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3 years ago

Convert 7.02 mol of NaCl to formula units??

1 answer:

8 0

<span>To convert the number of moles a compound to formula units, the conversion factor is Avogadro's number, 6.022 x10^23 formula units/ mol. In this case, we are given with 7.02 moles of sodium chloride. Hence the answer is 4.23 x 10^24 formula units.</span>

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Indicate how the concentration of each species in the chemical equation will change to reestablish equilibrium after increasing

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Aloiza [94]

The correct answer is C

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18. How many milliliters of water at 23 °C with a density of 1.00 g/mL must be mixed with 180 mL (about 6 oz) of coffee at 95 °C

sveticcg [70]

Answer:

The number of water that must be mixed in the solution is 170.27 mL

Explanation:

given information :

Temperature of water ( $T_{1}$ ) = $23^{o}C$

density of water (ρ) = 1.00 g/mL

Temperature of coffee ( $T_{1}$ ) = $95^{o}C$

volume of coffee ( $V_{2}$ ) = 180 mL

Mixed Temperature ( $T_{mix}$ ) = $60^{o}C$

to calculate the heat, w use the formula :

Q = m x c x ΔT

where

m = mass of the substance (g)

c = specific heat (J/ $g^{o}C$ )

ΔT = the temperature change ( $^{o}C$ )

in the mixture solution, the heat of the water ( $Q_{1}$ ) should be the same as the heat of coffee ( $Q_{2}$ ). Thus,

$Q_{1}$ = $Q_{2}$

$m_{1}$ x $c_{1}$ x Δ $T_{1}$ = $m_{2}$ x $c_{2}$ x Δ $T_{2}$

where

$m_{1}$ is the mass of water

$m_{2}$ is the mass of coffee

$c_{1}$ is the specific heat of water

$c_{2}$ is the specific heat of coffee

Assume coffee and water have the same specific heat. So,

$c_{1}$ = $c_{2}$ , we can remove it from the equation.

Hence.

$m_{1}$ x Δ $T_{1}$ = $m_{2}$ x Δ $T_{2}$

we know that

ρ = $\frac{m}{V}$

m = ρ x V, subtitute it to the equation:

ρ $_{1}$ x V $_{1}$ x Δ $T_{1}$ = ρ $_{2}$ x V $_{2}$ x Δ $T_{2}$

$V_{1}$ is the volume of water

coffee and water have the same density, so we can remove the formula

V $_{1}$ x Δ $T_{1}$ = V $_{2}$ x Δ $T_{2}$

V $_{1}$ = (V $_{2}$ x Δ $T_{2}$ ) / Δ $T_{1}$

V $_{1}$ = V $_{2}$ x ( $\frac{(T_{2} - T_{mix}) }{(T_{mix} - T_{1})}$

V $_{1}$ = 180 mL x $\frac{(95-60)^{o}C}{(60-23)^{o}C}$

V $_{1}$ = 180 mL x $\frac{(35)^{o}C}{(37)^{o}C}$

V $_{1}$ = 170.27 mL

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