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2 years ago

How many moles are in 45.8 g of calcium nitrate, Ca(NO3)2? PLEASE HELP

Chemistry

1 answer:

VLD [36.1K]2 years ago

6 0

Answer:

the number of moles is 0.279

Explanation:

The computation of the number of moles of calcium nitrate, is shown below;

We know that

Number of moles = mass of substance ÷ molecular weight

= 45.8 g ÷ 164.088

= 0.279moles

hence, the number of moles is 0.279

We simply applied the above formula so that the correct value could come

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3 years ago

Mount Everest rises to a height of 8,850 m above sea level. At a base camp on the mountain the atmospheric pressure is measured

yan [13]

Answer:

Water will boil at $76^{0}\textrm{C}$ .

Explanation:

According to clausius-clapeyron equation for liquid-vapor equilibrium:

$ln(\frac{P_{2}}{P_{1}})=\frac{-\Delta H_{vap}^{0}}{R}[\frac{1}{T_{2}}-\frac{1}{T_{1}}]$

where, $P_{1}$ and $P_{2}$ are vapor pressures of liquid at $T_{1}$ (in kelvin) and $T_{2}$ (in kelvin) temperatures respectively.

Here, $P_{1}$ = 760.0 mm Hg, $T_{1}$ = 373 K, $P_{2}$ = 314.0 mm Hg

Plug-in all the given values in the above equation:

$ln(\frac{314.0}{760.0})=\frac{-40.7\times 10^{3}\frac{J}{mol}}{8.314\frac{J}{mol.K}}\times [\frac{1}{T_{2}}-\frac{1}{373K}]$

or, $T_{2}=349 K$

So, $T_{2}=349K=(349-273)^{0}\textrm{C}=76^{0}\textrm{C}$

Hence, at base camp, water will boil at $76^{0}\textrm{C}$

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3 years ago

In the laboratory a student uses a "coffee cup" calorimeter to determine the specific heat of a metal. She heats 19.6 grams of z

storchak [24]

Answer:

The specific heat of zinc is 0.375 J/g°C

Explanation:

Step 1: Data given

Mass of zinc = 19.6 grams

Mass of water = 82.9 grams

Initial temperature of zinc T1= 98.37 °C

Initial temperature of water T1= 24.16 °C

Final temperature of water (and zinc) T2 = 25.70 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculate Specific heat of zinc

Q=m*c*ΔT

Qzinc = -Qwater

m(zinc)*C(zinc)*ΔT(zinc) = -m(water)*C(water)*ΔT(water)

⇒ with mass of water = 82.9 grams

⇒ with C(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2 - T1 = 25.70 - 24.16 = 1.54

⇒ with mass of zinc = 19.6 grams

⇒ with C(zinc) = TO BE DETERMINED

⇒ with ΔT(zinc) = T2 -T1 = 25.70 - 98.37 = -72.67°C

Qzinc = -Qwater

m(zinc)*c(zinc)* ΔT(zinc) = - m(water)*c(water)* ΔT(water)

19.6g* C(zinc) * (-72.67°C) = - 82.9g* 4.184 J/g°C * 1.54 °C

-1424.332*C(zinc) = -534.155

C(zinc) = 0.375 J/g°C

The specific heat of zinc is 0.375 J/g°C

4 0

3 years ago