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koban [17]
3 years ago
9

Which place would have the most air pressure

Chemistry
2 answers:
earnstyle [38]3 years ago
7 0

Answer:

Death Valley

Explanation:

navik [9.2K]3 years ago
6 0

Answer:

2, Death valley

Explanation:

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Based on the information that is given, which atom is the table ?
andriy [413]

Answer:

The one with the greatest mass would be the one that has the most things in the nucleus, protons and nutrons

Explanation:

3 0
3 years ago
At equilibrium at 2500K, [HCl]=0.0625M and [H2]=[Cl2]=0.00450M for the reaction H2+Cl2 ⇌ HCl.
ASHA 777 [7]

Answer:

a. H_2+Cl_2 \rightleftharpoons 2HCl

b. K = 192.9

c. Products are favored.

Explanation:

Hello!

a. In this case, according to the unbalanced chemical reaction we need to balance HCl as shown below:

H_2+Cl_2 \rightleftharpoons 2HCl

In order to reach 2 hydrogen and chlorine atoms at both sides.

b. Here, given the concentrations at equilibrium and the following equilibrium expression, we have:

K=\frac{[HCl]^2}{[H_2][Cl_2]}

Therefore, we plug in the data to obtain:

K=\frac{(0.0625)^2}{(0.00450)(0.00450)}\\\\K=192.9

c. Finally, we infer that since K>>1 the forward reaction towards products is favored.

Best regards!

8 0
2 years ago
Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.
Ad libitum [116K]

Explanation:

The given data is as follows.

    Concentration = 0.1 mol/dm^{3}

                             = 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

                             = 6.022 \times 10^{25} ions/m^{3}

               T = 30^{o}C = (30 + 273) K = 303 K

Formula for electric double layer thickness (\lambda_{D}) is as follows.

            \lambda_{D} = \frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}

where, n^{o} = concentration = 6.022 \times 10^{25} ions/m^{3}

Hence, putting the given values into the above equation as follows.

                 \lambda_{D} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}                    

                          = \sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}  

                         = 9.669 \times 10^{-10} m

or,                     = 9.7 A^{o}

                          = 1 nm (approx)

Also, it is known that \lambda_{D} = \sqrt \frac{1}{n^{o}}

Hence, we can conclude that addition of 0.1 mol/dm^{3} of KCl in 0.1 mol/dm^{3} of NaBr "\lambda_{D}" will decrease but not significantly.

7 0
3 years ago
how many grams of calcium chloride can be prepared from 60.4 G of calcium oxide and 69.0 G of hydrochloric acid and a double dis
Ksenya-84 [330]
Balanced equation: 
<span>CaO + 2 HCl --> CaCl2 + H2O </span>
<span>Calculate moles of each reactant: </span>
<span>60.4 g CaO / 56.08 g/mol = 1.08 mol CaO </span>
<span>69.0 g HCl / 36.46 g/mol = 1.89 mol HCl </span>

<span>Identify the limiting reactant: </span>
<span>Moles CaO needed to react with all HCl: </span>
<span>1.89 mol HCl X (1 mol CaO / 2 mol HCl) = 0.946 mol CaO </span>
<span>Because you have more CaO than that available, HCl is the limiting reactant. </span>

<span>Calculate moles and mass CaCl2: </span>
<span>1.89 mol HCl X (1 mol CaCl2 / 2mol HCl) X 111.0 g/mol = 105 g CaCl2</span>
8 0
3 years ago
What is the total number of joules required to completely melt 221 grams of ice?
DiKsa [7]

Answer:

6680

Explanation:

hope it help :)

7 0
3 years ago
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