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Irina18 [472]
2 years ago
6

How much energy is needed to raise the temperature of 15.21 g of a substance with a specific heat of 1.65 J/g°C from 137.0 °C to

234.3 °C?
Chemistry
1 answer:
Tamiku [17]2 years ago
7 0

Answer:

2441.9 Joules

Explanation:

This is a sensible heat change problem.

Energy = Specific Heat Capacity x Mass x Change in Temperature

To find the change in temperature, you need to use this formula: Temperature Final - Temperature Initial.

234.3 - 137.0 = 97.3

Energy = 1.65 x 15.21 x 97.3

Multiply that equation, and you get your answer! I hope this helps!

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Suppose 0.245 g of sodium chloride is dissolved in 50. mL of a 18.0 m M aqueous solution of silver nitrate.
Bezzdna [24]

Answer:

\large \boxed{\text{ 0.066 mol/L}}

Explanation:

We are given the amounts of two reactants, so this is a limiting reactant problem.

1. Assemble all the data in one place, with molar masses above the formulas and other information below them.

Mᵣ:       58.44  

            NaCl + AgNO₃ ⟶ NaNO₃ + AgCl

m/g:     0.245

V/mL:                 50.

c/mmol·mL⁻¹:       0.0180

2. Calculate the moles of each reactant  

\text{Moles of NaCl} = \text{245 mg NaCl} \times \dfrac{\text{1 mmol NaCl}}{\text{58.44 mg NaCl}} = \text{4.192 mmol NaCl}\\\\\text{ Moles of AgNO}_{3}= \text{50. mL AgNO}_{3} \times \dfrac{\text{0.0180 mmol AgNO}_{3}}{\text{1 mL AgNO}_{3}} = \text{0.900 mmol AgNO}_{3}

3. Identify the limiting reactant  

Calculate the moles of AgCl we can obtain from each reactant.

From NaCl:  

The molar ratio of NaCl to AgCl is 1:1.

\text{Moles of AgCl} = \text{4.192 mmol NaCl} \times \dfrac{\text{1 mmol AgCl}}{\text{1 mmol NaCl}} = \text{4.192 mmol AgCl}

From AgNO₃:  

The molar ratio of AgNO₃ to AgCl is 1:1.  

\text{Moles of AgCl} = \text{0.900 mmol AgNO}_{3} \times \dfrac{\text{1 mmol AgCl}}{\text{1 mmol AgNO}_{3}} = \text{0.900 mmol AgCl}

AgNO₃ is the limiting reactant because it gives the smaller amount of AgCl.

4. Calculate the moles of excess reactant

                   Ag⁺(aq)  +  Cl⁻(aq) ⟶ AgCl(s)

 I/mmol:      0.900        4.192            0

C/mmol:    -0.900       -0.900        +0.900

E/mmol:      0                3.292          0.900

So, we end up with 50. mL of a solution containing 3.292 mmol of Cl⁻.

5. Calculate the concentration of Cl⁻

\text{[Cl$^{-}$] } = \dfrac{\text{3.292 mmol}}{\text{50. mL}} = \textbf{0.066 mol/L}\\\text{The concentration of chloride ion is $\large \boxed{\textbf{0.066 mol/L}}$}

8 0
3 years ago
How many moles of atoms are there in 8.0 g of Mg?
JulsSmile [24]

Answer:

0.33

Explanation:

1mol=24

x=8

=0.33

5 0
4 years ago
Read 2 more answers
In nature, oxygen has three common isotopes. The atomic masses and relative abundances of these isotopes are given in the table
Vaselesa [24]

Answer: The average atomic mass of oxygen is 15.999 amu

Explanation:

Mass of isotope O-16 = 15.995 amu

% abundance of isotope O-16= 99.759 % = \frac{99.759}{100}=0.99759

Mass of isotope O-17 = 16.995 amu

% abundance of isotope O-17 = 0.037% = \frac{0.037}{100}=0.00037

Mass of isotope O-18 = 17.999 amu

% abundance of isotope O-18 = 0.204% = \frac{0.204}{100}=0.00204

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(15.995\times 0.99759)+(16.995\times 0.00037)+(17.999 \times 0.00204)]

A=15.999

Thus the average atomic mass of oxygen is 15.999 amu

5 0
3 years ago
A flower gardener discover that all the flower he planted could not grow. The farmer requested the services of an agro technicia
sp2606 [1]
I think it’s b because in the question it says that the agro technician ran some tests on the soil and found out that it’s acidic and then after that the farmer should discuss how to remove the acid so the soil can be fit for farming
6 0
3 years ago
All answers must have the correct unit! (g/mL) d= mass/volume, d= masa/volumen
Radda [10]

heh.......... sorry man............... but this was posted over 5 hours ago........ so nobody is gonna see it and you probably dont need the answer anymore..... so errrr..... imma justtttt..... take these points :D

8 0
3 years ago
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