Identify the total number of molecules in the chemical formula: 6CH4 a 6 b 10 c 24 d 30

What is the mass in grams of 0.700 moles of water, H2O

irga5000 [103]

Answer:

12.6

Explanation:

$no \: of \: mole= \frac{mass}{molar \: mass}$

0.7=mass/(2+16)

8 0

2 years ago

Determine the heat of reaction for the process TiO2(s) + 4HCl(g) TiCl4(l) + 2H2(g) + O2(g) using the information given below: Ti

Aleks [24]

Answer : The heat of reaction for the process is, 1374.7 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The main chemical reaction is,

$TiO_2(s)+4HCl(g)\rightarrow TiCl_4(l)+2H_2(g)+O_2(g)$ $\Delta H_{rxn}=?$

The intermediate balanced chemical reaction will be,

(1) $Ti(s)+O_2(g)\rightarrow TiO_2(s)$ $\Delta H_1=-939.7kJ$

(2) $2HCl(g)\rightarrow H_2(g)+Cl_2(g)$ $\Delta H_2=-184.6kJ$

(3) $Ti(s)+2Cl_2(g)\rightarrow TiCl_4(l)$ $\Delta H_3=-804.2kJ$

We reversing reaction 1, 3 and multiplying reaction 2 by 2 and then adding all the equations, we get :

(1) $TiO_2(s)\rightarrow Ti(s)+O_2(g)$ $\Delta H_1=939.7kJ$

(2) $4HCl(g)\rightarrow 2H_2(g)+2Cl_2(g)$ $\Delta H_2=2\times (-184.6kJ)=-369.2kJ$

(3) $TiCl_4(l)\rightarrow Ti(s)+2Cl_2(g)$ $\Delta H_3=804.2kJ$

The expression for heat of reaction for the process is:

$\Delta H_{rxn}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H_{rxn}=(939.7kJ)+(-369.2kJ)+(804.2kJ)$

$\Delta H_{rxn}=1374.7kJ$

Therefore, the heat of reaction for the process is, 1374.7 kJ

3 0

3 years ago

Determine what mass of carbon monoxide and what mass of hydrogen are required to form 6.0 kg of methanol by the reaction CO(g) +

siniylev [52]

Answer:

5250 grams or 5.25 kg of carbon monoxide and 375 grams of hydrogen are required to form 6 kg of methanol.

Explanation:

The balanced reaction:

CO (g) + 2 H₂ (g) -> CH₃OH (l)

By stoichiometry of the reaction, the following amounts of moles of each compound participate in the reaction:

CO: 1 mole
H₂: 2 moles
CH₃OH: 1 mole

Being the molar mass of each compound:

CO: 28 g/mole
H₂: 1 g/mole
CH₃OH: 32 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

CO: 1 mole* 28 g/mole= 28 grams
H₂: 2 moles* 1 g/mole= 2 grams
CH₃OH: 1 mole* 32 g/mole= 32 grams

Being 6 kg equivalent to 6000 grams (1 kg= 1000 grams), you can apply the following rules of three:

If by stoichiometry 32 grams of methanol are formed from 28 grams of carbon monoxide, 6000 grams of methanol are formed from how much mass of carbon monoxide?

$mass of carbon monoxide=\frac{6000 grams of methanol*28 grams of carbon monoxide}{32 grams of methanol}$

mass of carbon monoxide= 5250 grams= 5.25 kg

If by stoichiometry 32 grams of methanol are formed from 2 grams of hydrogen, 6000 grams of methanol are formed from how much mass of hydrogen?

$mass of hydrogen=\frac{6000 grams of methanol*2 grams of hydrogen}{32 grams of methanol}$

mass of hydrogen= 375 grams

5250 grams or 5.25 kg of carbon monoxide and 375 grams of hydrogen are required to form 6 kg of methanol.

8 0

3 years ago

In 1961, the reference for the atomic mass unit (amu or u) was changed from naturally occurring oxygen with a value of 16.000 u

vovangra [49]

To determine the mass of an atom in terms of atomic mass unit (a.m.u.) the standard should have the fixed mass otherwise the calculated mass of the other atoms will get error. Initially oxygen was used as the reference standard for the calculation of other atoms mass. But after the invention of different isotopes of oxygen i.e. $O^{17}$ and $O^{18}$ which ratios with respect to $O^{16}$ is 1.002 the mass of standard oxygen became unstable. Thus the mass of the new atom have error. Thus in 1961 the standard reference is canged to carbon-12. After that onward the mass of the other atoms are measured with respect to carbon.

3 0

3 years ago

the ph of a solution of HCl in water is found to be 2.50. what volume of water would you add to 1.00L of this solution to raise

morpeh [17]

Answer:

To raise the pH of the solution to 3.10 we have to add 2.34 L of water.

Explanation:

Given that the pH of the solution of HCl in water is 2.5. Here the solution’s pH is changing from 2.5 to 3.10 which means the acidic nature of the solution is decreasing here on dilution. $[H^+]$ ions contribute to a solution’s acidic nature and $[OH^-]$ contribute to a solution’s basic nature.