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Lunna [17]
3 years ago
11

A pound of margarine costs $1.39. What is the cost of a kilogram of margarine

Chemistry
1 answer:
den301095 [7]3 years ago
8 0

Answer:

              3.064 $

Solution:

First of all we will find the number of pounds contained by one Kg. Therefore, it is found that,

                                    1 Kilogram  = 2.20462 Pounds

Hence, as given,

                        1 Pound of Margarine costs  =  1.39 Dollars

So,

                  2.20462 Pounds Margarine will cost  =  X Dollars

Solving for X,

                     X  =  (2.20462 Pounds × 1.39 Dollars) ÷ 1 Pound

                     X =  3.064 Dollars

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Based on these data, what is the value of the formation constant, Kf, of [Cu(NH3)4]2+? [Cu 2+ ]=6.47x10 -15 kf=
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A sample of an unknown gas at STP has a density of 0.630 gram per liter. What is the gram molecular mass of this gas?
jeyben [28]
<span>STP means standard temperature and pressure at 0°C (273K) and 1 atm (atmosphere). The density of the unknown gas is 0.63 gram per liter. The deal gas equation is PV = nRT. The n is the numer of moles and can be represented as mass of the gas, m, divided by the molar mass, c.  so we have,</span>  

PV = nRT
PV = (m/c)RT
Since the density is d = m/V
Pc = (m/V)RT
Pc = dRT
c = drT/P  

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3 0
3 years ago
What is the new pressure of 150 mL of a gas that is compressed to 50 mL when the original pressure was 2.0 atm and the temperatu
tatyana61 [14]

Answer:

1) 6.0 atm.

2) 2.066 atm.

Explanation:

  • From the general law of ideal gases:

<em>PV = nRT.</em>

where, P is the pressure of the gas.

V is the volume of the container.

n is the no. of moles of the gas.

R is the general gas constant.

T is the temperature of the gas (K).

<em>1) What is the new pressure of 150 mL of a gas that is compressed to 50 mL when the original pressure was 2.0 atm and the temperature is held constant?</em>

  • At constant T and at two different (P, and V):

<em>P₁V₁ = P₂V₂.</em>

P₁ = 2.0 atm, V₁ = 150.0 mL.

P₂ = ??? atm, V₂ = 50.0 mL.

<em>∴ P₂ = P₁V₁/V₂</em> = (2.0 atm)(150.0 mL)/(50.0 mL) = <em>6.0 atm.</em>

<em>2. A sample of a gas in a rigid container at 30.0°C and 2.00 atm has its temperature increased to 40.0°C. What will be the new pressure?</em>

<em></em>

  • Since the container is rigid, so it has constant V.
  • At constant V and at two different (P, and T):

<em>P₁/T₁ = P₂/T₂.</em>

P₁ = 2.0 atm, T₁ = 30.0°C + 273 = 303 K.

P₂ = ??? atm, T₂ = 40.0°C + 273 = 313 K.

<em>∴ P₂ = P₁T₂/T₁ </em>= (2.0 atm)(313.0 K)/(303.0 K) =<em> 2.066 atm.</em>

3 0
3 years ago
Consider the following reaction: SO2Cl2(g)⇌SO2(g)+Cl2(g) A reaction mixture is made containing an initial [SO2Cl2] of 2.2×10−2M
lukranit [14]
I think you want to ask about Keq. At equilibrium, we can know [SO2Cl2] is 2.2*10-2 M -1.3*10-2M=9*10^-3 M. And [SO2]=[Cl2]. So the Keq=1.88*10^-2.
5 0
3 years ago
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