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2 years ago

Convert a pressure of 0.0248 mm Hg to the equivalent pressure in pascals (Pa). How did you get your answer ?

Chemistry

1 answer:

Rudik [331]2 years ago

5 0

Hope this helps :) remember your conversions and just practice it's fairly easy:)

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2 years ago

Which of the following describes the electron sharing between hydrogen and fluorine?

EastWind [94]

Hello
We know both the elements involved in bonding are non-metals and the primary type of bonds involved in non-metals are covalent bonds. Covalent bonds are formed when two atoms share one or more electrons; thus we know that whatever the number of electrons shared, it has to be equal for both. We can eliminate choices A and B.
Next, we understand that it is easier for one atom to be shared among the two, rather than the 7. First, because Hydrogen needs only 1 electron to be stable and would require energy to compensate the remaining 6. Second, electrons are attracted towards the nucleus so it is inefficient to try and share 7 electrons when 1 provides the same amount of stability to Fluorine.
Therefore, the answer is C.

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2 years ago

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1.000 mile is equal to 1609 meters. A truck is traveling at 42.00 km/h. What is the speed of the truck in miles per hour?

bija089 [108]

Around 24.85 miles per hour

, divide the length value by 1.60

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2 years ago

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4.82 g of an unknown metal is heated to 115.0∘C and then placed in 35 mL of water at 28.7∘C, which then heats up to 34.5∘C. What

nikitadnepr [17]

<u>Answer:</u> The specific heat of metal is 2.34 J/g°C

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 35 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{35mL}\\\\\text{Mass of water}=(1g/mL\times 35mL)=35g$

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of metal = 4.82 g

$m_2$ = mass of water = 35 g

$T_{final}$ = final temperature = 34.5°C

$T_1$ = initial temperature of metal = 115°C

$T_2$ = initial temperature of water = 28.7°C

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$4.82\times c_1\times (34.5-110)=-[35\times 4.186\times (34.5-28.7)]$

$c_1=2.34J/g^oC$

Hence, the specific heat of metal is 2.34 J/g°C

4 0

2 years ago

When 10 g of diethyl ether is converted to vapor at its boiling point, about how much heat is absorbed? (c4h10o, δhvap = 15.7 kj

vagabundo [1.1K]

Answer:

ΔH= 3KJ

Explanation:

The total heat absorbed is the total energy in the process, and that is in form of entalpy.

ΔH = q + ΔHvap, where q is the heat necessary for elevate the temperature of dietil ether. Suppose the initial temperature is room temperature (25ºC=298 K), then

q= 10g x2.261 J/gK x(310 K - 298K)= 271.32 J= 0.3 kJ

Then

ΔHvap = 10g C4H10O x (1 mol C4H10O/74.12 g C4H10O) x( 15.7 KJ/ 1 mol C4H10O) = 2.12 KJ

ΔH= 2.5KJ ≈ 3KJ

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2 years ago