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balandron [24]
3 years ago
5

What type of chemical hazard results from incomplete combustion of petroleum-based chemicals and other organic substances?

Chemistry
1 answer:
kiruha [24]3 years ago
5 0

Answer:

the answer is the option "a"

Explanation:

in petroleum polycyclic aromatic hydrocarbons are found. These organic compounds are composed of aromatic rings. PAH (Polycyclic Aromatic Hydrocarbons), are formed during the incomplete combustion of any type of organic matter. In general, exposure will be to a mixture of PAHs.

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if the theoretical yield of a reaction is 26.0 grams and you actually recovered 22.0 grams what is the precent yield
netineya [11]
Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100  

∴ if theoretical yield is 26 g, but only 22.0 is recovered from the reaction, 
then Percentage Yield = (22 g ÷ 26 g) × 100  
                                       =  84.6 %
8 0
3 years ago
Communication is everything... If you want a relationship to work. Y'all gotta be grown and talk about everything and get an und
VladimirAG [237]

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hello pls help me in this question

With determination and hard work, people can overcome difficult challenges. How do the people of Greensburg, Kansas, and Chicago, Illinois, illustrate this idea? Use details from both texts to support your response. i cant do it

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5 0
2 years ago
What is the mass of 0.28 mole of iron?
vredina [299]

Answer:

Your answer should be 15.68 grams.

Explanation:

Seeing as 1 mole has a mass of 56 g, 56*0.28 would get you 15.68 g.

6 0
3 years ago
The Mass of the mineral is 9.6 grams . The mineral is placed in a graduated cylinder containing 8.0 ml of water . The water rise
ryzh [129]

Answer:

1.2 g/ ml

Explanation:

The volume of the mineral = increase in volume of the water whuich is 16 - 8 = 8mls.

Therefore the  mineral's density = 9.6 / 8

=  1.2 g/ ml  answer

7 0
3 years ago
1. The heat of fusion for the ice-water phase transition is 335 kJ/kg at 0°C and 1 bar. The density of water is 1000 kg/m3 at th
vodomira [7]

Answer:

Expression for the change of melting temperature with pressure..> T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa), Freezing Point = 0°C

Explanation:

Derivation from state postulate

Using the state postulate, take the specific entropy,  , for a homogeneous substance to be a function of specific volume  and temperature  .

ds = (partial s/partial v)(t) dv + (partial s/partial T)(v) dT

During a phase change, the temperature is constant, so

ds = (partial s/partial v)(T)  dv

Using the appropriate Maxwell relation gives

ds = (partial P/partial T)(v) dv

s(β) – s(aplαha) = dP/dT (v(β) – v(α))

dP/dT = s(β) – s(α)/v(β) – v(α) = Δs/Δv

Here Δs and Δv are respectively the change in specific entropy and specific volume from the initial phase α to the final phase β.

For a closed system undergoing an internally reversible process, the first law is

du = δq – δw = Tds - Pdv

Using the definition of specific enthalpy, h and the fact that the temperature and pressure are constant, we have

du + Pdv = dh Tds,

ds = dh/T,

Δs = Δh/T = L/T

After substitution of this result into the derivative of the pressure, one finds

dp/dT = L/TΔv

<u>This last equation is the Clapeyron equation.</u>

a)

(dP/dT) = dH/TdV => dP/dlnT = dH/dV

=> dP/dlnT = dH/dV = [H(liquid) - H(solid)]/[V(liquid) - V(solid)]

= [335,000 J/kg]/[1000⁻¹ - 915⁻¹ m³/kg]

= -3.61x10⁹ J/m³ = -3.61x10⁹ Pa

=> P₂ = P₁ - 3.61x10⁹ ln(T₂/T₁) Pa

or

T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa)

b) if the pressure in Denver is 84.6 kPa:

T₂(freezing) = 273.15exp[-(84,600-100,000)/(3.61x10⁹)]

≅ 273.15 = 0°C T₁(freezing) essentially no change

5 0
3 years ago
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