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2 years ago

A substance with a mass of 50.0 g has a temperature change of 40.0

Chemistry

2 answers:

grigory [225]2 years ago

7 0

Answer:

Mass=50.0g

H=670J

change in temperature=40

using. c=h÷m×change in temperature

c=670÷50×40

C=670÷2000

C=0.335jkg-1k-1

Alex2 years ago

6 0

Answer: The specific heat of a substance with a mass of 50. 0g and a temperature change of 40. 0 degrees whne it absorbs 670 J of heat is 337 J/kg. K

Explanation:

Specific heat of a substance is said to be the energy required for an increase in temperature by 1°C in the unit of J/Kg. K

Formula of Specific heat:

Q = mcΔθ

Q represents the quantity of heat

m represents the mass of the substance in Kilograms = 50. 0g÷ 1000 = 0.05kg

c represents the specific heat capacity of the substance = ?

θ represents the temperature change = 40° C

Specific heat, c = Q ÷ mΔθ

= 670 ÷ 0.05 × 40

= 670 ÷ 2

= 335 J/ Kg. K

Therefore, the specific heat of the substance is 335 J/kg. K

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Answer:

The final temperature of sulfur dioxide gas is 215.43 C

Explanation:

Gay Lussac's Law establishes the relationship between the temperature and the pressure of a gas when the volume is constant. This law says that if the temperature increases the pressure increases, while if the temperature decreases the pressure decreases. In other words, the pressure and temperature are directly proportional quantities.

Mathematically, the Gay-Lussac law states that, when a gas undergoes a transformation at constant volume, the quotient of the pressure exerted by the temperature of the gas remains constant:

$\frac{P}{T}=k$

Assuming you have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment, by varying the temperature to a new value T2, then the pressure will change to P2, and it will be true:

$\frac{P1}{T1} =\frac{P2}{T2}$

The reference temperature is the absolute temperature (in degrees Kelvin)

In this case:

P1= 0.450 atm
T1= 20 C= 293.15 K (being 0 C= 273.15 K)
P2=0.750 atm
T2= ?

Replacing:

$\frac{0.450atm}{293.15 K} =\frac{0.750 atm}{T2}$

Solving:

$T2 =\frac{0.750 atm}{\frac{0.450atm}{293.15 K} }$

$T2=\frac{0.750 atm}{0.450 atm} *293.15K$

T2=488.58 K

Being 273.15 K= 0 C, then 488.58 K= 215.43 C

<u><em>The final temperature of sulfur dioxide gas is 215.43 C</em></u>

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3 years ago

What is the electron configuration of Iron

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The electron configuration of Iron is:

[Ar] 3d⁶4s²

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3 years ago

For each of the following sets of elements, circle the element expected to be most electronegative and draw a box which is expec

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Answer: The element expected to be most electronegative is Ca.

The element expected to be least electronegative is K.

Explanation:

Electronegativity is defined as the property of an element to attract a shared pair of electron towards itself.

Down the group:

The size of an atom increases as we move down the group because a new shell is added and electron gets added up.

As, the size of an element increases, the valence electrons gets away from the nucleus. So, the attraction between the nucleus and the shared pair of electrons decreases

Hence, electronegativity decreases moving from top to bottom down a group

Across a period:

The size of an atom decreases as we move across the period because the electrons get added to the same shell and the nuclear charge keeps on increasing. Thus the electrons get more tightly held by the nucleus.

As, the size of an element decreases, the valence electrons come near to the nucleus. So, the attraction between the nucleus and the shared pair of electrons increases.

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Thus as K, Sc and Ca are arranged across a period, the electronegativity order is K< Sc < Ca.

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3 years ago

150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reac

Murljashka [212]

Answer:

$0.175\; \rm mol \cdot L^{-1}$ .

Explanation:

Magnesium chloride and silver nitrate reacts at a $2:1$ ratio:

$\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s)$ .

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

$\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}$ .

The precipitate silver chloride $\rm AgCl$ is insoluble in water and barely ionizes. Hence, $\rm AgCl\!$ isn't rewritten as ions.

Net ionic equation:

$\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}$ .

Calculate the initial quantity of nitrate ions in the mixture.

$\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}$ .

Since nitrate ions $\rm {NO_3}^{-}$ do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

$n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol$ .

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Explanation:

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