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3 years ago

6

Consider this equilibrium: COCl2(g) CO(g) + Cl2(g) Keq = 8.1 x 10-4. What is true for this system? [COCl2] < [CO][Cl2] [COCl2

] = [CO][Cl2] [COCl2] > [CO][Cl2]

2 answers:

const2013 [10]3 years ago

5 0

Answer is: [COCl₂] > [CO][Cl₂].

Chemical reaction: COCl₂(g) ⇄ CO(g) + Cl₂(g); Keq = 8.1 x 10⁻⁴.

Keq = [CO] · [Cl₂] / [COCl₂]; equilibrrium constant of chemical reaction.

[CO] · [Cl₂] / [COCl₂] = 0.00081.

Equilibrium product concentration is much more less than equilibrium concentration of reactant.

Zepler [3.9K]3 years ago

4 0

I'm pretty sure the answer here is C.

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A chemist has 3.55x1022 molecules of nitrogen monoxide. How

Alina [70]

Answer:

<h2>0.059 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{3.55 \times {10}^{22} }{6.02 \times {10}^{23} } \\ = 0.05897...$

We have the final answer as

<h3>0.059 moles</h3>

Hope this helps you

6 0

3 years ago

A 360. g iron rod is placed into 750.0 g of water at 22.5°C. The water temperature rises to 46.7°C. What was the initial tempera

Grace [21]

Answer: The initial temperature of the iron was $515^0C$

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of iron = 360 g

$m_2$ = mass of water = 750 g

$T_{final}$ = final temperature = $46.7^0C$

$T_1$ = temperature of iron = ?

$T_2$ = temperature of water = $22.5^oC$

$c_1$ = specific heat of iron = $0.450J/g^0C$

$c_2$ = specific heat of water= $4.184J/g^0C$

Now put all the given values in equation (1), we get

$-360\times 0.450\times (46.7-x)=[750\times 4.184\times (46.7-22.5)]$

$T_i=515^0C$

Therefore, the initial temperature of the iron was $515^0C$

4 0

3 years ago

If i combined 15.0 grams of calcium hydroxide with 75.0 ml of 0.500 m hcl, how many grams of calcium chloride would be formed?

Zina [86]

The equation is:
Ca(OH)₂(s) + 2 HCl(aq) → CaCl₂(aq) + 2 H₂<span>O(l)
</span>

n=mass in g/M.M

15 g Ca(OH)₂ is n=15 g/ 74.1 g/mol=0.2024 mol of Ca(OH)₂

no. of mol of HCl:

n=0.5 mol/L*0.075L=0.0375 mol

This could react with 0.0375/2= 0.01875 mol of Ca(OH)₂ We have a lot more than that.

Therefore, HCl is the limiting reagent and determines how much CaCl₂ forms.

Based on the balanced reaction, 2 moles of HCl gives 1 mole of CaCl₂

no. of mol of CaCl₂= 0.0375/2= 0.01875 mol

mass in g=n*MM= 0.01875*111= 2.08 g

7 0

3 years ago

30 POINTS PLEASE HELP ASAP

lina2011 [118]

Hello!

We have the following data:

f (radiation frequency) = $3.0*10^{19}\:Hz$

v (speed of light) = $3.0*10^8\:m/s$

λ (wavelength) = ? (in m)

Let's find the wavelength, let's see:

$f = \dfrac{v}{\lambda}$

$3.0*10^{19} = \dfrac{3.0*10^8}{\lambda}$

$\lambda = \dfrac{3.0*10^8}{3.0*10^{19}}$

$\boxed{\boxed{\lambda = 1*10^{-11}\:m}}\Longleftarrow(wavelenght)\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR! =)

4 0

3 years ago

What is the molecular formula for a compound with a molar mass of 42.08g/mol and an empirical formula of ch2

yan [13]

Answer:

$C_3H_6$

Explanation:

Hello!

In this case, since the empirical formula is the smallest representation of the molecular formula, it is known that the times in which the empirical formula is into the molecular formula is a whole number and is computed by dividing the molar mass of the molecular formula by that of the empirical formula as shown below:

$\frac{42.08g/mol}{12.01+1.01x2}= \frac{42.08g/mol}{14.03g/mol}=3$

Thus, the molecular formula times the empirical formula by 3 to obtain:

$C_3H_6$

Regards!

3 0

2 years ago