All categories

1 year ago

Part B

Chemistry

1 answer:

dezoksy [38]1 year ago

3 0

Answer:

5.83g C4H10 x (1 mol C4H10/58.05 g (molar mass of C4H10) x (10 mol H2O/ 2 mol C4H10) x (18.002 g H2O (molar mass of H2O)/ 1 mol H2O=

Answer: 9.04 g of H2O

Explanation:

First set up equation: C4H10 (g)+ O2(g) -> CO2(g) + H2O(g)

Next balance it: 2C4H10 (g)+ 13O2(g) -> 8CO2(g) + 10H2O (g)

Use equation to get moles and plug given

5.83g C4H10 x (1 mol C4H10/58.05 g (molar mass of C4H10) x (10 mol H2O/ 2 mol C4H10) x (18.002 g H2O (molar mass of H2O)/ 1 mol H2O

You might be interested in

What charge does 3protons 3neutrons 2electrons

cluponka [151]

The charge that 3 protons, 3 neutrons and 2 electrons have is that it will be a cation of Li with a charge of +1.

Li^+1.

6 0

3 years ago

Calculate the solubility of benzene in water at 25 c in ppm. the required henry's law constant is 5.6 bar/mol/kg and benzene's s

kap26 [50]

The relationship between pressure and solubility of the gas is given by Henry's law as:

$S_g = kP_g$

where,

$S_g$ is the solubility of the gas.

$k$ is proportionality constant i.e. Henry's constant.

$P_g$ is pressure of the gas.

$k = 5.6 bar/mol/kg$ (given)

$P_g = 0.13 bar$ (given)

Substituting the values,

$S_g = 5.6 bar/mol/kg\times 0.13 bar = 0.728 mole/kg$

To convert $mole/kg$ to $g/kg$ :

Molar mass of benzene, $C_6H_6$ = $6\times 12+6\times 1 = 78 g/mol$

$0.728\times 78 = 56.784 g/kg$

Now for converting into $ppm$ :

Since, $1 ppm = 0.001 g/kg$

So, $56.784\times 1000 = 56784 ppm$ .

Hence, the solubility of benzene in water at $25^{o} C$ in $ppm$ is $56784 ppm$ .

7 0

3 years ago

Classify each element. Note that another term for main group is representative, another term for semimetal is metalloid, and the

NikAS [45]

The question is incomplete, here is the complete question:

Classify each element. Note that another term for main group is representative, another term for semi-metal is metalloid, and the inner transition metals are also called the lanthanide and actinide series.

Hf, Am, In, Ta, As, Se, Rn

Answer:

Hafnium and tantalum are transition elements.

Americium is a inner transition element.

Indium, Selenium and Radon are main group elements.

Arsenic is a metalloid.

Explanation:

Main group elements are the elements which belong to s block and p block. They are also known as representative elements.

S-block elements are defined as the elements whose last electron enters s-sub shell. The general electronic configuration of these elements is $ns^{1-2}$

P-block elements are defined as the elements whose last electron enters p-sub shell. The general electronic configuration of these elements is $np^{1-6}$

Metalloids are defined as the elements which show intermediate properties between metals and non-metals. There are 7 metalloids in the periodic table. They are: Boron, Silicon, germanium, Arsenic, Antimony, Tellurium and Polonium.

Transition elements are known as d-block elements. D block elements are defined as the elements whose last electron enters d sub shell. The general electronic configuration of these elements is $[(n-1)d^{1-10}ns^{0-2}]$

Inner transition elements are known as (f block) elements. (F block) elements are defined as the elements whose last electron enters (f subshell). The general electronic configuration of these elements is $[(n-2)f^{1-14}(n-1)d^{0-1}ns^{2}]$ . They are also known as lanthanide and actinide series.

For the given elements:

Option 1: Hf

Hafnium is the 72nd element of the periodic table having electronic configuration of $[Xe]4f^{14}5d^26s^2$

As, the last electron is entering the d subshell, it is a transition element.

Option 2: Am

Americium is the 95th element of the periodic table having electronic configuration of $[Rn]5f^{7}6d^07s^2$

As, the last electron is entering the (f subshell), it is a inner transition element.

Option 3: In

Indium is the 49th element of the periodic table having electronic configuration of $[Kr]5s^25p^1$

As, the last electron is entering the p subshell, it is a main group element.

Option 4: Ta

Tantalum is the 73rd element of the periodic table having electronic configuration of $[Xe]4f^{14}5d^56s^2$

As, the last electron is entering the d subshell, it is a transition element.

Option 5: As

Arsenic is the 33rd element of the periodic table having electronic configuration of $[Ar]4s^24p^3$

As, the last electron is entering the p subshell, it is a main group element. It shows an intermediate property of metal and non-metal. Thus, it is a metalloid.

Option 6: Se

Selenium is the 34th element of the periodic table having electronic configuration of $[Ar]4s^24p^4$

As, the last electron is entering the p subshell, it is a main group element.

Option 7: Rn

Radon is the 86th element of the periodic table having electronic configuration of $[Xe]4f^{14}5d^{10}6s^26p^6$

As, the last electron is entering the p subshell, it is a main group element.

5 0

3 years ago

6 At STP, which element is a good conductor of electricity?(1) chlorine (3) silver

alexgriva [62]

For an object to conduct electricity it should have free or delocalised electrons that are free to pass the charge and hence take part in conducting electricity.
From the given choices
Chlorine is a halogen existing as a diatomic gas. Iodine too is a halogen and 2 Iodine atoms held together by covalent bond. Cl - Cl bonds and I-I bonds are covalent bonds. the outer electrons of Cl and I take part in covalent bonds therefore they are fixed and not free to move about. therefore no free electrons to conduct electricity.
Sulfur is a solid that too is held together by covalent bonds so it does not have free electrons to conduct electricity.

Silver is a metal and a general property of metals are their ability to conduct electricity.
metal structure are metal ions tightly packed together. when the metal atoms are tightly packed their valence electrons are removed and delocalised. Positively charged metal ions are embedded in a sea of delocalised electrons.
therefore there are delocalised electrons that can conduct electricity

answer is 3) silver

3 0

3 years ago

Read 2 more answers

What is the molar ratio for the following equation after it has been properly balanced?

slava [35]

The answer
the molar ratio for the following equation
____C3H8 + ____O2 Imported Asset ____CO2 + ____ H2O

after it has been properly balanced:
__1_C3H8 + ____5O2 Imported Asset ____3CO2 + ____ 4H2O

proof:
number of C =3 (C3H8; 3CO2)
number of H =8 (C3H8 ; 4H2O)
number of O = 10(5x2) or (2x3+4) (5O2;4H2O)

the answer is
Reactants: C3H8 = 1, O2 = 8; Products: CO2 = 3 and H2O = 4

4 0

2 years ago