85\times 10^{-2} mol}{1 L}=3.85\times 10^{-2} M$

Concentration of NOBr gas = $[Br_2]=\frac{9.56\times 10^{-2} mol}{1 L}=9.56\times 10^{-2} M$

The reaction quotient is given as:

$Q=\frac{[NOBr]^2}{[NO]^2[Br_2]}=\frac{(9.56\times 10^{-2} M)^2}{(2.69\times 10^{-2} M)^2\times 3.85\times 10^{-2} M}$

$Q=328.06$

$Q>K$

The reaction will go in backward direction in order to achieve an equilibrium state.

1. In order to reach equilibrium NOBr (g) must be produced. False

2. In order to reach equilibrium K must decrease. False

3. In order to reach equilibrium NO must be produced. True

4. Q. is less than K . False

5. The reaction is at equilibrium. No further reaction will occur. False

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3 years ago

How much heat energy is required to raise the temperature of 0.368 kg of copper from 23.0 ∘C to 60.0 ∘C? The specific heat of co

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Calculate the entropy change for the surroundings of the reaction below at 350K: N2(g) + 3H2(g) -> 2NH3(g) Entropy data: NH3

krek1111 [17]

Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K

Explanation :

We have to calculate the entropy change of reaction $(\Delta S^o)$ .

$\Delta S^o=S_{product}-S_{reactant}$

$\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

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$\Delta S^0{(H_2)}$ = standard entropy of $H_2$

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

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3 years ago

What is the correct answer?!????

ddd [48]

Answer:

B

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4 0

4 years ago