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zmey [24]
2 years ago
12

Magnesium (24.30g) reacts with hydrogen chloride (xg) to produce hydrogen gas(2.04g) and magnesium chloride (96.90g). How much h

ydrogen chloride was used in the reaction
Chemistry
2 answers:
weqwewe [10]2 years ago
6 0

Answer:

Answer is 74.42 if you are on cK-12.

Explanation:

xz_007 [3.2K]2 years ago
5 0

Answer: 74.46

Explanation:

Is what I got, for ck12

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What is the name of an acid with the formula HNO2?
Step2247 [10]
HNO2 is the formula for Nitrous Acid
8 0
3 years ago
Which part of a feedback mechanism causes change to make up for the departure from the set point? A. sensor B. effector C. respo
koban [17]
I would have to say C. Response.
5 0
3 years ago
Read 2 more answers
Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =
Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Wherem  

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314 J / K mol  

Thus, given that, E_a = ?

k_2=4.0\times 10^3s^{-1}

k_1=1.5\times 10^3s^{-1}  

T_1=5\ ^0C  

T_2=25\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (5 + 273.15) K = 278.15 K  

T = (25 + 273.15) K = 298.15 K  

T_1=278.15\ K

T_2=298.15\ K

So,

\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)

E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}

E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}

E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}

E_a=33813.28\ J/mol

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0
2 years ago
In the titration of HCl with NaOH, the equivalence point is determined
kondaur [170]

Answer:

In the titration of HCl with NaOH, the equivalence point is determined from the point where the phenolphthalein turns pink and then remains pink on swirling.

Explanation:

The equivalence point is the point at which exactly enough titrant (NaOH) has been added to react with all of the analyte (HCl). Up to the equivalence point, the solution will be acidic because excess HCl remains in the flask.

Phenolphtalein is chosen because it changes color in a pH range between 8.3 – 10. Phenolphthalein is naturally colorless but turns pink in alkaline solutions. It remains colorless throughout the range of acidic pH levels, but it begins to turn pink at a pH level of 8.3 and continues to a bright purple in stronger alkalines.

It will appear pink in basic solutions and clear in acidic solutions.

The more NaOH added, the more pink it will be. (Until pH≈ 10)

In strongly basic solutions, phenolphthalein is converted to its In(OH)3− form, and its pink color undergoes a rather slow fading reaction and becomes completely colorless above 13.0 pH

a. from the point where the pink phenolphthalein turns colorless and then remains colorless on swirling.

⇒ the more colorless it turns, the more acid the solution. (More HCl than NaOH)

b. from the point where the phenolphthalein turns pink and then remains pink on swirling.

The equivalence point is the point where phenolphtalein turns pink and remains pink ( Between ph 8.3 and 10). (

Although, when there is hydrogen ions are in excess, the solution remains colorless. This begins slowely after ph= 10 and can be noticed around ph = 12-13

c. from the point where the pink phenolphthalein first turns colorless and then the pink reappears on swirling.

Phenolphthalein is colorless in acid solutions (HCl), and will only turn pink when adding a base like NaOH

d. from the point where the colorless phenolphthalein first turns pink and then disappears on swirling

Phenolphthalein is colorless in acid or neutral solutions. Once adding NaOH, the solution will turn pink. The point where the solution turns pink, and stays pink after swirling is called the equivalence point. When the pink color disappears on swirling, it means it's close to the equivalence point but not yet.

3 0
2 years ago
Cesium has a radius of 272 pm and crystallizes in a body-centered cubic structure. What is the edge length of the unit cell
olchik [2.2K]

Answer: Edge length of the unit cell = 628pm

Explanation: For a body centred cubic structured system, the relationship between the edge length of the unit cell and radius of the atoms in the structure is

Edge length of Unit cell (a) = (4R)/(√3)

R = 272pm = (272 × (10^-12))m = (2.72 × (10^-10))m

a = (4 × (2.72 × (10^-10)))/(√3)

a = (6.28157 × (10^-10))m = 628pm

4 0
2 years ago
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