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Free_Kalibri [48]
3 years ago
11

The specific heat of a certain type of metal is 0.128j/(g.c). What is the final temperature if 305.J of heat is added to 72.7g o

f the metal initially at 20.0°C?
Chemistry
1 answer:
expeople1 [14]3 years ago
3 0
E answer is -60.57 = -60.6 KJ.
CaC2(s) + 2 H2O(l) ---> Ca(OH)2(s) +C2H2(g) H= -127.2 KJ
Hf C2H2 = 226.77
Hf Ca(OH)2 = -986.2
<span>Hf H2O = -285.83
Now,

</span><span>add them up. 226.77 - 986.2 + (2*285.83) = -187.77 
</span><span>Add back the total enthalpy that is given in the question
 -187.77+127.2 = -60.57 

</span>
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Carbon monoxide (CO) reacts with hydrogen (H2) to form methane (CH4) and water (H2O).
Artemon [7]

Answer:

5.9x10^-2 M

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Concentration of CO, [CO] = 0.30 M

Concentration of H2, [H2] = 0.10 M

Concentration of H2O, [H2O] = 0.020 M

Equilibrium constant, K = 3.90

Concentration of CH4, [CH4] =..?

Step 2:

The balanced equation for the reaction. This is given below:

CO(g) + 3H2(g) <=> CH4(g) + H2O(g)

Step 3:

Determination of the concentration of CH4.

The expression for equilibrium constant of the above equation is given below:

K = [CH4] [H2O] / [CO] [H2]^3

3.9 = [CH4] x 0.02/ 0.3 x (0.1)^3

Cross multiply to express in linear form

[CH4] x 0.02= 3.9 x 0.3 x (0.1)^3

Divide both side by 0.02

[CH4] = 3.9 x 0.3 x (0.1)^3 /0.02

[CH4] = 5.9x10^-2 M

Therefore, the equilibrium concentration of CH4 is 5.9x10^-2 M

5 0
3 years ago
What is the study of acid-base chemistry called in the environment
Charra [1.4K]

Answer:

An acid is a substance that donates protons (in the Brønsted-Lowry definition) or accepts a pair of valence electrons to form a bond (in the Lewis definition). A base is a substance that can accept protons or donate a pair of valence electrons to form a bond. Bases can be thought of as the chemical opposite of acids.

8 0
3 years ago
The heat of vaporization of 1-pentanol is 55.5 kJ/mol, and its entropy of vaporization is 148 J/K.mol. What is the approximate b
____ [38]

Answer:

Approximately 100 °C.

Explanation:

Hello,

In this case, since the entropy of vaporization is computed in terms of the heat of vaporization and the temperature as:

\Delta S_{vap}=\frac{\Delta H_{vap}}{T}

We can solve for the temperature as follows:

T=\frac{\Delta H_{vap}}{\Delta S_{vap}}

Thus, with the proper units, we obtain:

T=\frac{55500J/mol}{148J/(mol*K)} =375K\\\\T=102 \°C

Hence, answer is approximately 100 °C.

Best regards.

6 0
3 years ago
A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipita
mestny [16]

Answer:

The answer is:

(a) NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)

(b) NaCl

(c) 0.211 g

Explanation:

Given:

The mass of NaCl,

= 0.0860 g

The molar mass of NaCl,

= 58.44 g/mol

The volume of AgNO_3,

= 30.0 ml

or,

= 0.030 L

Molarity of AgNO_3,

= 0.050 M

Moles of NaCl will be:

= \frac{Given \ mass}{Molar \ mass}

= \frac{0.0860}{58.44}

= 0.00147 \ mol

now,

Moles of AgNO_3 will be:

= Molarity\times Volume

= 0.050\times 0.030

=0.0015 \ mol

(a)

The reaction is:

⇒ NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)

(b)

1 mole of NaCl react with,

= 1 mol of AgNO_3

0.0015 mol AgNO_3 needs,

= 0.00150 \ mol \ NaCl

Available mol of NaCl < needed amount of NaCl

So,

The limiting reagent is "NaCl".

(c)

The precipitate formed,

= 0.00147\times \frac{1}{1}\times \frac{143.32}{1}

= 0.211 \ g \ AgCl

4 0
3 years ago
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