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3 years ago

Which of the following are pure substances? (Select all that apply.)

1 answer:

riadik2000 [5.3K]3 years ago

6 0

Elements and compounds are both examples of pure substances. Compounds are substances that are made up of more than one type of atom. Elements are the simplest substances made up of only one type of atom. i hope that helps:)

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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?

Nastasia [14]

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.

5 0

3 years ago

If 20.0 g acetylene is allowed to completely react with oxygen, how many grams of o2 are reacted? 2c2h2(g 5o2(g → 4co2(g 2h2o(g

KATRIN_1 [288]

The balanced chemical reaction is written as:

2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)

We are given the amount of acetylene to be used in the reaction. This will be the starting point for the calculations.

20.0g C2H2 ( 1 mol / 26.04 g) ( 5 mol O2 / 2 mol C2H2 ) ( 32 g / 1 mol ) = 61.44 g O2 is needed

7 0

3 years ago

2NaOH + H2SO4 --> Na2SO4 + 2H2O

iren2701 [21]

Answer:

Sodium hydroxide, a base, reacts with sulfuric acid to form sodium sulfate (complete neutralization) and water. If you write the balanced equation, you will see that 2 moles of sodium hydroxide react with 1 mole of sulfuric acid to form 1 mole of sodium sulfate and 2 moles of water. Therefore it would take .25 moles of sulfuric acid to react with .5 moles of sodium hydroxide.

Explanation:

4 0

3 years ago

Read 2 more answers

Section I: Experimental Overview

Dafna11 [192]

Answer:

Give me a minutes

Explanation:

7 0

2 years ago

Determine the volume of hydrogen collected at STP and SATP, and the molar volume at STP and SATP

mario62 [17]

The other given data are:
Room temperature (°C) 22.0; pressure (kPa) 100.5; Water vapor pressure at 22 °C (2.60 kPa); Mass of Mg ribbon (0.05g); Volume of hydrogen gas (mL) 48.3

Vstp = 22.4 L/mol, and Vsatp = 24.8 L/mol Vstp:0 degrees Celsius and 101.325 kPa while Vsatp: 25 degrees Celsius and 100 kPa. 100.5 kPa x 760 mm Hg/101.325 kP = 753.8119911 mmHg

2HCl(g) + Mg(s) -> H2(g) + MgCl2(aq) nMg = 0.0020571899 mol nH2 = 0.0020571899 mol Wet H2 pressure = 100.5 kPa Dry H2 pressure = 97.9 kPa

So to determine the volume of hydrogen gas collected at STP: V2 = (753.8119911 mmHg)(48.3 mL)(273 K) / (295 K)(760 mmHg) = 44.33403003 mL.

Molar volume at STP: mol/L = 44.33403003 mL / 0.0020571899 mol
= 21550.77176 mL/mol
= 21.55077176 L/mol
To determine volume of hydrogen gas collected at SATP:
V2 = (100.5 kPa)(48.3 mL)(298 K) / (295K)(100kPa) = 49.03514237 mL

Molar volume at SATP: mol/L = 49.03514237 mL / 0.0020571899 mol
= 23835.98246 mL/mol
= 23.8359824 L/mol

4 0

3 years ago