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For the following reaction, 35.4 grams of zinc oxide are allowed to react with 6.96 grams of water . zinc oxide(s) + water(l) --

IRISSAK [1]

Answer:

$m_{Zn(OH)_2}=38.4g$

Explanation:

Hello!

In this case, for the undergoing chemical reaction:

$ZnO(s)+H_2O(l)\rightarrow Zn(OH)_2$

We evaluate the yielded moles of zinc hydroxide by each reactant as shown below:

$n_{Zn(OH)_2}^{by ZnO}=35.4gZnO*\frac{1molZnO}{81.38gZnO}*\frac{1molZn(OH)_2}{1molZnO} =0.435molZn(OH)_2\\\\n_{Zn(OH)_2}^{by H_2O}=6.96gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{1molZn(OH)_2}{1molH_2O} =0.386molZn(OH)_2$

In such a way, since the water yields a smaller amount of zinc hydroxide we conclude it is the limiting reactant so the maximum mass is computed below:

$m_{Zn(OH)_2}=0.386molZn(OH)_2*\frac{99.424 gZn(OH)_2}{1molZn(OH)_2} \\\\m_{Zn(OH)_2}=38.4g$

Because the water limits the yielded amount of zinc hydroxide.

Best regards!

5 0

3 years ago

Why is salt water intrusion a big problem in florida?

Schach [20]

Its leading to contaminated drinking water.

7 0

3 years ago

Read 2 more answers

Two objects are brought into contact Object 1 has mass 0.76 kg, specific heat capacity 0.87) g'c and initial temperature 52.2 'C

taurus [48]

Answer:

$T_F=77.4\°C$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible to set up the following energy equation for both objects 1 and 2:

$Q_1=-Q_2$

In terms of mass, specific heat and temperature change is:

$m_1C_1(T_F-T_1)=-m_2C_2(T_F-T_2)$

Now, solve for the final temperature, as follows:

$T_F=\frac{m_1C_1T_1+m_2C_2T_2}{m_1C_1+m_2C_2}$

Then, plug in the masses, specific heat and temperatures to obtain:

$T_F=\frac{760g*0.87\frac{J}{g\°C} *52.2\°C+70.7g*3.071\frac{J}{g\°C}*154\°C}{760g*0.87\frac{J}{g\°C} +70.7g*3.071\frac{J}{g\°C}} \\\\T_F=77.4\°C$

Yet, the values do not seem to have been given correctly in the problem, so it'll be convenient for you to recheck them.

Regards!

4 0

3 years ago

Pls this need this I am doing a test and I need to get this

Alexxx [7]

Answer:

nib7tttugi

Explanation:

ilbbu

6 0

3 years ago

The addition of an inert gas has no effect on the equilibrium position of a gaseous reaction because ______. Multiple choice que

boyakko [2]

Inert gas does not affect the equilibrium position:

It is because the partial pressures of the reaction components remain the same.

What is Inert Gas?

Under a given set of conditions, an inert gas is a gas that does not undergo chemical reactions.
The noble gases (helium, neon, argon, krypton, xenon, and radon) were previously known as "inert gases" due to their perceived lack of involvement in any biochemical processes.
Because inert gases are non-reactive, they do not affect equilibrium partial pressures and thus do not affect volume.
An inert gas does not react with the reactants or products; it does not change the concentration of the products and reactants. Furthermore, because the volume is constant, the concentrations are unaffected. As a result, this does not affect equilibrium.

The equilibrium position won't change if an inert gas is added. A volume change won't change the equilibrium position if the total moles of gas in the products and reactants are the same. When the volume is reduced, the process changes to create fewer moles of gas.

Learn more about the inert gas here,

brainly.com/question/15909389

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5 0

2 years ago

I WILL MARK BRAINLIEST!!!!! !! (picture included with one question ^^^^^^^)