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DedPeter [7]
3 years ago
12

What is the answer, with the correct number of decimal places, for this problem? 4.392 g+ 102.40 g + 2.51 g=

Chemistry
1 answer:
blondinia [14]3 years ago
4 0
The correct answer is C
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Hiii pls help me to write out the ionic equation ​
emmasim [6.3K]

Answer:

<u>STEP I</u>

This is the balanced equation for the given reaction:-

2KOH_{(aq)} + H_2SO_4{}_{(aq)}   \rightarrow K_2SO_4{}_{(aq)} + 2H_2O_{(l)}

<u>STEP II</u>

The compounds marked with (aq) are soluble ionic compounds. They must be

broken into their respective ions.

see, in the equation KOH, H2SO4, and K2SO4 are marked with (aq).

On breaking them into their respective ions :-

  • 2KOH -> 2K+ + 2OH-
  • H2SO4 -> 2H+ + (SO4)2-
  • K2SO4 -> 2K+ + (SO4)2-

<u>STEP III</u>

Rewriting these in the form of equation

\underline{\pmb{2K^+} }+ 2OH^- + 2H^+ + \pmb{\underline{{SO_4{}^{2-}}} \: \rightarrow \:  \underline{\pmb{2K^+}}} + \underline{\pmb{SO_4{}^{2-}}} + 2H_2O

<u>STEP </u><u>IV</u>

Canceling spectator ions, the ions that appear the same on either side of the equation

<em>(note: in the above step the ions in bold have gotten canceled.)</em>

\boxed{ \mathfrak{ \red{ 2OH^-{}_{(aq)} + 2H^+{(aq.)} \rightarrow H_2O{}_{(l)}}}}

This is the net ionic equation.

____________________________

\\

\mathfrak{\underline{\green{ Why\: KOH \:has\:  been\: taken\: as\: aqueous ?}}}

  • KOH has been taken as aqueous because the question informs us that we have a solution of KOH. by solution it means that KOH has been dissolved in water before use.

[Alkali metal hydroxides are the only halides soluble in water ]

4 0
2 years ago
Suppose you have 75 gas-phase molecules of methanol (CH3OH) at T = 470 K. These molecules are contained in a spherical container
Katarina [22]

Answer:

The average pressure in the container due to these 75 gas molecules is P=9.72 \times 10^{-16} Pa

Explanation:

Here Pressure in a container is given as

P=\frac{1}{3} \rho

Here

  • P is the pressure which is to be calculated
  • ρ is the density of the gas which is to be calculated as below

                                         \rho =\frac{mass}{Volume of container}

        Here

                mass is to be calculated for 75 gas phase molecules as

                      m=n_{molecules} \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=75 \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=3.98 \times  10^{-21} g

              Volume of container is 0.5 lts

     So density is given as

                         \rho =\frac{mass}{Volume of container}\\\rho =\frac{3.98 \times 10^{-21} \times 10^{-3} kg}{0.5 \times 10^{-3} m^3}\\\rho =7.97 \times 10^{-21}\, kg/m^3

  • is the mean squared velocity which is given as

                                        =RMS^2

      Here RMS is the Root Mean Square speed given as 605 m/s so

                                      =RMS^2\\=(605)^2\\=366025

Substituting the values in the equation and solving

P=\frac{1}{3} \rho \\P=\frac{1}{3} \times 7.97 \times 10^{-21} \times 366025\\P=9.72 \times 10^{-16} Pa

So the average pressure in the container due to these 75 gas molecules is P=9.72 \times 10^{-16} Pa

6 0
3 years ago
Which groups of the periodic table want to gain electrons? What kind(s) of elements are these groups
lukranit [14]

Answer:

Elements that are metals tend to lose electrons and become positively charged ions called cations. Elements that are nonmetals tend to gain electrons and become negatively charged ions called anions. Metals that are located in column 1A of the periodic table form ions by losing one electron.

Explanation:

6 0
3 years ago
Na and Cl are chemical symbol
Sliva [168]

Answer:

Cl is chloride I think, and NaCl is a salt molecule

8 0
3 years ago
Which of the following is an example of a micro-habitat?
Anna11 [10]
I believe it is d, a microhabitat is a small area which differs somehow from the surrounding habitat. Its unique conditions may be home to unique species that may not be found in the larger region.
3 0
3 years ago
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