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2 years ago

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I've been stuck on this assignment for 3 weeks now, help please.-Pics included-

1 answer:

Bond [772]2 years ago

3 0

1. 2NaN3 —> 2Na + 3N2
2. 2Pb + 2H2O + O2 —> 2Pb(OH)2
3. 2C4H12 + 13O2 —> 8CO2 + 10H2O
4. 4Fe + 3Sn(NO3)4 —> 4Fe(NO3)3 + 3Sn
5. Fe(NO3)3 + 3NaOH —> Fe(OH)3 + 3NaNO3

Balancing equations depends on the atom numbers you start with.

If you are having trouble you can look up ‘chemical equation balancer’ and type in the starting elements and simply replace the arrow with an = equal sign and then enter the reactants. It will show you the balanced equation
But please— make sure you actually understand the process of balancing because it’s important for future chem classes

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Determine the concentration of sulfuric acid that needed 47 mL of 0.39M potassium hydroxide solution to neutralize a 25 mL sampl

77julia77 [94]

Answer:

<u></u>

<u>0.37M</u>

Explanation:

Since sulfuric acid, H₂SO₄, is a diprotic acid and potassum hydroxide, KOH, contains one OH⁻ in the formula, the number of moles of potassium hydroxide must be twice the number of moles of sulfuric acid.

<u>1. Determine the number of moles of KOH in 47mL of 0.39M potassium hydroxide solution</u>

number of moles = molarity × volume in liters

number of moles = 0.39M × 47mL × 1liter/1,000 mL = 0.1833mol

<u>2. Determine the number of moles of sulfuric acid needed</u>

number of moles of H₂SO₄ = number of moles of KOH/2 = 0.1833/2 = 0.009165mol

<u>3. Determine the concentration that contains 0.009165 mol in 25mL of the acid.</u>

Molarity = number of moles / volume in liters

M = 0.009165mol/(25mL) × (1,000mL/liter) = 0.3666M

Round to two significant figures: 0.37M

7 0

3 years ago

How much energy would be released as an electron moved from the n=4 to the n=3 energy level?

Thepotemich [5.8K]

The energy released when electron move from n=4 to n=3 is 0.66 eV

We know that in an atom energy of nth state is

$E_n = -13.6/n^{2}$ eV

where n is the energy level

Therefore,

$E_4 = -13.6/4^{2} \\E_3 = -13.6/3^{2}$

Thus, $E_4$ = -0.85eV

$E_3$ = -1.51eV

Therefore, total mount of energy released in moving electron from n=4 to n=3 is given by -

$E_4 -E_3$

= -0.85 - ( -1.51)

= 0.66eV

To know more about energy released in electron transition

brainly.com/question/8384785

#SPJ4

8 0

2 years ago

Write a balanced chemical equation for the following: magnesium metal is burned in oxygen to produce magnesium oxide.

KiRa [710]

Answer:

2Mg (s) + $O_{2}$ (g) ---> 2MgO (s)

Explanation:

8 0

2 years ago

Need Answer ASAP. Scientists design drugs alongside which type of engineers? Structural, civil, chemical, aerospace

Scilla [17]

Answer:

Chemical

Explanation:

3 0

3 years ago

What is the total pressure of air in lungs of an individual with oxygen at 100. mmHg, nitrogen at 573 mmHg, carbon dioxide at 0.

ryzh [129]

Answer: The total pressure of air in lungs of an individual is 760.28 mm Hg

Explanation:

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_A+p_B+p_C...$

Given : $p_{total}$ =total pressure of gases = ?

$p_{O_2}$ = partial pressure of oxygen = 100 mm Hg

$p_{N_2}$ = partial pressure of nitrogen = 573 mm Hg

$p_{CO_2}$ = partial pressure of Carbon dioxide = 0.053 atm = 40.28 mm Hg(1 atm = 760 mmHg)

$p_{H_2O}$ = partial pressure of water vapor = 47 torr = 47 mm Hg (1torr=1 mm Hg)

putting in the values we get:

$p_{total}=(100+573+40.28+47)mmHg$

$p_{total}=760.28mmHg$

Thus the total pressure of air in lungs of an individual is 760.28 mm Hg

5 0

3 years ago