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2 years ago

10

Isopropanol burns in oxygen to yield carbon dioxide, water, and heat. Which word equation best represents this chemical reaction

?

2 answers:

melomori [17]2 years ago

8 0

C3H8O + O2 → CO2 + H2O

emmainna [20.7K]2 years ago

7 0

Answer:

isopropanol + oxygen (arrow) carbon dioxide + water + heat

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The specific heat of aluminum is s 9.00 J/g C how much energy is needed to raise the temperature of 20.0 g block of aluminum fro

Nimfa-mama [501]

E=MC(delta)T
=20.0g x 9.00J/g x (94.4-22.8) C
= 12,924.0 J

7 0

2 years ago

What’s the number 9.1x10^3 in standard form

mafiozo [28]

Answer:

.0091

Explanation:

3 0

3 years ago

What is the volume of 14.0g of nitrogen gas at STP?

lozanna [386]

Answer:

<em>The volume of 14.0 g of nitrogen gas at STP is </em><u><em>11.2 liter.</em></u>

Explanation:

STP stands for standard pressure and temperature.

The International Institute of of Pure and Applied Chemistry, IUPAC changed the definition of standard temperature and pressure (STP) in 1982:

Before the change, STP was defined as a temperature of 273.15 K and an absolute pressure of exactly 1 atm (101.325 kPa).

After the change, STP is defined as a temperature of 273.15 K and an absolute pressure of exactly 105 Pa (100 kPa, 1 bar).

Using the ideal gas equation of state, PV = nRT you can calculate the volume of one mole (n = 1) of gas. With the former definition, the volume of a mol of gas at STP, rounded to 3 significant figures, was 22.4 liter. This is classical well known result.

With the later definition, the volume of a mol of gas at STP is 22.7 liter.

I will use the traditional measure of 22.4 liter per mole of gas.

<u>1) Convert 14.0 g of nitrogen gas to number of moles:</u>

n = mass in grams / molar mass

Atomic mass of nitrogen: 14.0 g/mol

Nitrogen gas is a diatomic molecule, so the molar mass of nitrogen gas = molar mass of N₂ = 14.0 × 2 g/mol = 28.0 g/mol

n = 14.0 g / 28.0 g/mol = 0.500 mol

<u>2) Set a proportion to calculate the volume of nitrogen gas:</u>

22.4 liter / mol = x / 0.500 mol

Solve for x: x = 0.500 mol × 22.4 liter / mol = 11.2 liter.

<u>Conclusion:</u> the volume of 14.0 g of nitrogen gas at STP is 11.2 liter.

6 0

3 years ago

What two general products do combustion reactions create? *

Dimas [21]

It would be carbon dioxide and water.

hope this helps you

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2 years ago

How many moles of gas X are present if the gas has a volume of 2dm³ at room temperature and pressure? Give your answer to 2 deci

bezimeni [28]

Answer:

Approximately $0.08\; \rm mol$ , assuming that this gas is an ideal gas.

Explanation:

Look up the standard room temperature and pressure: $25\; \rm ^{\circ}C$ and $P = 101.325 \; \rm kPa$ .

The question states that the volume of this gas is $V = 2\; \rm dm^{3}$ .

Convert the unit of all three measures to standard units:

$\begin{aligned} T &= 25\; \rm ^{\circ}C \\ &= (25 + 273.15)\; \rm K \\ &= 293.15\; \rm K\end{aligned}$ .

$\begin{aligned}P &= 101.325\; \rm kPa \\ &= 101.325 \; \rm kPa \times \frac{10^{3}\; \rm Pa}{1\; \rm kPa} \\ &= 1.01325 \times 10^{5}\; \rm Pa\end{aligned}$ .

$\begin{aligned}V &= 2\; \rm dm^{3} \\ &= 2 \; \rm dm^{3} \times \frac{1\; \rm m^{3}}{10^{3}\; \rm dm^{3}} \\ &= 2 \times 10^{-3}\; \rm m^{3}\end{aligned}$ .

Look up the ideal gas constant in the corresponding units: $R \approx 8.31\; \rm m^{3}\cdot Pa \cdot mol^{-1} \cdot K^{-1}$ .

Let $n$ denote the number of moles of this gas in that $V = 2\; \rm dm^{3}$ . By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:

$P \cdot V = n \cdot R \cdot T$ .

Rearrange this equation and solve for $n$ :

$\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &\approx \frac{1.01325 \times 10^{5}\; {\rm Pa} \times 2 \times 10^{-3}\; {\rm m^{3}}}{8.31 \; {\rm m^{3} \cdot Pa \cdot mol^{-1} \cdot K^{-1}} \times 293.15\; {\rm K}} \\ &\approx 0.08\; \rm mol\end{aligned}$ .

In other words, there is approximately $2\; \rm mol$ of this gas in that $V = 2\; \rm dm^{3}$ .

6 0

2 years ago