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sertanlavr [38]
2 years ago
5

One of the basic building blocks of matter, which cannot be broken down by chemical means is a(n)______ , of which there are 92

that occur naturally.
Chemistry
1 answer:
Nadusha1986 [10]2 years ago
6 0

One of the basic building blocks of matter, which cannot be broken down by chemical means is an element.

The first chemical element is hydrogen (atomic number is 1) and the last is oganesson (atomic number is 118).

Elements are scheduled in Periodic table, ordered by their atomic number.

Other example, krypton is a chemical element with symbol Kr and atomic number 36.

Krypton has 36 electron and 36. It is noble gas (group 18).

Noble gases are in group 18: helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe) and radon (Rn). They have very low chemical reactivity.

More about chemical element: brainly.com/question/28376204

#SPJ4

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Write the formula for the polyatomic ion in KOH. Express as an ion..
Bad White [126]
KOH is a compound containing two ions, K+ and OH-. 
<span>The polyatomic ion present is OH- which is called hydroxide. </span>
<span>The compound is named potassium hydroxide.</span>
6 0
4 years ago
Read 2 more answers
A solution of NaOH is titrated with H2SO4. It is found that 20.05 mL of 0.3564 M H2SO4 solution is equivalent to 43.42 mL of NaO
Darya [45]

Answer : The concentration of NaOH is, 0.336 M

Explanation:

To calculate the concentration of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=0.3564M\\V_1=20.05mL\\n_2=1\\M_2=?\\V_2=43.42mL

Putting values in above equation, we get:

2\times 0.3564M\times 20.05mL=1\times M_2\times 43.42mL

M_2=0.336M

Thus, the concentration of NaOH is, 0.336 M

3 0
3 years ago
What is the mole ratio of N2 to H2 to NH3?
Alexeev081 [22]

Answer:

the answer is 1:3:2

Hope this helps, let me know if you need any other help, Stoichiometry is hard

4 0
4 years ago
Calculate the standard free-energy change at 25 ∘C for the following reaction:
lianna [129]

Answer:

Standard free-energy change at 25^{0}\textrm{C} is -3.80\times 10^{2}kJ/mol

Explanation:

Oxidation: Mg(s)-2e^{-}\rightarrow Mg^{2+}(aq.)

Reduction: Fe^{2+}(aq.)+2e^{-}\rightarrow Fe(s)

--------------------------------------------------------------------------------------

Overall: Mg(s)+Fe^{2+}(aq.)\rightarrow Mg^{2+}(aq.)+Fe(s)

Standard cell potential, E_{cell}^{0}=E_{Fe^{2+}\mid Fe}^{0}-E_{Mg^{2+}\mid Mg}^{0}

So, E_{cell}^{0}=(-0.41V)-(-2.38V)=1.97V

We know, standard free energy change at 25^{0}\textrm{C}(\Delta G^{0}): \Delta G^{0}=-nFE_{cell}^{0}

where, n is number of electron exchanged during cell reaction, 1F equal to 96500 C/mol

Here n = 2

So, \Delta G^{0}=-(2)\times (96500C/mol)\times (1.97V)=-380210J/mol=-380.21kJ/mol=-3.80\times 10^{2}kJ/mol

8 0
3 years ago
Read 2 more answers
8. __H2 + __O2-&gt; __H2O
Rus_ich [418]

Anser:

Explanation:

hope this helps

8 0
3 years ago
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