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3 years ago

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Chemistry

1 answer:

coldgirl [10]3 years ago

4 0

Answer:

0.895 g/mL

Explanation:

Step 1: Given data

Mass of liquid Z (m): 2.763 lb

Volume of liquid Z (V): 5.93 cups

Step 2: Convert "m" to grams

We will use the relationship 1 lb = 453.59 g.

$2.763 lb \times \frac{453.59 g}{1lb} = 1,253g$

Step 3: Convert "V" to milliliters

We will use the relationship 1 cup = 236.59 mL.

$5.93cup \times \frac{236.59mL}{1cup} = 1.40 \times 10^{3} mL$

Step 4: Calculate the density of the liquid Z

The density (ρ) of the liquid Z is equal to its mass divided by its volume.

$\rho = \frac{m}{V} = \frac{1,253g}{1.40 \times 10^{3}mL } = 0.895 g/mL$

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Q #4 please help if you are sure

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Nuclear decay rate refers to the constant ratio of the number of atoms of radioactive nucleus that decay during a certain interval of time to the total number of radioactive atoms at the beginning of the time interval. Nuclear decay rates are constant and do not vary with the conditions of the reaction.

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A weighed piece of magnesium ribbon is added to a dried crucible, which is reweighed and heated in air to form the compound MgO.

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Answer:

One of the errors for low percentage of magnesium could be because not all the magnesium may have reacted.

Explanation:

During the heating process, if the magnesium have not reacted completely, it can lead to low percentage of magnesium in the oxide formed. The product may still look a bit greyish rather than whitish after the heating process.

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3 years ago

A gas mixture with 4 mol of Ar, x moles of Ne, and y moles

maks197457 [2]

Answer:

a) $\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

b) $x=4$

c) $\Delta G_{max}=-2721.9 J/mol$

Explanation:

Gas mixture:

$n_{Ar}= 4 mol$

$n_{Ne}= x mol$

$n_{Xe}= y mol$

$n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}$

$n_{Ne} + n_{Xe}=2*n_{Ar}$

$x + y=8 mol$

$y=8 mol- x$

Mol fractions:

$x_{Ar}=\frac{4 mol}{12 mol}=1/3$

$x_{Ne}=\frac{x mol}{12 mol}=x/12$

$x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12$

Expression of $\Delta G_{mixing}$

$\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)$

$\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]$

$\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

Expression of $\Delta G_{max}$

$\frac{d \Delta G_{mixing}}{dx}=0$

$\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]$

$0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)$

$0=[ln (x/12)-ln ((8-x)/12)$

$ln (x/12)=ln ((8-x)/12)$

$x=(8-x)$

$x=4$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]$

$\Delta G_{max}=-2721.9 J/mol$

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3 years ago

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Dafna1 [17]

Answer:

The answer to your question is Q = 355.64 J

Explanation:

Data

Heat = Q = ?

Temperature 1 = T1 = 20°C

Temperature 2 = T2 = 37°C

mass = m = 5 g

Specific heat = Cp = 4.184 J/g°C

Formula

Q = mCp(T2 - T1)

-Substitution

Q = (5)(4.184)(37 - 20)

-Simplification

Q = (5)(4.184)(17)

-Result

Q = 355.64 J

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