What are the things to consider in classifying the different kinds of mixture?

Water has a density of 0.997 g/cm^3 at 25 degrees C; ice has a density of 0.917 g/cm^3 at -10 degrees C. (question part a) If a

Luden [163]

Mass of water added:
0.997 x 1500
= 1495.5 grams

a) Volume = mass / density
Volume = 1495.5 / 0.917
Volume = 1630 cm³ = 1.63 L

b) The ice cannot be contained in the bottle as its volume exceeds that of the bottle.

4 0

2 years ago

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what is an empirical formulaWhat is the percent composition by mass of nitrogen in (NH4)2CO3 (gram-formula mass = 96.0 g/mol)?

Margarita [4]

Answer:

Percentage composition = 14.583%

Explanation:

In chemistry, the emprical formular of a compound is the simplest formular a compound can have. It shows the simplest ratio in which the elements are combined in the compound.

Percentage composition by mass of Nitrogen

Nitrogen = 14g/mol

In one mole of the compound;

Mass of Nitrogen = 1 mol * 14g/mol = 14g

Mass of compound = 1 mol * 96.0 g/mol = 96

Percentage composition of Nitrogen = (Mass of Nitrogen / Mass of compound) * 100

percentage composition = 14/96 * 100

Percentage composition = 0.14583 * 100

Percentage composition = 14.583%

3 0

2 years ago

Calculate the [OH-] given pH of 9.9

Juli2301 [7.4K]

Answer

pOH = 4.1

Explanation

<em>Given:</em>

pH = 9.9

<em>Required</em>: The concentration of OH-

Solution

pH + pOH = 14

9.9 + pOH = 14

pOH = 14-9.9

pOH = 4.1

3 0

1 year ago

What volume (in L) will a 32 g sample of butane gas, C4H10(g), occupy at a temperature of 45.0 oC and a pressure of 728 mm Hg?

larisa86 [58]

Answer:

15.0 L

Explanation:

To find the volume, you need to use the Ideal Gas Law:

PV = nRT

In this equation,

-----> P = pressure (mmHg)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas constant (62.36 L*mmHg/mol*K)

-----> T = temperature (K)

To calculate the volume, you need to (1) convert grams C₄H₁₀ to moles (via the molar mass), then (2) convert the temperature from Celsius to Kelvin, and then (3) calculate the volume (via the Ideal Gas Law).

Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)

Molar Mass (C₄H₁₀): 58.124 g/mol

32 grams C₄H₁₀ 1 moles
------------------------- x ----------------------- = 0.551 moles C₄H₁₀
58.124 grams

P = 728 mmHg R = 62.36 L*mmHg/mol*K

V = ? L T = 45.0 °C + 273.15 = 318.15 K

n = 0.551 moles

PV = nRT

(728 mmHg)V = (0.551 moles)(62.36 L*mmHg/mol*K)(318.15 K)

(728 mmHg)V = 10922.7632

V = 15.0 L

6 0

1 year ago

1.How many mL of 0.401 M HI are needed to dissolve 5.97 g of BaCO3?

garri49 [273]

Answer:

The answer to your question is:

1.- volume = 0.151 l or 151 ml

2.- 0.241 l or 241 ml of NaOH

Explanation:

1.-

Data

V = ? HI = 0.401 M

BaCO3 = 5.97 g

2HI(aq) + BaCO3(s) ⇒ BaI2(aq) + H2O(l) + CO2(g)

MW BaCO3 = 137 + 12 + 48 = 197 g

197 g of BaCO3 ----------------- 1 mol

5.97 g ----------------- x

x = (5.97 x 1) /197

x = 0.03 mol of BaCO3

2 moles of HI ---------------- 1 mol of BaCO3

x ---------------- 0.03 mol of BaCO3

x = (0.03 x 2) / 1

x = 0.060 mol of HI

Molarity = moles / volume

volume = moles / molarity

volume = 0.060 / 0.401

volume = 0.151 l or 151 ml

2.-

V = ? NaoH 0.757 M

Co⁺² Volume = 167 ml 0.548 M

CoSO4(aq) + 2NaOH(aq) ⇒ Co(OH)2(s) + Na2SO4(aq)

Moles of Co = Molarity x volume

Moles of Co = 0.548 x 0.167

Moles of Co = 0.092

1 mol of CoSO4 -------------- 2 moles of NaOH

0.092 moles --------------- x

x = (0.092 x 2) /1

x = 0.183 moles of NaOH

Volume of NaOH = moles / molarity

= 0.183 / 0.757

= 0.241 l or 241 ml of NaOH

6 0

3 years ago