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3 years ago

Explain step by step, please!

Chemistry

1 answer:

konstantin123 [22]3 years ago

4 0

Answer:

10. Stacey

11. 7.86 g/cm^3 (3dp)

OR 786 kg/m^3 (SI units)

Explanation:

10. The correct exact answer for 10 is 26.169 (6.71x3.9). While Sam was correct in the precision of the number (2dp just like the given measurements), he was incorrect in the number itself. While Stacy's number was not as precise as Sam's (0dp), it is correctly rounded to the nearest whole number, therefore Stacy is correct.

11. The formula for density is mass/volume. The mass here is 264g and the volume is 33.6ml, therefore the density is 7.86 (3 sig figs because given values had 3) g/cm^3, because the mass was given in grams and 1ml = 1cm^3. Converted to SI units it is 786 kg/m^3.

Hope this helped!

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describe how knowledge of the periodic table would be important in three different careers, based on what you’ve read.

ANEK [815]

The knowledge of periodic table would be important in these three different careers such as medicine, pharmacology and engineering.

<h3>What is the importance of periodic table in following careers?</h3>

Periodic table is organised the whole elements vertically in groups and horizontally in periods with increasing atomic number.

The knowledge of arrangement of these elements serves alot of purposes in the following three different careers:

Medicine: The element, iron is an important metal in the periodic table used in treatment of low blood levels.

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brainly.com/question/15987580

5 0

3 years ago

A 0.879 g sample of a CaCl2 ∙ 2 H2O / K2C2O4 ∙ H2O solid salt mixture is dissolved in 150 mL of deionized water. A precipitate f

Alecsey [184]

Answer:

a. CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)

b. Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted = 0.326 g

e. Moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted = 0.408 g

g. mass of K₂C₂O₄.H₂O remaining unreacted = 0.145 g

h. Percent by mass CaCl₂.2 H₂O = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 62.9%

Explanation:

a. Molecular equation of the reaction is given below :

CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)

b. The net ionic equation is given below

Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. mass CaC₂O₄ produced = 0.284 g, molar mass of CaC₂O₄ = 128 g/mol

moles CaC₂O₄ produced = 0.284 g / 128 g/mol = 0.00222 moles

Mole ratio of CaC₂O₄ and CaCl₂.2 H₂O is 1 : 1, therefore moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted in the mixture = number of moles × molar mass

Molar mass of CaCl₂.2 H₂O = 147 g/mol

Mass of CaCl₂.2 H₂O reacted = 0.00222 moles × 147 g/mol = 0.326 g

e. Mole ratio of K₂C₂O₄.2 H₂O and CaC₂O₄ is 1 : 1, therefore, moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted in the mixture = number of moles × molar mass

Molar mass of K₂C₂O₄.H₂O = 184 g/mol

grams K2C2O4-H2O reacted = 0.00222 moles 184 g/mole = 0.408 g

g. Mass of sample = 0.879 g

mass of CaCl₂.2 H₂O in sample completely used up = 0.326 g

mass of K₂C₂O₄.H₂O in sample = 0.879 g - 0.326 g = 0.553 g

mass of K₂C₂O₄.H₂O remaining unreacted = 0.553 g - 0.408 g = 0.145 g

h. Percent by mass CaCl₂.2 H₂O = 0.326 /0.879 x 100% = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 0.553/0.879 × 100% = 62.9%

6 0

3 years ago

How many moles of water would you have if you had 3x10^24 molecules of

goldfiish [28.3K]

I am going to be honest i’m good at chemistry but i need the point cause i need help on my history work

5 0

3 years ago

The equilibrium constant, Kc, for the following reaction is 1.29E-2 at 600 K. COCl2(g) CO(g) Cl2(g) Calculate the equilibrium co

RSB [31]

Answer:

At Equilibrium

[COCl₂] = 0.226 M

[CO] = 0.054 M

[Cl₂] = 0.054 M

Explanation:

Given that;

equilibrium constant Kc = 1.29 × 10⁻² at 600k

the equilibrium concentrations of reactant and products = ?

when 0.280 moles of COCl2(g) are introduced into a 1.00 L vessel at 600 K. [COCl²]

Concentration of COCl₂ = 0.280 / 1.00 = 0.280 M

COCl₂(g) ----------> CO(g) + Cl₂(g)

0.280 0 0 ------------ Initial

-x x x

(0.280 - x) x x ----------- equilibrium

we know that; solid does not take part in equilibrium constant expression

KC = [CO][Cl₂] / COCl₂

we substitute

1.29 × 10⁻² = x² / (0.280 - x)

0.0129 (0.280 - x) = x²

x² = 0.003612 - 0.0129x

x² + 0.0129x - 0.003612 = 0

x = -b±√(b² - 4ac) / 2a

we substitute

x = [-(0.0129)±√((0.0129)² - 4×1×(-0.003612))] / [2 × 1 ]

x = [-0.0129 ± √( 0.00017 + 0.01445)] / 2

x = [-0.0129 ± 0.1209] / 2

Acceptable value of x =[ -0.0129 + 0.1209] / 2

x = 0.108 / 2

x = 0.054

At equilibrium

[COCl₂] = (0.280 - x) = 0.280 - 0.054 = 0.226 M

[CO] = 0.054 M

[Cl₂] = 0.054 M

7 0

3 years ago

If have 72 Liters of a gas held at a pressure of 3.4 atm and a temperature of 225K how many moles

tekilochka [14]

Answer:

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2 years ago