Balance the following chemical equations:

29.6 g of Carbon (MM = 12.01 g/mol) is reacted with 88.2 g of TiO2 (MM= 79.87 g/mol). 42.9 grams of Titanium (MM = 47.87) are ob

Basile [38]

Answer:

MM=23.45

Explanation:

5 stars if i am right

6 0

3 years ago

Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79

BARSIC [14]

Answer : The mole fraction and partial pressure of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of $CH_4$ = 1.79 mole

Moles of $CO_2$ = 1.20 mole

Moles of $He$ = 3.71 mole

Now we have to calculate the mole fraction of $CH_4,CO_2$ and $He$ gases.

$\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267$

and,

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179$

and,

$\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554$

Thus, the mole fraction of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of $CH_4,CO_2$ and $He$ gases.

According to the Raoult's law,

$p_i=X_i\times p_T$

where,

$p_i$ = partial pressure of gas

$p_T$ = total pressure of gas = 5.78 atm

$X_i$ = mole fraction of gas

$p_{CH_4}=X_{CH_4}\times p_T$

$p_{CH_4}=0.267\times 5.78atm=1.54atm$

and,

$p_{CO_2}=X_{CO_2}\times p_T$

$p_{CO_2}=0.179\times 5.78atm=1.03atm$

and,

$p_{He}=X_{He}\times p_T$

$p_{He}=0.554\times 5.78atm=3.20atm$

Thus, the partial pressure of $CH_4,CO_2$ and $He$ gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0

3 years ago

Al colocar unos trocitos de zinc en un tubo de ensayo, agregamos 2 mL de acido clorhídrico. El Zinc sustituye el hidrogeno del a

Rudik [331]

Answer:

DICLORURO DE ZINC

Explanation:

La ecuación es:

Zn (s) + 2HCl (aq) → ZnCl₂ (aq) +H₂ (g)

El zinc reacciona en una reacción redox con el acido clorhídrico para formar una sal y liberar el hidrogeno gaseoso.

Se trata de una reacción redox porque el hidrógeno se reduce (disminuye el numero de oxidación, de -1 a 0, en su estado fundamental), mientras que el Zn se oxida (el numero de oxidación aumenta de 0 a +2).

Como la sal se forma con dos atomos de cloro, se denomina dicloruro por lo que el nombre correcto de la sal en nomenclatura sistemática es:

- DICLORURO DE ZINC

8 0

3 years ago

2. A 237 g piece of molybdenum, initially at 100.0 °C, was dropped into 244 g of

Sophie [7]

Answer:

$\large\boxed{C = 0.270J/g\textdegree C}$

Explanation:

1. Energy balance

By the first law of thermodynamics, considering the system is closed and isolated, the heat released by the 237 g piece of molybdenum equals the heat absorbed by the 244 g of water.

2. Heat equation

The heat released or absorbed by a substance is proportional to the product of the mass, the specific heat and the change in temperature

Q = m × C × ΔT

3. Heat released by the 237 g piece of molybdenum

At equilibrium:

Q₁ = 237 g × C × (100.0ºC - 15.3ºC)

4. Heat absorbed by the 244 g of water

At equilibrium:

Q₂ = 244 g × 4.184 J/gºC × (15.3ºC - 10.0ºC)

5. Solve for C from Q₁ = Q₂

237 g × C × (100.0ºC - 15.3ºC) = 244 g × 4.184 J/gºC × (15.3ºC - 10.0ºC)

20,073.9 × C = 5,410.7488 J/gºC

C = 0.2695 J/gºC

The result must be reported with 3 significant figures: C = 0.270 J/gºC.

7 0

3 years ago

For the set of ionic compounds, CaSO4, CaCO3, Ca(OH)2, choose the correct characterization of their solubilities in water from t

maxonik [38]

Answer:

None of the three salts are soluble.

Explanation:

According to the solubility rule, the carbonates and sulphates of group two elements are insoluble in water.

All three substances mentioned possess very low solubility in water and can be said to be slightly soluble in water. If we compare them with other ionic substances that dissolve readily in water, we can rightly say that they are insoluble in water.

Hence all three substances are insoluble in water.

7 0

4 years ago