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VLD [36.1K]
3 years ago
7

Naomi is investigating the properties of a solid material. It takes 120 joules to raise the temperature of 10 grams of the mater

ial by 5 degrees. What is the specific heat of the material?
Chemistry
1 answer:
MrRissso [65]3 years ago
3 0
When heat energy is supplied to a material it can raise the temperature of mass of the material.
Specific heat is the amount of energy required by 1 g of material to raise the temperature by 1 °C.
equation is 
H = mcΔt
H - heat energy 
m - mass of material 
c - specific heat of the material 
Δt - change in temperature
substituting the values in the equation 
120 J = 10 g x c x 5 °C
c = 2.4 Jg⁻¹°C⁻¹
You might be interested in
Find the equilibrium value of [CO] if Kc=14.5 : CO (g) + 2H2 (g) ↔ CH3OH (g) Equilibrium concentrations: [H2] = 0.322 M and [CH3
Annette [7]

Answer:

The equilibrium value of [CO] is 1.04 M

Explanation:

Chemical equilibrium is the state to which a spontaneously evolving  chemical system, in which a reversible chemical reaction takes place.  When this situation is reached, it is observed that the  concentrations of substances, both reagents and reaction products,  they remain constant over time. That is, the rate of reaction of reagents to products is the same as that of products to reagents.

Reagent concentrations  and products in equilibrium are related by the equilibrium constant Kc. Being:

aA + bB ⇔ cC + dD

Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b} }

Then this constant Kces equals the multiplication of the concentrations of the products raised to their stoichiometric coefficients between the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

In this case:

Kc=\frac{[CH_{3}OH ]}{[CO]*[H_{2} ]^{2} }

You know:

  • Kc= 14.5
  • [H₂]= 0.322 M
  • [CH₃OH] =1.56 M

Replacing:

14.5=\frac{1.56}{[CO]*0.322^{2} }

Solving:

[CO]=\frac{1.56}{14.5*0.322^{2} }

[CO]= 1.04 M

The equilibrium value of [CO] is 1.04 M

8 0
3 years ago
25 g of a compound is added to 500 mL of water if the freezing point of the resulting solution is
vlabodo [156]

Answer:

a)   119 g/mol

Explanation:

-We apply the formula for freezing point depression to obtain the molality of the solution:

\bigtriangleup T_f=K_fm, \  \ K_f=1.36\textdegree C/m\\\\\therefore m=\frac{\bigtriangleup T_f}{K_f}\\\\=\frac{0.57\textdegree C}{1.36\textdegree C}\\\\=0.4191\ mol/Kg\\\\

#We use the molality above to calculate the molar mass:

m=\frac{0.4191\ mol}{1\ Kg}=\frac{25\ g}{0.5\ Kg}\\\\\therefore 1 \ mol=\frac{25\ g}{0.5\ Kg}\times\frac{1\ Kg}{ 0.4191}\\\\=119.3033\approx 119\ g/mol

Hence, the molar mass of the compound is 119 g/mol

5 0
4 years ago
Compare which element would have larger first ionization energy: an alkali metal in Period 2 or an alkali metal in Period 4?
maria [59]

Answer:

An alkali metal present in period 2 have larger first ionization energy.

Explanation:

Ionization energy:

The amount of energy required to remove the electron from the atom is called ionization energy.

Trend along period:

As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.

Trend along group:

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

As the size of atom increases the ionization energy from top to bottom also  decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus.  Thus alkali metal present in period 2 have larger ionization energy because of more nuclear attraction as compared to the alkali metal present in period 4.

6 0
3 years ago
How is this model useful?
const2013 [10]

Answer:

A) It shows how electrons are distributed in the Shells of an Iron atom.

Explanation: Took the test on Edge

6 0
3 years ago
Locations Cost of land Cost of equipment Cost of mining and reclamation Time taken to mine the area 1 $100,000 $10,000 $5,000 pe
mario62 [17]

Answer:

Location 2 will cost the least to the company to mine.

Explanation:

a) Data and Calculations:

Locations  Cost of land     Cost of       Cost of mining      Time taken to

                                        equipment   and reclamation    mine the area

1                 $100,000      $10,000         $5,000 per day        30 days

2                 $35,000       $6,000          $4,500 per day        45 days

3                 $30,000       $7,500          $3,500 per day       120 days

4                 $40,500       $8,000          $7,000 per day        65 days

Locations  Cost of land     Cost of       Cost of mining                      Total

                                        equipment   and reclamation                    Costs

1                 $100,000      $10,000         $150,000 ($5,000 * 30)  $260,000

2                 $35,000       $6,000         $202,500 ($4,500 * 45)   $243,500

3                 $30,000       $7,500         $420,000 ($3,500 * 120)  $457,500

4                 $40,500       $8,000         $445,000 ($7,000 * 65)   $503,500

7 0
3 years ago
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