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podryga [215]
2 years ago
15

Y

Chemistry
1 answer:
Sindrei [870]2 years ago
7 0

Answer:

TIX. y) -> (5x 271

Explanation:

You might be interested in
Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6
liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 137ml+137ml=274ml

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 274 g

T_{final} = final temperature of water = 325.8 K

T_{initial} = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

q=274g\times 4.18J/g^oC\times (325.8-298)K

q=31839.896J=31.84KJ

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

3 0
3 years ago
What mass in grams would 5.7L of hydrogen gas occupy at STP?​
tekilochka [14]

Answer:  The correct answer is:  " 0.54 g " .

__________________________________________

Explanation:

Note that "hydrogen gas" is:  

H₂ (g)  ;   that is:  a "diatomic element" (diatomic gas) ;

_________________________________________

The molecular weight of "H" is:  1.00794 g ;

   (From the Periodic Table of Elements).

So, the molecular weight of:  H₂ (g)  is:

    " 1.00794 g * 2  = 2.01588 g ; {use calculator) ;

_________________________________________

Note the conversion for a gas at STP:

______

  1 mol of a gas = 22.4 L gas;

___

i.e. " 1 mol / 22.4 L " ;

____

So:     " 5.7 L H₂ (g)  *  \frac{1 mol H_{2} }{22.4 L} *\frac{2.01588 g}{mol} =? ;

The "L" ("literes" cancel out to "1" ;  since "L/L = 1 ;

The "mol" (moles) cancel out to "1" ; since "mol/mol = 1 ;

____

and we are left with:

____

 [5.7 * 2.104588 g ] / 22.4  =  ?  g ;

______________________

→ [ 11.9961516  g ] / 22.4 =

          0.53554248214  g ;l

_____________________________

We round this value to:  " 0.54 g " ;

 → since "5.7 L " has 2 (two) significant figures;  

     22.4 is an exact number conversion;

     and "5.7 L" has fewer significant figures than:

    " 2.104588 " ; or:  " 1.00794 " .

  → as such: We round to "2 (two) significant figures."

______________________________

Hope this is helpful.  Wishing you the best in your academic endeavors!

_______________________________

8 0
2 years ago
If we start with 8000 atoms of radium-226 how much would remain after 3200 years?
olya-2409 [2.1K]
<h3>Answer:</h3>

2000 atoms

<h3>Explanation:</h3>

We are given the following;

Initial number of atoms of radium-226 as 8000 atoms

Time taken for the decay 3200 years

We are required to determine the number of atoms that will remain after 3200 years.

We need to know the half life of Radium

  • Half life is the time taken by a radio active material to decay by half of its initial amount.
  • Half life of Radium-226 is 1600 years
  • Therefore, using the formula;

Remaining amount = Original amount × 0.5^n

where n is the number of half lives

n = 3200 years ÷ 1600 years

 = 2

Therefore;

Remaining amount = 8000 atoms × 0.5^2

                                = 8000 × 0.25

                                 = 2000 atoms

Thus, the number of radium-226 that will remain after 3200 years is 2000 atoms.

6 0
3 years ago
Who is dalton and what is daltons theory
ludmilkaskok [199]

Answer:

Dalton's atomic theory proposed that all matter was composed of atoms, indivisible and indestructible building blocks. While all atoms of an element were identical, different elements had atoms of differing size and mass.

7 0
2 years ago
Which statements about moon phases In a lunar month are true?
saveliy_v [14]

Answer:

1. The same pattern of phases repeats monthly

2. Waxing moon phases are visible between the new and full moon

6. A full moon appears toward the middle of a lunar month

Explanation:

Moon has eight phases which is used by the ancient people for the identification of date of the month. Phases of moon tells us about the starting, middle and end of the month. At the starting of the month, waxing crescent appears while at the middle of the month, the moon gains its full shape. Phases which occurs between new and full moon is known as waxing.

6 0
2 years ago
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