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4 years ago

Ray puts a glass of ice water on the table. One hour later he observes that only water is left in the glass. What happened?

Chemistry

2 answers:

tankabanditka [31]4 years ago

6 0

Answer:

The ice in the glass melted into liquid water (a physical change, known and change of phase)

Explanation:

Heat flows from warmer substances to cooler substances.

Ice water contents ice (solid water) and liquid water.

Ice freezes at 0°C and below.

Ice is the solid state of water; at this stage, the water molecules have low kinetic energy and the remain vibrating around a relatively fixed position in a crystal structure.

The liquid water around the solid ice cubes has a larger kinetic energy than the cubes. Also, the environment is at a higher temperature than the content of the glass.

Hence, both heat from the room and heat from the liquid water will flow toward the cooler ice cubes. The molecules that form the ice cubes will gain kinetic energy, causing the temperature of the ice cubes increase; the molecules will vibrate faster until they can move and slide past each other, which means that they have become liquid (melted).

This is a physical change known as change of phase, specifically melting.

vekshin14 years ago

4 0

Answer:

Ice Water

Explanation:

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Answer:

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Explanation:

3 0

3 years ago

Read 2 more answers

2. Calculate the pl of the following amino acids(use their Pka values) a. Arginine b. Glutamic acid of water an c. Asparagine d.

Savatey [412]

<h2>♨ANSWER♥</h2>

pl (25*C)

Arginine -----> 10.76

Glutamic -----> 3.08

Asparagine -----> 5.43

Tyrosine -----> 5.63

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5 0

2 years ago

Seawater has a ph of 8.1. what is the concentration of oh–?

lara31 [8.8K]

PH scale is used to determine how acidic or basic a solution is.
pH can be calculated as follows;
by knowing the ph we can calculate pOH
pH + pOH = 14
pOH = 14 - 8.1
pOH = 5.9
pOH is used to calculate the hydroxide ion concentration
pOH = -log[OH⁻]
[OH⁻] = antilog(-pOH)
[OH⁻] = 1.26 x 10⁻⁶ M
therefore hydroxide ion concentration is 1.26 x 10⁻⁶ M

4 0

3 years ago

2. Iron reacts with oxygen gas according to the following equation:

erastovalidia [21]

Answer:

2. a) 2.67 mol.

b) 1.33 mol.

3. 4.35 g.

4. 8.67 g.

5. a) 143.86 L.

b) 251.75 L.

Explanation:

2. Iron reacts with oxygen gas according to the following equation:

4Fe + 3O₂ → 2Fe₂O₃ , If 2 moles of oxygen gas is used in the reaction,

Fe reacts with O₂ according to the balanced equation:

4Fe + 3O₂ → 2Fe₂O₃,

It is clear that 4 mole of Fe react with 3 mole of O₂ to produce 2 moles of Fe₂O₃.

a) how many moles of iron, Fe, will be required?

using cross multiplication:

3 mol of O₂ require → 4 mol of Fe, from the stichiometry.

2 mol of O₂ require → ??? mol of Fe.

∴ The no. of moles of of Fe are required = (2 mol)(4mol)/(3 mol) = 2.67 mol.

(b) how many moles of iron(III) oxide, Fe₂O₃, will be produced?

using cross multiplication:

3 mol of O₂ produce → 2 mol of Fe₂O₃, from the stichiometry.

2 mol of O₂ require → ??? mol of Fe₂O₃.

∴ The no. of moles of of Fe₂O₃ are produced = (2 mol)(2 mol)/(3 mol) = 1.33 mol.

3. Potassium sulfate can be prepared by the reaction between dilute sulfuric acid and potassium carbonate.

H₂SO₄ + K₂CO₃ → K₂SO₄ + CO₂ + H₂O,

Calculate the mass of potassium sulfate that can be prepared from 3.45 g of potassium carbonate.

H₂SO₄ reacts with K₂CO₃ according to the balanced equation:

H₂SO₄ + K₂CO₃ → K₂SO₄ + CO₂ + H₂O,

It is clear that 1 mole of H₂SO₄ reacts with 1 mole of K₂CO₃ to produce 1 mole of K₂SO₄, 1 mole of CO₂ and 1 mole of H₂O.

Firstly, we need to calculate the no. of moles of 3.45 g of K₂CO₃:

no. of moles of K₂CO₃ = mass/molar mass = (3.45 g)/(138.205 g/mol) = 0.025 mol.

using cross multiplication:

1 mol of K₂CO₃ produce → 1 mol of K₂SO₄, from the stichiometry.

∴ 0.025 mol of K₂CO₃ produce → 0.025 mol of K₂SO₄.

∴ The mass of K₂SO₄ are produced = (no. of moles of K₂SO₄ produced)(molar mass of K₂SO₄) = (0.025 mol)(174.259 g/mol) = 4.35 g.

4. The reaction between zinc and aqueous chromium(III) nitrate can be represented by the following equation:

3Zn(s) + 2Cr(NO₃)₃ → 3Zn(NO₃)₂ + 2Cr

If 16.25 g of zinc is used to react with chromium(III) nitrate, calculate the mass of chromium that will be produced.

Zn reacts with Cr(NO₃)₃ according to the balanced equation:

3Zn(s) + 2Cr(NO₃)₃ → 3Zn(NO₃)₂ + 2Cr ,

It is clear that 3 mole of Zn reacts with 2 mole of Cr(NO₃)₃ to produce 3 mole of Zn(NO₃)₂ and 2 mole of Cr.

Firstly, we need to calculate the no. of moles of 16.25 g of Zn:

no. of moles of Zn = mass/atomic mass = (16.25 g)/(65.38 g/mol) = 0.25 mol.

using cross multiplication:

3 mol of Zn produce → 2 mol of Cr, from the stichiometry.

∴ 0.25 mol of Zn produce → ??? mol of Cr.

∴ The no. of moles of Cr are produced = (2 mol)(0.25 mol)/(3 mol) = 0.167 mol.

∴ The mass of Cr are produced = (no. of moles of Cr produced)(atomic mass of Cr) = (0.167 mol)(51.9961 g/mol) = 8.67 g.

5. Ethane, C₂H₆, burns in oxygen gas according to the following equation:

2 C₂H₆ + 7 O₂ → 4 CO₂ + 6H₂O,

If 72 dm³ of ethane gas is completely burnt in oxygen, calculate

(a) the volume of carbon dioxide, measured at room temperature and

pressure produced.

Firstly, we can calculate the no. of moles of 72 dm³ ethane at room temperature and pressure using the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 1.0 atm).

V is the volume of the gas in L (V = 72.0 dm³ = 72.0 L).

n is the no. of moles of the gas in mol (n = ??? mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 298.0 K, room temperature).

∴ n of ethane = PV/RT = (1.0 atm)(72.0 L)/(0.0821 L.atm/mol.K)(298.0 K) = 2.94 mol.

So, we can calculate the no. of moles of CO₂:

using cross multiplication:

2 mol of C₂H₆ produce → 4 mol of CO₂, from the stichiometry.

∴ 2.94 mol of C₂H₆ produce → ??? mol of CO₂.

∴ The no. of moles of CO₂ are produced = (2.94 mol)(4.0 mol)/(2 mol) = 5.88 mol.

∴ The volume of moles of CO₂ are produced = nRT/P = (5.88 mol)(0.0821 L.atm/mol.K)(298.0 K)/(1.0 atm) = 143.86 L.

(b) the volume of oxygen, measured at room temperature and pressure

required.

using cross multiplication:

2 mol of C₂H₆ require → 7 mol of O₂, from the stichiometry.

∴ 2.94 mol of C₂H₆ require → ??? mol of O₂.

∴ The no. of moles of O₂ are required = (2.94 mol)(7.0 mol)/(2 mol) = 10.29 mol.

∴ The volume of moles of O₂ are produced = nRT/P = (10.29 mol)(0.0821 L.atm/mol.K)(298.0 K)/(1.0 atm) = 251.75 L.

8 0

3 years ago

A chemist must prepare 550.0 ml of hydrochloric acid solution with a pH of 1.60 at 25°C . He will do this in three steps: Fill a

horsena [70]

Answer:

Vol concentrated HCl needed = 1.73 ml

Explanation:

Prep 550 ml of HCl(aq) solution with pH = 1.60 at 25°C using 8M stock concentrate.

[HCl] = [H⁺] = 10⁻¹°⁶⁰M = 0.0251M HCl(aq)

Molarity Concentrate x Volume Concentrate = Molarity Dilute x Volume Dilute

8M x Vol Conc. = 0.0251M x 550ml

Vol Conc needed = 0.0251M x 550 ml / 8M = 1.73 ml of concentrate

Mixing => to the quantity of water in the mixing flask transfer 1.73 ml of the 8M HCl concentrate solution and dilute up to but not to exceed the calibration mark on the mixing flask.

4 0

3 years ago