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kap26 [50]
3 years ago
5

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

Chemistry
1 answer:
algol133 years ago
4 0

Answer:

Whether the mixture can be separated

You might be interested in
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
Read 2 more answers
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
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