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faltersainse [42]
3 years ago
12

How many molecules of HCl would react with 1 mole of Mg0? Mg0 +2HCI MgCl, + H,0

Chemistry
1 answer:
Len [333]3 years ago
5 0

Answer:

12.044 ×10²³ molecules of HCl

Explanation:

Given data:

Number of moles of MgO = 1 mol

Number of molecules of HCl react = ?

Solution:

Chemical equation:

MgO + 2HCl       →     MgCl₂  + H₂O

with 1 mole of MgO 2 moles of HCl are react.

Number of molecules of HCl react:

1 mole contain 6.022×10²³ molecules

2 mol × 6.022×10²³ molecules / 1 mol

12.044 ×10²³ molecules

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Alinara [238K]

Answer:

Silver Nitrate

Explanation:

5 0
3 years ago
In a sample containing a mixture of only these gases at exactly one atmosphere pressure, the partial pressures of carbon dioxide
Black_prince [1.1K]

Answer:

Explanation:

The pressure of a gaseous mixture is equal to the sum of the partial pressures of the individual gases:

ΣP_g_a_s = P_1+P_2+P_3+...+P_n

The prompt is trying to confuse you, but it actually tells us the pressure of the mixture to be 1 atm, but this can be converted to torr. Furthermore, we are informed only three gases are in the mixture: diatomic nitrogen, diatomic oxygen, and carbon dioxide:

P_g_a_s=1 \ atm = 760 \ torr= P_N_2+P_O_2+P_C_O_2\\760 \ torr = 582.008 \ torr + P_O_2 \ + 0.285 \ torr

Solve for Po2:

P_o_2=(760-582.008-0.285) \ torr = 177.707 \ torr

Thus, the partial pressure of diatomic oxygen is 177.707 torr.

<u><em>If you liked this solution, hit Thanks or give a Rating!</em></u>

4 0
3 years ago
How many grams of a stock solution that is 92.5 percent H2SO4 by mass would be needed to make 250 grams of a 35.0 percent by mas
IrinaVladis [17]

94.6 g.  You must use 94.6 g of 92.5 % H_2SO_4 to make 250 g of 35.0 % H_2SO_4.

We can use a version of the <em>dilution formula</em>

<em>m</em>_1<em>C</em>_1 = <em>m</em>_2<em>C</em>_2

where

<em>m</em> represents the mass and

<em>C</em> represents the percent concentrations

We can rearrange the formula to get

<em>m</em>_2= <em>m</em>_1 × (<em>C</em>_1/<em>C</em>_2)

<em>m</em>_1 = 250 g; <em>C</em>_1 = 35.0 %

<em>m</em>_2 = ?; _____<em>C</em>_2 = 92.5 %

∴ <em>m</em>_2 = 250 g × (35.0 %/92.5 %) = 94.6 g

4 0
4 years ago
Given the balanced equation below, calculate the moles of aluminum that are needed to react completely with 28.7 moles of FeO. Y
Alexxx [7]

Answer:

43.05 moles of Al needed to react with 28.7 moles of FeO.

Explanation:

Given data:

Moles of FeO = 28.7 mol

Moles of Al needed to react with FeO = ?

Solution:

Chemical equation:

2Al + 3FeO → 3Fe + Al₂O₃

Now we will compare the moles of Al with FeO.

                            FeO        :           Al

                             2            :            3

                          28.7          :         3/2×28.7 = 43.05 mol

Thus 43.05 moles of Al needed to react with 28.7 moles of FeO.

8 0
3 years ago
Which of the following statements is a true statement regarding a solution with [H1+] =1x10-5 M and [OH1-]= 1x10-9 M?
zheka24 [161]

Answer:

B. The [H1+] >[OH1-] and the solution is acidic

4 0
3 years ago
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