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Naily [24]
2 years ago
8

which of the following correctly shows a prefix used in naming a binary molecular compound along with its corresponding number?

Penta-,6; hexa-,4; deca-,8; nona-,9
Chemistry
2 answers:
anyanavicka [17]2 years ago
4 0

Answer : The correct option is, nona-, 9

Explanation :

The naming of binary molecular compound is given by:

(1) The first element or less electronegative element is written first.

(2) The second element or more electronegative element is written then, and a suffix is added with it. The suffix added is '-ide'.

(3) If atoms of an element is greater than 1, then prefixes are added which are 'mono' for 1 atom, 'di' for 2 atoms, 'tri' for 3 atoms and so on..

(4) The prefixes used as 'meth' for 1, 'eth' for 2, 'prop' for 3, 'but' for 4, 'pent' for 5, 'hex' for 6, 'sept' for 7, 'oct' for 8, 'nona' for 9 and 'deca' for 10.

Hence, the correct option is, nona-, 9

aivan3 [116]2 years ago
3 0
The answer is Nona-9, penta- is five, hexa- is six, and deca- is ten.
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The volume of a sample gas, initially at 25 C and 158 mL, increased to 450 mL. What is the final temperature of the sample of ga
Rashid [163]

Answer:

Final temperature of the gas is  576 ^{0}\textrm{C}.

Explanation:

As the amount of gas and pressure of the gas remains constant therefore in accordance with Charles's law:

                                       \frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}

where V_{1} and V_{2} are volume of gas at T_{1} and T_{2} temperature (in kelvin scale) respectively.

Here V_{1}=158mL , T_{1}=(273+25)K=298K and V_{2}=450mL

So  T_{2}=\frac{V_{2}T_{1}}{V_{1}}=\frac{(450mL)\times (298K)}{(158mL)}=849K 

849 K = (849-273) ^{0}\textrm{C} = 576 ^{0}\textrm{C}

So final temperature of the gas is  576 ^{0}\textrm{C}.

3 0
3 years ago
How many milliliters of 0.0050 N KOH are required to neutralize 41 mL of 0.0050 M H2SO4?
Kipish [7]

Answer:

V KOH = 41 mL

Explanation:

for neutralization:

  • ( V×<em>C </em>)acid = ( V×<em>C </em>)base

∴ <em>C </em>H2SO4 = 0.0050 M = 0.0050 mol/L

∴ V H2SO4 = 41 mL = 0.041 L

∴ <em>C</em> KOH = 0.0050 N = 0.0050  eq-g/L

∴ E KOH = 1 eq-g/mol

⇒ <em>C</em> KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L

⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH

⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)

⇒ V KOH = 0.041 L

4 0
2 years ago
Sodium carbonate reacts with silver nitrate according to the following balanced equation: Na2CO3 (s) + 2 AgNO3 (aq) → Ag2CO3 (s)
klemol [59]

Answer:

a) 2.01 g

Explanation:

  • Na₂CO₃ (s) + 2AgNO₃ (aq) → Ag₂CO₃ (s) + 2NaNO₃

First we <u>convert 0.0302 mol AgNO₃ to Na₂CO₃ moles</u>, in order to <em>calculate how many Na₂CO₃ moles reacted</em>:

  • 0.0302 mol AgNO₃ * \frac{1molNa_2CO_3}{2molAgNO_3}  = 0.0151 mol Na₂CO₃

So the remaining Na₂CO₃ moles are:

  • 0.0340 - 0.0151 = 0.0189 moles Na₂CO₃

Finally we <u>convert Na₂CO₃ moles into grams</u>, using its <em>molar mass</em>:

  • 0.0189 moles Na₂CO₃ * 106 g/mol = 2.003 g Na₂CO₃

The closest answer is option a).

8 0
3 years ago
Acetic acid has a Kb of 2.93 °C/m and a normal boiling point of 118.1 °C. What would be the boiling point of a solution made by
alexdok [17]

Boiling point elevation is given as:

ΔTb=iKbm

Where,

ΔTb=elevation in the boiling point

that is given by expression:

ΔTb=Tb (solution) - Tb (pure solvent)

Here Tb (pure solvent)=118.1 °C

i for CaCO3= 2

Kb=2.93 °C/m

m=Molality of CaCO₃:

Molality of CaCO₃=Number of moles of CaCO₃/ Mass of solvent (Kg)

=(Given Mass of CaCO3/Molar mass of CaCO₃)/ Mass of solvent (Kg)

=(100.0÷100 g/mol)/0.4

= 2.5 m

So now putting value of m, i and Kb in the boiling point elevation equation we get:

ΔTb=iKbm

=2×2.93×2.5

=14.65 °C

boiling point of a solution can be calculated:

ΔTb=Tb (solution) - Tb (pure solvent)

14.65=Tb (solution)-118.1

Tb (solution)=118.1+14.65

=132.75

3 0
3 years ago
Help me please guys :)
nika2105 [10]

Explanation:

1. Methane

2. Diamond

3. not sure : )

6 0
3 years ago
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