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Setler79 [48]
2 years ago
15

A blacksmith heated an iron bar to 1445 °C. The blacksmith then tempered the metal by dropping it into 42,800 mL of

Chemistry
1 answer:
Wittaler [7]2 years ago
8 0

Answer:

6626 g

Explanation:

Given that:

Density of water = 1.00 g/ml, volume of water = 42800 ml.

Since density = mass/ volume

mass of water = volume of water * density of water = 42800 ml * 1 g/ml = 42800 g

Initial temperature of water = 22°C and final temperature of water = 45°C.

specific heat capacity for water = 4.184 J/g°C

ΔT water = 45 - 22 = 23°C

For iron:

mass = m,  

specific heat capacity for iron  = 0.444 J/g°C

Initial temperature of iron = 1445°C and final temperature of water = 45°C.

ΔT iron = 45 - 1445 = -1400°C

Quantity of heat (Q) to raised the temperature of a body is given as:

Q = mCΔT

The quantity of heat required to raise the temperature of water is equal to the temperature loss by the iron.

Q water (gain) + Q iron (loss) = 0

Q water = - Q iron

42800 g ×  4.184 J/g°C × 23°C = -m × 0.444 J/g°C × -1400°C

m = 4118729.6/621.6

m = 6626 g

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Suppose a current of is passed through an electroplating cell with an aqueous solution of agno3 in the cathode compartment for s
WARRIOR [948]

The given question is incomplete. The complete question is:

Suppose a current of 0.920 A is passed through an electroplating cell with an aqueous solution of agno3 in the cathode compartment for 47.0 seconds. Calculate the mass of pure silver deposited on a metal object made into the cathode of the cell.

Answer:   0.0484 g

Explanation:

Q=I\times t

where Q= quantity of electricity in coloumbs

I = current in amperes = 0.920 A

t= time in seconds = 47.0 sec

Q=0.920A\times 47.0s=43.24C

AgNO_3\rightarrow Ag^++NO_3^-

Ag^++e^-\rightarrow Ag

96500 Coloumb of electricity electrolyzes 1 mole of Ag

43.24 C of electricity deposits =\frac{1}{96500}\times 43.24=0.00045moles of Ag

\text{ mass of Ag}={\text{no of moles}\times {\text{Molar mass}}=0.00045mol\times 108g=0.0484g

Thus the mass of pure silver deposited on a metal object made into the cathode of the cell is 0.0484 g

6 0
2 years ago
A 500 ml aqueous solution of na3po4 was prepared using 82g of the solute. what is the molarity​
liraira [26]

Answer:

1 M

Explanation:

The molarity of a solution, M, is a measure of the concentration of that solution and it refers to the number of moles of solute (mol) per liter (L) of solution. The molarity (M) can be calculated using the formula:

M = number of moles (n) /volume (V)

In this question, a 500 ml aqueous solution of Na3PO4 was prepared using 82g of the solute.

Molar mass of Na3PO4 = 23(3) + 15 + 16(4)

= 69 + 31 + 64

= 164g/mol

Mole = mass/molar mass

mole = 82/164

mole = 0.5 mol

Volume in Litres (L) = 500 ml ÷ 1000 = 0.500L

Therefore, Molarity (M) = 0.5/0.500

Molarity = 1 M or 1 mol/L

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2 years ago
What is common to all elements in the same group on the periodic table
marin [14]
They all have the same number of electrons.
3 0
2 years ago
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PLEASE HELP!!!!!!!
dsp73

Answer: 1. 2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu

2. 3 moles of CuCl_2 : 2 moles of Al

3. 0.33 moles of CuCl_2 : 0.92 moles of Al

4. CuCl_2 is the limiting reagent and Al is the excess reagent.

5. Theoretical yield of AlCl_3 is 29.3 g

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Al=\frac{25.0g}{27g/mol}=0.92moles

\text{Moles of} CuCl_2=\frac{45.0g}{134g/mol}=0.33moles

The balanced chemical equation is:

2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu  

According to stoichiometry :

3 moles of CuCl_2 require = 2 moles of Al

Thus 0.33 moles of CuCl_2 will require=\frac{2}{3}\times 0.33=0.22moles  of Al

Thus CuCl_2 is the limiting reagent as it limits the formation of product and Al is the excess reagent.

As 3 moles of CuCl_2 give = 2 moles of AlCl_3

Thus 0.33 moles of CuCl_2 give =\frac{2}{3}\times 0.33=0.22moles  of AlCl_3

Theoretical yield of AlCl_3=moles\times {\text {Molar mass}}=0.22moles\times 133.34g/mol=29.3

Thus 29.3 g of aluminium chloride is formed.

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Alkanes are hydrocarbons that contain what type of bonds
olga2289 [7]
Carbon carbon triple bonds
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