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Semmy [17]
3 years ago
7

What can you infer about copper's properties?

Chemistry
1 answer:
tiny-mole [99]3 years ago
7 0

Answer:

good electrical conductor

lustre in appearance

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Se requieren preparar 1500 ml de una solución al 55% en masa de hidróxido de calcio, sabiendo que la densidad de dicha solución
Masja [62]

Respuesta:

968 g Ca(OH)₂

Explicación:

Paso 1: Calcular la masa de solución

Tenemos 1500 mL de una solución cuya densidad es 1.17 g/mL, es decir, 1 mL de solución tiene una masa de 1.17 g.

1500 mL × 1.17 g/mL = 1.76 × 10³ g

Paso 2: Calcular la masa de hidróxido de calcio en 1.76 × 10³ g de solución

La solución tiene una concentración de 55% en masa de hidróxido de calcio, es decir, cada 100 gramos de solución hay 55 gramos de hidróxido de calcio.

1.76 × 10³ g Solución × 55 g Ca(OH)₂/100 g Solución = 968 g Ca(OH)₂

6 0
2 years ago
PLZ HELP ME!!!!! Identify the compound that plants make to store as energy.
aliya0001 [1]

Answer:

Glucose

Explanation:

Plants make glucose to store as energy.

4 0
2 years ago
Read 2 more answers
What is the correct name of the compound Mn3(PO4)2? A. manganese phosphate B. manganese(I) phosphate C. manganese(III) phosphate
Sergio039 [100]
The correct name of the compound Mn3(PO4)2 is  definitely the last option represented above <span>D. manganese(II) phosphate. I am pretty sure this answer will help you

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3 0
3 years ago
A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical
Katarina [22]

Answer:

empirical formula: H_2SO_4

2 g H

32.7 g S

65.3 g O

Explanation:

Like the problem said, the first thing we can do is calculate the mass of each of the 3 elements in a 100-gram sample:

- 2.00% * 100g = 2 g H

- 32.7% * 100g = 32.7 g S

- 65.3% * 100g = 65.3 g O

Now we need to find the empirical formula from these. To do so, convert all of those masses into moles by using the molar mass for each element:

- the molar mass of H is 1.01 g/mol

- the molar mass of S is 32.06 g/mol

- the molar mass of O is 16 g/mol

2 g H ÷ 1.01 g/mol = 1.98 mol H

32.7 g S ÷ 32.06 g/mol = 1.02 mol S

65.3 g O ÷ 16 g/mol = 4.08 mol O

Our ratio of H : S : O is now:

1.98 mol : 1.02 mol : 4.08 mol

Divide them all by the smallest number, which is 1.02:

1.98/1.02  :  1.02/1.02  :  4.08/1.02

1.94 : 1 : 4

1.94 ≈ 2

So:

2 : 1 : 4

Thus, the empirical formula is: H_2SO_4.

7 0
3 years ago
In one step in the synthesis of the insecticide Sevin, naphthol reacts with phosgene as shown.
Serhud [2]

Answer:

Explanation:

Chemical equation:

C₁₀H₈O + COCl₂  → C₁₁H₇O₂Cl + HCl

A. How many kilograms of C₁₁H₇O₂Cl form from 2.5×10*2 kg of naphthol?

Given data:

Mass of naphthol = 2.5 ×10² kg ( 250×1000 = 250000 g)

Mass of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol = mass/ molar mass

Number of moles of naphthol = 250000 g/ 144.17 g/mol

Number of moles of naphthol = 1734.1 mol

Now we will compare the moles of naphthol with C₁₁H₇O₂Cl.

                     C₁₀H₈O       :         C₁₁H₇O₂Cl

                        1               :                1

                       1734.1         :             1734.1

Mass of C₁₁H₇O₂Cl:

Mass = number of moles × molar mass

Mass = 1734.1 mol × 206.5 g/mol

Mass = 358091.65 g

Gram to kilogram:

1 kg×358091.65 g/ 1000 g  = 358.1 kg

B. If 100. g of naphthol and 100. g of phosgene react, what is the theoretical yield of C11H7O2Cl?

Given data:

Mass of naphthol = 100 g

Mass of COCl₂ = 100 g

Theoretical yield of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol:

Number of moles of naphthol = mass/ molar mass

Number of moles of naphthol = 100 g/ 144.17 g/mol

Number of moles of naphthol = 0.694 mol

Number of moles of phosgene:

Number of moles  = mass/ molar mass

Number of moles =  100 g/ 99 g/mol

Number of moles = 1.0 mol

Now we will compare the moles of naphthol and phosgene with C₁₁H₇O₂Cl.

                     C₁₀H₈O        :         C₁₁H₇O₂Cl

                        1                :                1

                       0.694        :              0.694

                    COCl₂          :             C₁₁H₇O₂Cl

                        1                :                1

                       1.0              :              1.0

The number of moles of C₁₁H₇O₂Cl produced by C₁₀H₈O are less so it will limiting reactant and limit the yield of  C₁₁H₇O₂Cl.

Mass of C₁₁H₇O₂Cl:

Mass = number of moles × molar mass

Mass =  0.694 mol × 206.5 g/mol

Mass = 143.3 g

Theoretical yield  =  143.3 g

C. If the actual yield of C11H7O2Cl in part b is 118 g, what is the percent yield?

Given data:

Actual yield of C₁₁H₇O₂Cl = 118 g

Theoretical yield = 143.3 g

Percent yield = ?

Solution:

Formula :

Percent yield = actual yield / theoretical yield × 100

Now we will put the values in formula.

Percent yield = 118 g/ 143.3 g × 100

Percent yield = 0.82 × 100

Percent yield = 82%

5 0
2 years ago
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