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ziro4ka [17]
3 years ago
10

A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?

Chemistry
1 answer:
Nastasia [14]3 years ago
5 0

Answer:

At -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

                                          \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

where P_{1} and P_{2} are initial and final pressure respectively.

           V_{1}  and V_{2} are initial and final volume respectively.

           T_{1} and T_{2} are initial and final temperature in kelvin scale respectively.

Here P_{1}=150.0kPa , V_{1}=1.75L , T_{1}=(273-23)K=250K, P_{2}=210.0kPa and V_{2}=1.30L

Hence    T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}

            \Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}

            \Rightarrow T_{2}=260K

            \Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}

So at -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

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An iron nail rusts when exposed to oxygen. According to the following reaction, how many moles of iron(III) oxide will be formed
Zigmanuir [339]

Answer:

0.453 moles

Explanation:

The balanced equation for the reaction is:

2Fe(s) + 3O2(g)  ==>  2Fe2O3

From the equation,  mass of O2 involved = 16 x 2 x 3 = 96g

                                 mass of Fe2O3 involved = [(2x26) + 3 x 16] x 2

                                                                            = 100g

                Therefore 96g of O2 produced 100g of Fe2O3

                                  32.2g of O2 Will produce   100x32.2/96

                                                   = 33.54g of Fe2O3

Converting it to mole using   number of mole = mass/molar mass

but molar mass of Fe2O3 = 26 + (16 X 3)

                                           = 74g/mole

Therefore number of mole of 33.54g of Fe2O3 = 33.54/74

                                                                           = 0.453 moles

5 0
3 years ago
Based on the diagram below, how much of the excess reactant is left over? *
Alexxandr [17]

Answer:

3 of lunchmeat and 2 slices of cheese

Explanation:

From the question given,

Each sandwich contains:

2 bread + 3 lunch meat + 1 cheese.

Now, the limiting reactant is the slice of bread.

We can determine the leftover as follow:

1. For the lunchmeat.

From the simple equation above,

2 bread requires 3 lunchmeat.

Therefore, 6 bread will require = (6 x 3)/2 = 9 lunchmeat.

Lunchmeat given = 12

Lunchmeat required = 9

Leftover lunchmeat = 12 – 9 = 3

Therefore, 3 lunchmeat is leftover.

2. For cheese.

From the simple equation above,

2 bread requires 1 cheese.

Therefore, 6 bread will require = 6/2 = 3 cheese.

Cheese given = 5

Cheese required = 3

Leftover cheese = 5 – 3 = 2

Therefore, 2 cheese is leftover.

From the simple illustrations above,

3 lunchmeat and 2 cheese are leftover.

7 0
3 years ago
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