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ozzi
2 years ago
15

Gold is one of the densest substances known, with a density of 19.3 g/cm3. If the gold in the crown was mixed with a less-valuab

le metal like bronze or copper, how would that affect its density?
Chemistry
1 answer:
Andrews [41]2 years ago
4 0

Answer:

Explanation:

Density of gold is 19.3 g / cm³

Density of copper is 8.96 g / cm³

Density of bronze is 8.7  g / cm³

Hence when the gold and copper or bronze are mixed , the density of gold will be reduced due to less density of copper and bronze in comparison to that of gold.

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What is the total amount of heat required to completely melt 347 grams of ice at its melting point
jolli1 [7]
In order to calculate the amount of heat energy required to melt 347 grams of ice, we need to will apply the equation:
Q = ml, where m is the mass of substance and l is its latent heat of fusion. For ice, the latent heat of fusion is 334 joules per gram. Therefore:
Q = 347 x 334
Q = 112,558 joules of energy
8 0
2 years ago
Read 2 more answers
The combustion of ethyne, shown below unbalance, produces heat which can be used to weld metals:
Andreyy89

Answer:

3.69 g

Explanation:

Given that:

The mass m = 325 g

The change in temperature ΔT = ( 1540 - 165)° C

= 1375 ° C

Heat capacity c_p = 0.490 J/g°C

The amount of heat required:

q = mcΔT

q =  325 × 0.490 × 1375

q = 218968.75 J

q = 218.97 kJ

The equation for the reaction is expressed as:

C_2H_{2(g)} + 5O_{2(g)} \to 2CO_{2(g)} + H_2O_{(g)}   \ \   \ \ \  \Delta H^o_{reaction} = -1544 \ kJ

Then,

1 mole of the ethyne is equal to 26 g of ethyne required for 1544 kJ heat.

Thus, for 218.97 kJ, the amount of ethyne gas required will be:

= \dfrac{26 \ g}{1544 \ kJ} \times 218.97 \ kJ

= 3.69 g

3 0
2 years ago
Calculate the cell potential E at 25°C for the reaction 2 Al(s) + 3 Fe2+(aq) → 2 Al3+(aq) + 3 Fe(s) given that [Fe 2+] = 0.020 M
Elodia [21]

Answer:

1.18 V

Explanation:

The given cell is:

Al(s)/Al^{3+}(0.10M)||Fe^{2+}(0.020M)/Fe(s)

Half reactions for the given cell follows:

Oxidation half reaction: Al(s)\rightarrow Al^{3+}(0.10M)+2e^-;E^o_{Al^{3+}/Al}=-1.66V

Reduction half reaction: Fe^{2+}(0.020M)+2e^-\rightarrow Fe(s);E^o_{Fe^{2+}/Fe}=-0.45V

Multiply Oxidation half reaction by 2 and Reduction half reaction by 3

Net reaction: 2Al(s)+3Fe^{2+}(0.020M)\rightarrow 2Al^{3+}(0.10M)+3Fe(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=-0.45-(-1.66)=1.21V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}

where,

E_{cell} = electrode potential of the cell = ?V

E^o_{cell} = standard electrode potential of the cell = +1.21 V

n = number of electrons exchanged = 6

Putting values in above equation, we get:

E_{cell}=1.21-\frac{0.059}{6}\times \log(\frac{0.10^2}{0.020^3})\\\\E_{cell}=1.18V

5 0
2 years ago
If 500.0 mL of 0.10 M Ca2+ is mixed with 500.0 mL of 0.10 M SO42−, what mass of calcium sulfate will precipitate? Ksp for CaSO4
statuscvo [17]

Answer:

The mass of calcium sulfate that will precipitate is 6.14 grams

Explanation:

<u>Step 1:</u> Data given

500.0 mL of 0.10 M Ca^2+ is mixed with 500.0 mL of 0.10 M SO4^2−

Ksp for CaSO4 is 2.40*10^−5

<u>Step 2:</u> Calculate moles of Ca^2+

Moles of Ca^2+ = Molarity Ca^2+ * volume

Moles of Ca^2+ = 0.10 * 0.500 L

Moles Ca^2+ = 0.05 moles

<u>Step 3: </u>Calculate moles of SO4^2-

Moles of SO4^2- = 0.10 * 0.500 L

Moles SO4^2- = 0.05 moles

<u>Step 4: </u>Calculate total volume

500.0 mL + 500.0 mL = 1000 mL = 1L

<u>Step 5: </u> Calculate Q

Q = [Ca2+] [SO42-]  

[Ca2+]= 0.050 M   [O42-]

Qsp = (0.050)(0.050 )=0.0025 >> Ksp

This means precipitation will occur

<u> Step 6:</u> Calculate molar solubility

Ksp = 2.40 * 10^-5 = [Ca2+][SO42-] =(x)(x)

2.40 * 10^-5 = x²

x = √(2.40 * 10^-5)

x = 0.0049 M = Molar solubility

<u> Step 7:</u> Calculate total CaSO4 dissolved

total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 mol/L = 0.667 g

<u>Step 8:</u> Calculate initial mass of CaSO4

Since initial moles CaSo4 = 0.050

Initial mass of CaSO4 = 0.050 * 136.14 g/mol

Initial mass of CaSO4 = 6.807 grams

<u>Step 9:</u> Calculate mass precipitate

6.807 - 0.667 = 6.14 grams

The mass of calcium sulfate that will precipitate is 6.14 grams

5 0
2 years ago
How many moles of KCl will be<br> produced from 15.0 g of KClO3?
slega [8]

Answer:

0.12 mol KCl

Explanation:

2 KClO3 (s)   2 KCl (s) + 3 O2 (g)

15 g                x mol

x g KCl = 15 g KClO3 x[ (1 mol KClO3)/ (122.5 g KClO3) ] x [(2 mol KCl)/ (2 mol KClO3)]

x g KCl = 0.12 mol KCl

7 0
2 years ago
Read 2 more answers
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