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1 year ago

1a) 2 properties that glass and plastic always share

1 answer:

4 0

4)There's no single substance called "plastic." That term covers many materials made from an array of organic and inorganic compounds. Substances are often added to plastic to help shape or stabilize it. Two of these plasticizers arebisphenol-A (BPA), added to make clear, hard plasticphthalates, added to make plastic soft and flexible

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How many liters of CO2 are in 4.76 moles? (at STP)

dem82 [27]

Answer: The volume of carbon dioxide gas at STP for given amount is 106.624 L

Explanation:

We are given:

Moles of carbon dioxide = 4.76 moles

At STP:

1 mole of a gas occupies a volume of 22.4 Liters

So, for 4.76 moles of carbon dioxide gas will occupy a volume of = $\frac{22.4L}{1mol\times 4.76mol=106.624L$

Hence, the volume of carbon dioxide gas at STP for given amount is 106.624 L

6 0

1 year ago

Upon heating with acid salicylic acid can form a polymer. what is its structure likely to be

Gnesinka [82]

Take a look at the attached picture. This is not the reaction of polymerization of salicylic acid. It just shows the structure of salicylic acid and the structure when it's polymerized. The polymerization is done by connecting the salicylic acids where you produce one molecule of water.

4 0

2 years ago

A compound that contains only carbon, hydrogen, and oxygen is 48.64% C and 8.16% H by mass. What is the empirical formula of thi

TiliK225 [7]

Answer:

The empirical formula of this substance is:

$C_3H_6O_2$

Explanation:

To find the empirical formula of this substance we need the molecular weight of the elements Carbon, Hydrogen and Oxygen, we can find this information in the periodic table:

- C: 12.01 g/mol

- H: 1.00 g/mol

- O: 15.99 g/mol

With the information in this exercise we can suppose in 100 g of the substance we have:

C: 48.64 g

H: 8.16 g

O: 43.2 g (100 g - 48.64g - 8.16g= 43.2 g)

Now, we need to divide these grams by the molecular weight:

$C=\frac{48.64g}{12.01 g/mol} =4.05 mol\\H=\frac{8.16g}{1.00g/mol}= 8.16 mol\\O=\frac{43.2g}{15.99 g/mol} = 2.70 mol$

We need to divide these results by the minor result, in this case O=2.70 mol

$C=\frac{4.05mol}{2.70mol}= 1.5\\H= \frac{8.16mol}{2.70 mol} = 3.02 \\O=\frac{2.70mol}{2.70mol} = 1$

We need to find integer numbers to find the empirical formula, for this reason we multiply by 2:

$C= 1.5*2=3\\H= 3.02*2= 6.04 \\O= 1*2=2$

This numbers are very close to integer numbers, so we can find the empirical formula as subscripts in the chemical formula:

$C_3H_6O_2$

6 0

1 year ago

A piece of iron metal is heated to 155 degrees C and placed into a calorimeter that contains 50.0 mL of water at 18.7 degrees C.

Korvikt [17]

Answer:

D = 28.2g

Explanation:

Initial temperature of metal (T1) = 155°C

Initial Temperature of calorimeter (T2) = 18.7°C

Final temperature of solution (T3) = 26.4°C

Specific heat capacity of water (C2) = 4.184J/g°C

Specific heat capacity of metal (C1) = 0.444J/g°C

Volume of water = 50.0mL

Assuming no heat loss

Heat energy lost by metal = heat energy gain by water + calorimeter

Heat energy (Q) = MC∇T

M = mass

C = specific heat capacity

∇T = change in temperature

Mass of metal = M1

Mass of water = M2

Density = mass / volume

Mass = density * volume

Density of water = 1g/mL

Mass(M2) = 1 * 50

Mass = 50g

Heat loss by the metal = heat gain by water + calorimeter

M1C1(T1 - T3) = M2C2(T3 - T2)

M1 * 0.444 * (155 - 26.4) = 50 * 4.184 * (26.4 - 18.7)

0.444M1 * 128.6 = 209.2 * 7.7

57.0984M1 = 1610.84

M1 = 1610.84 / 57.0984

M1 = 28.21g

The mass of the metal is 28.21g

3 0

1 year ago

The oxidation numbers of nitrogen in NH3, HNO3, and NO2 are, respectively: -3, -5, +4 +3, +5, +4 -3, +5, -4 -3, +5, +4

Evgesh-ka [11]

In NH3 , let oxidation number of N be x

x + (+1)3 = 0

x = -3

In HNO3 , let oxidation number of N be x

1 + x + (-2)3 = 0

x = +5

In NO2 , let oxidation number of N be x

x + (-2)2 = 0

x = +4

5 0

1 year ago