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2 years ago

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The amount of heat involved in the synthesis of 1 mole of a compound from its elements, with all substances in their standard st

ates at 25°C, is called _____.
A) standard heat of formation

B) heat of solidification

C) enthalpy

D) heat of reaction

2 answers:

Pachacha [2.7K]2 years ago

8 0

Answer:

A

Explanation:

goblinko [34]2 years ago

6 0

Answer:

A) standard heat of formation

Explanation:

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Dan described two layers of Earth's atmosphere.

saul85 [17]

There are following layers in earth atmosphere

a) Troposhere: the innermost layer or lowest part of atmosphere

b)stratosphere: next to troposhere

c)mesosphere:above stratosphere, here lowest temperature is observed

d)Thermosphere and Ionosphere: This layer allows long distance radio communication

e)Exosphere: this layer separates outer space from earth's surface.

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2 years ago

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1(30%). A thin plate black body, insulated at the bottom is placed faced up in a room with wall temperatures of 30 C. Air at 0 C

ch4aika [34]

Explanation:

The given data is as follows.

Heat transfer coefficient (h) = 12 $W/m^{2}.C$

Plate temperature ( $T_{1}$ ) = $30 ^{o}C$ = 303 K

Steady state temperature ( $T_{2}$ ) = ?

Hence, formula applied for steady state is as follows.

$(h \times A \times \Delta T)_{air}$ = $\sigma \times A \times (T^{4}_{1} - T^{4}_{2})_{plate}$

Putting the given values into the above formula as follows.

$(h \times A \times \Delta T)_{air}$ = $\sigma \times A \times (T^{4}_{1} - T^{4}_{2})_{plate}$

$12 \times (T_{2} - 273)$ = $5.67 \times 10^{-8} \times [(30 + 273)^{4} - T^{4}_{2}]$

$T_{2}$ = 282.66 K

= (282.66 -273) $^{o}C$

= 9.66 $^{o}C$

Thus, we can conclude that the steady state temperature will be 9.66 $^{o}C$ .

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2 years ago

What is the mass of 6.78x1050 molecules of NaCl

ryzh [129]

Answer:

Explanation:

As

Molar mass of NaCl= 58.44 gm

So we got it that,

58.44 mass of NaCl contain = 6.023*10^23 molecules of Na Cl

x mass will contain = 6.78*10^50

by cross multiplication , we get

x= 58.44*(6.78*10^50)/ 6.023*10^23

x= 65.785 * 10^27 molecules

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2 years ago

SVETLANKA909090 [29]

D- Physical

Explanation:

A physical property is anything that has characteristics associated with a change in it's chemical composition

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2 years ago

At 25.0°c, a solution has a concentration of 3.179 m and a density of 1.260 g/ml. the density of the solution at 50.0°c is 1.249

oksano4ka [1.4K]

Answer: -

3.151 M

Explanation: -

Let the volume of the solution be 1000 mL.

At 25.0 °C, Density = 1.260 g/ mL

Mass of the solution = Density x volume

= 1.260 g / mL x 1000 mL

= 1260 g

At 25.0 °C, the molarity = 3.179 M

Number of moles present per 1000 mL = 3.179 mol

Strength of the solution in g / mol

= 1260 g / 3.179 mol = 396.35 g / mol (at 25.0 °C)

Now at 50.0 °C

The density is 1.249 g/ mL

Mass of the solution = density x volume = 1.249 g / mL x 1000 mL

= 1249 g.

Number of moles present in 1249 g = Mass of the solution / Strength in g /mol

= $\frac{1249 g}{396.35 g/mol}$

= 3.151 moles.

So 3.151 moles is present in 1000 mL at 50.0 °C

Molarity at 50.0 °C = 3.151 M

7 0

2 years ago