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3 years ago

Solve these Please. (15 points)

Chemistry

1 answer:

Ghella [55]3 years ago

3 0

Answer:

1) 4Fe + 3O2 → 2Fe2O3

2) H2 + Cl2 → 2HCl

3) 2Ag + H2S → Ag2S + H2

4) CH4 + 2O2 → CO2 + 2H2O

5) 2HgO → 2Hg + O2

6) 2Co + 3H2O → Co2O3 + 3H2

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In what sequence do electrons fill the atomic orbitals within a sub level

harina [27]

Each orbital must contain a single electron before any orbital contains two electrons.

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3 years ago

You believe you have cracked a secret code that uses elemental symbols to spell words. The code uses numbers to designate the el

Stells [14]

Explanation:

Atomic numbers for neutral atoms are the same as the number of electrons. The number of electrons are used in writing the subshell notation where we can see the highest princpal quantum number(n).

In order to solve this problem, a good knowledge of the periodic table and shell notation is required.

For example;

Given a number 5;

We know the number is made of both the atomic number of the element and the highest quantum number.

It is easier for us to know the highest quantum number from the shell notation:

shell notation 1s²2s²2p¹

highest principal quantum number is 2

We are left with atomic number 5-2, which is 3

The element is therefore Li.

Part A. 5, 30, 58, 56, 99, 79, 19, 98, 9

Li Fe I Sb U Ta S Pa N

Part B. 9, 99, 30, 95, 19, 47, 79

N U Fe Ra S Mo Ta

Learn more:

s-orbital brainly.com/question/9288609

#learnwithBrainly

4 0

3 years ago

Which of the following compounds does NOT contain molecules?<br> O CO2<br> ОН2<br> O Naci<br> OH20

wel

Answer:

Explanation:

NaCl does not contain molecules

7 0

3 years ago

Read 2 more answers

What is the answer to it

mr_godi [17]

what is the answer to what there is nothing here

8 0

3 years ago

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H2SO4(aq)? For your ans

Alik [6]

Question is incomplete, complete question is;

A 34.8 mL solution of $H_2SO_4$ (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

$H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)$

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of $H_2SO_4(aq)$ ? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

Answer:

0.044 M is the molarity of $H_2SO_4$ (aq).

Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, 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 $H_2SO_4$

$n_1,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=?\\V_1=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL$

Putting values in above equation, we get:

$2\times M_1\times 34.8 mL=1\times 0.15 M\times 20.4 mL\\\\M_1=\frac{1\times 0.15 M\times 20.4 mL\times 10}{2\times 34.8 mL}=0.044 M$

0.044 M is the molarity of $H_2SO_4$ (aq).

4 0

3 years ago