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Delicious77 [7]
3 years ago
6

Which of the following statements is not true regarding acids?

Chemistry
2 answers:
Romashka [77]3 years ago
7 0
"Active metals will react with acids in double replacement reactions" is the statement among the following statements that <span>is not true regarding acids. The correct option among all the options that are given in the question is the last option or option "D". I hope that this answer has helped you.</span>
shepuryov [24]3 years ago
3 0
I believe the statement that is not true regarding acids is most likely D. Active metals will react with acids in double replacement reactions. It should be that reactive active metals will react with acids in single replacement reactions.
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“Dry ice” is the solid form of carbon dioxide. Determine the number of mass of CO2 gas in grams that are present in 61.8 L of CO
Fofino [41]

Answer:

Mass  = 121 g

Explanation:

Given data:

Mass in gram of CO₂ = ?

Volume = 61.8 L

Pressure = standard = 1 atm

Temperature = 273.15 K

Solution:

Formula:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

1 atm × 61.8 L = n ×0.0821 atm.L/ mol.K   × 273.15 k

61.8 L.atm = 22.42 atm.L/ mol × n

n = 61.8 L.atm /22.42 atm.L/ mol

n = 2.76 mol

Mass in gram:

Mass =  number of moles × molar mass

Mass = 2.76 mol × 44 g/mol

Mass  = 121 g

6 0
3 years ago
If a reaction is reversible what are the relative amounts of reactant and product
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3 0
3 years ago
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Aneli [31]
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A 1.28-kg sample of water at 10.0 °C is in a calorimeter. You drop a piece of steel with a mass of 0.385 kg at 215 °C into it. A
Kryger [21]

Answer:

T_{2}=16,97^{\circ}C

Explanation:

The specific heats of water and steel are  

Cp_{w}=4.186 \frac{KJ}{Kg^{\circ}C}

Cp_{s}=0.49 \frac{KJ}{Kg^{\circ}C}

Assuming that the water and steel are into an <em>adiabatic calorimeter</em> (there's no heat transferred to the enviroment), the temperature of both is identical when the system gets to the equilibrium T_{2}_{w}= T_{2}_{s}  

An energy balance can be written as

m_{w}\times Cp_{w}\times (T_{2}- T_{1})_{w}= -m_{s}\times Cp_{s}\times (T_{2}- T_{1})_{s}  

Replacing

1.28Kg\times 4.186\frac{KJ}{Kg^{\circ}C}\times (T_{2}-10^{\circ}C)= -0.385Kg\times 0.49 \frac{KJ}{Kg^{\circ}C} \times (T_{2}-215^{\circ}C)

Then, the temperature T_{2}=16,97^{\circ}C

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3 years ago
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vampirchik [111]

Answer:

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