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2 years ago

How many grams are in 3.3 moles of potassium sulfide , k2s

1 answer:

7 0

In 3.3 moles of potassium sulphide (K2S), there are 363.99 g. Let's first calculate the molar mass of K2S (Mr) which is the sum of atomic masses (Ar) of its elements. According to the periodic table, Ar(K) = 39.1 g/mol and Ar(S) = 32.1 g/mol. Mr(K2S) = 2Ar(K) + Ar(S) = 2 * 39.1 + 32.1 = 110.3 g/mol. Thus, there are 110.3 g per 1 mol. There will be x grams in 3.3 moles. 110.3 g : 1 mol = x : 3.3 mol. x = 110.3 g * 3.3 mol : 1 mol = 363.99 g.

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The crystalline hydrate cd(no3)2 ⋅ 4h2o(s) loses water when placed in a large, closed, dry vessel at room temperature: cd(no3)2⋅

Sever21 [200]

The given dehydration equation is,

$Cd(NO_{3})_{2}. 4H_{2}O (s) ---> Cd(NO_{3})_{2}(s) + 4 H_{2}O(g)$

Cadmiumnitrate tetrahydrate when heated dehydrates releasing the combined water as water vapor. The reaction produces 4 moles of gaseous product water vapor. So, the degree of disorder or randomness increases. Hence, the sign of change in entropy is positive.

This reaction is spontaneous at room temperature even if it is endothermic as the sign of change in entropy is positive.

8 0

2 years ago

Read 2 more answers

A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percent composition, the student measures out 5.8

ra1l [238]

Answer:

1. Net ionic equation:

Cl⁻(aq) + Ag⁺(aq) → AgCl(s)

2. Volume of 1.0M AgNO₃

41ml

Explanation:

1. Net ionic equation for the reaction of NaCl with AgNO₃.

i) Molecular equation:

It is important to show the phases:

(aq) for ions in aqueous solution
(s) for solid compounds or elements
(g) for gaseous compounds or elements

NaCl(aq) + AgNO₃(aq) → AgCl(s) + NaNO₃(aq)

ii) Dissociation reactions:

Determine the ions formed:

NaCl(aq) → Na⁺(aq) + Cl⁻(aq)

AgNO₃(aq) → Ag⁺(aq) + NO₃⁻(aq)

NaNO₃(aq) → Na⁺(aq) + NO₃⁻(aq)

iii) Total ionic equation:

Substitute the aqueous compounds with the ions determined above:

Na⁺(aq) + Cl⁻(aq) + Ag⁺(aq) + NO₃⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq)

iv) Net ionic equation

Remove the spectator ions:

Cl⁻(aq) + Ag⁺(aq) → AgCl(s) ← answer

2. How many mL of 1.0 M AgNO₃ will be required to precipitate 5.84 g of AgCl

i) Determine the number of moles of AgNO₃

The reaction is 1 to 1: 1 mole of AgNO₃ produces 1 mol of AgCl

The number of moles of AgCl is determined using the molar mass:

number of moles = mass in grams / molar mass

molar mass of AgCl = 143.32g/mol

number of moles = 5.84g / (143.32g/mol) = 0.040748 mol

ii) Determine the volume of AgNO₃

molarity = number of moles of solute / volume of solution in liters

1.0M = 0.040748mol / V

V = 0.040748mol / (1.0M) = 0.040748 liter

V = 0.040748liter × 1,000ml / liter = 40.748 ml

Round to two significant figures: 41ml ← answer

4 0

2 years ago

In a laboratory setting, concentrations for solutions are measured in molarity, which is the number of moles per liter (mol/L).

slamgirl [31]

Answer:

im pretty sure its A or C im leaning more toward A tho

Explanation:

6 0

1 year ago

Read 2 more answers

Lithium chloride forms three hydrates. They are LiCl.H2O, LiCl.2H2O and LiCl.3H2O.

Stels [109]

Answer:

The answer is LiCl.2H2O

Explanation:

Li=7

Cl=35.5

O=16

LiCl.H2O

7+35.5+16+2

60.5

%comp=60.5/78.5×100

22.9

LiCl.2H20

7+35.5+2(2+16)

42.5+36

78.5

%comp=36/78.5×100

45.9

LiCl.3H20

7+35.5+3(2+16)

42.5+54

96.5

54/96.5×100

56.0

7 0

2 years ago

What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x101 J. (A)

Delvig [45]

Explanation:

The given data is as follows.

Energy of radiation absorbed by the electron in hydrogen atom = $1.08 \times 10^{-17} J$

As energy is absorbed as a photon. Hence, frequency will be calculated will be as follows.

E = $h \nu$

$1.08 \times 10^{-17} J$ = $6.626 \times 10^{-34} Js \times \nu$

$\nu$ = $0.163 \times 10^{17} s^{-1}$

or, $\nu$ = $1.63 \times 10^{16} s^{-1}$

It is known that, $\nu = \frac{c}{\lambda}$

$1.63 \times 10^{16} s^{-1} = \frac{3 \times 10^{8} m/s}{\lambda}$

$\lambda$ = $1.84 \times 10^{-8} m$

And, according to De-Broglie equation $\lambda = \frac{h}{p}$

as, p = $m \times \nu$

So, $\lambda = \frac{h}{m \times \nu}$

$m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m}$

= $3.6 \times 10^{-26} J/m$

Now, on squaring both the sides we get the following.

$(m \times \nu)^{2}$ = $(3.6 \times 10^{-26} J/m)^{2}$

= $12.96 \times 10^{-52}$

$m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

where, m = mass of electron

So, $m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

= $\frac{12.96 \times 10^{-52}}{9.1 \times 10^{-31}}$

= $1.42 \times 10^{-21}$ J

Since, K.E = $\frac{1}{2}m \nu^{2}$

= $\frac{1.42 \times 10^{-21} J}{2}$

= $0.71 \times 10^{-21} J$

Thus, we can conclude that kinetic energy acquired by the electron in hydrogen atom is $7.1 \times 10^{-22} J$ .

4 0

2 years ago