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tia_tia [17]
3 years ago
9

HELP PLEASE!!

Chemistry
1 answer:
AleksandrR [38]3 years ago
8 0

Answer:

The mass of the object would be 491.55g/mL. Hope this helped! :)

Explanation:

D= 7.25g/mL                                                           M=D•V

V= 67.8mL

M=7.25g/mL•67.8mL

M= 491.55g

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Coffee is a solution of organic substances in water
Eduardwww [97]

Answer:

yes it is

Explanation:

8 0
2 years ago
The boiling point of chloroform is 61.7 C. The enthalapy of vaporization is 31.4 kj/mol. Caculate the entropy of vaporization. D
dedylja [7]

Answer:

Δ S = 93.8 J/mol-K

Explanation:

Given,

Boiling point of chloroform = 61.7 °C

                                             = 273 + 61.7 = 334.7 K.

Enthalapy of vapourization = 31.4 kJ/mol.

Using Gibbs free   energy equation

Δ G = Δ H - T (ΔS)

at equilibrium (when the liquid is boiling), Δ G = 0

so,  0 = ΔH - T (Δ S)

      T (Δ S) = Δ H

and ΔS = ΔH / T

Δ S = (31400 J/mol.) / 334.7 K

Δ S = 93.8 J/mol-K

6 0
3 years ago
In which set do all elements tend to form anions in binary ionic compounds? li, na, k n, o, i c, s, pb k, fe, br?
pav-90 [236]

Atoms or molecule after gaining of electron possesses negative charge and is known as anion.

For the given sets:

  • Li, Na, K

The given elements are alkali metals and have tendency to lose electrons easily and form cations.

  • N, O, I

The given elements are non-metals and are electronegative. So, they gain electrons easily and form anion.

  • C, S, Pb

Carbon has tendency to form bond by sharing of electrons, Sulfur has tendency to gain electrons and form anion whereas Lead has tendency to lose electron.

  • K, Fe, Br

Potassium and Iron has tendency to lose electron and form cation whereas Bromine has tendency to gain electron to form anion.

Hence, from the given sets, all elements of set: N, O, I have tendency to form anions in binary ionic compounds.

8 0
3 years ago
Read 2 more answers
A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.
Lubov Fominskaja [6]

Answer:

Ag_2SO_4

Explanation:

Formula for the calculation of no. of Mol is as follows:

mol=\frac{mass\ (g)}{molecular\ mass}

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

Ag, \frac{0.05305}{0.02657} \approx 2

S, \frac{0.02657}{0.02657} \approx 1

O, \frac{0.10594}{0.02657} \approx 4

Therefore, empirical formula of the compound = Ag_2SO_4

7 0
3 years ago
Read 2 more answers
Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:
Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: Al_{2}S_{3} +H_{2}O  _\to  Al(OH)_{3} + H_{2}S

First, balance the chemical equation

Al_{2}S_{3} + 6H_{2}O  \to 2Al(OH)_{3} + 3H_{2}S

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

n_{Al_{2}S_{3}  } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3}  } = \frac{158g}{150.16g/mol}   \\n_{Al_{2}S_{3} }=1.05 mol

From the chemical reaction , the ratio of molar is 3mol H_{2}S/1 mol Al_{2}S_{3}. So, the moles of hydrogen sulfide are:

n_{H_{2} O} =\frac{131g}{18.02g/mol}

       = 7.26mol

From the chemical reaction, the molar ratio is 3 mol H_{2}S/6 mol H_{2}O. So, the moles of hydrogen sulfide are:

Moles of H_{2}S formed = 7.26 mol H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }

Th liming reactant isAl_{2}S_{3} beacuse the mass of Al_{2}S_{3} forms less product than water. Therefore, the maximum number of moles of H_{2}S is 3.15 mol.  We know that molar mass of H_{2}S is 34.10g/mol. So, the maximum mass of H_{2}S formed is,

m_{H_{2}S } = n_{H_{2}S } \times Molar mass of H_{2}S

         = 3.15 mol \times 34.10g/mol

         = 107.4g

Now, multiplying the number of moles of Al_{2}S_{3} by the molar ratio between Al_2S_3 and H_2O which is 6mol H_2O/1mol Al_2S_3 we get the number of moles of H_2O reacted.

Moles of H_2O reacted = 1.05 mol Al_{2}S_3 \times \frac{6 mol H_2O}{1 mol Al_2S_3}

                                     = 6.31 mol H_2O

The mass of H_2O is,

m_{H_{2} O} = 6.31 mol \times 18.02g/ mol

          = 114g

On subtracting, the mass of H_2O reacted from the given mass of H_2O is,

m_{H_2O} = (131-114)g

         = 17g

Hence, the excess remaining reactant is 17g

To learn more about Chemical Reaction

brainly.com/question/11231920

#SPJ4

 

8 0
2 years ago
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