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dalvyx [7]
3 years ago
14

What is the gasoline doing to our environment?

Chemistry
1 answer:
luda_lava [24]3 years ago
5 0
Gasoline use contributes to air pollution
Gasoline is a toxic and highly flammable liquid. The vapors given off when gasoline evaporates and the substances produced when gasoline is burned (carbon monoxide, nitrogen oxides, particulate matter, and unburned hydrocarbons) contribute to air pollution. Burning gasoline also produces carbon dioxide, a greenhouse gas.
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Which of the following materials is useful for making molds because it has a low melting point?
inessss [21]
<span>Molds are created to achieve a specific design of a material. These materials either came from a process of having a higher or lower temperature. Therefore, the molder must have thermal resistant properties. Low melting points means that the material to be shaped came from a cooler process. Wood and metal have higher thermal conductivity and therefore can easily be cooled. The wax can turn really hard and can be unbreakable when present in colder materials due to the lipids present in it. Clay however can become a mold because of its low melting point.</span>
7 0
3 years ago
Read 2 more answers
The state of matter of the interior of the sun and other stars​
Sphinxa [80]

Answer: <u>It is, as all stars are, a hot ball of gas made up mostly of Hydrogen. The Sun is so hot that most of the gas is actually plasma, the fourth state of matter. ... As we heat up liquid, the liquid turns to gas. Gas is the third state of matter</u>

Explanation:

7 0
3 years ago
Read 2 more answers
PbSO4 has a Ksp = 1.3 * 10-8 (mol/L)2.
Oduvanchick [21]

i. The dissolution of PbSO₄ in water entails its ionizing into its constituent ions:

\mathrm{PbSO_{4}}(aq) \rightleftharpoons \mathrm{Pb^{2+}}(aq)+\mathrm{SO_4^{2-}}(aq).

---

ii. Given the dissolution of some substance

xA{(s)} \rightleftharpoons yB{(aq)} + zC{(aq)},

the Ksp, or the solubility product constant, of the preceding equation takes the general form

K_{sp} = [B]^y [C]^z.

The concentrations of pure solids (like substance A) and liquids are excluded from the equilibrium expression.

So, given our dissociation equation in question i., our Ksp expression would be written as:

K_{sp} = \mathrm{[Pb^{2+}] [SO_4^{2-}]}.

---

iii. Presumably, what we're being asked for here is the <em>molar </em>solubility of PbSO4 (at the standard 25 °C, as Ksp is temperature dependent). We have all the information needed to calculate the molar solubility. Since the Ksp tells us the ratio of equilibrium concentrations of PbSO4 in solution, we can consider either [Pb2+] or [SO4^2-] as equivalent to our molar solubility (since the concentration of either ion is the extent to which solid PbSO4 will dissociate or dissolve in water).

We know that Ksp = [Pb2+][SO4^2-], and we are given the value of the Ksp of for PbSO4 as 1.3 × 10⁻⁸. Since the molar ratio between the two ions are the same, we can use an equivalent variable to represent both:

1.3 \times 10^{-8} = s \times s = s^2 \\s = \sqrt{1.3 \times 10^{-8}} = 1.14 \times 10^{-4} \text{ mol/L}.

So, the molar solubility of PbSO4 is 1.1 × 10⁻⁴ mol/L. The answer is given to two significant figures since the Ksp is given to two significant figures.

8 0
3 years ago
Will mark Brainliest!!! PLEASE HELP
GalinKa [24]
1= D
2= B
3= E
4= F
5= A
8 0
3 years ago
A sample of ammonium nitrate, weighing 25.2 g, was added to a 150.0 mL container of H2O at 20.0 °C, and then is thoroughly disso
OLga [1]

Explanation:

It is known that the relation between heat energy, specific heat and change in temperature is as follows.

                            q = m \times C \times \Delta T

where,         q = heat released or absorbed

                    m = mass of substance

                    C = specific heat

             \Delta T = change in temperature

It is given that mass is 25.2 g and it is added to 150 ml container so total mass will be (25.2 g + 150 g) = 175.2 g, T_{1} is 20^{o}C, T_{2} is 8.5^{o}C, and specific heat of water is 4.184 J/g ^{o}C.

Hence, putting the given values into the above formula as follows.

                  q = m \times C \times \Delta T

                     = 175.2 g \times 4.184 J/g^{o}C \times (8.5 - 20)^{o}C

                      = 175.2 g \times 4.184 J/g^{o}C \times -11.5^{o}C

                     = -8430 J

As, 1 J = 0.001 kJ. Hence, -8430 J will also be equal to -8.43 kJ. The negative sign indicates that heat is being lost.

Thus, we can conclude that heat (kJ) was lost by the solution (surroundings) is 8.43 kJ.

6 0
4 years ago
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