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jenyasd209 [6]
2 years ago
9

In two or more complete sentences explain how to balance the chemical equation and classify its reaction type. ___P4 + ___O2 ⟶ _

__P2O3
Chemistry
2 answers:
Karolina [17]2 years ago
6 0

Answer:

In two or more complete sentences explain how to balance the chemical equation and classify its reaction type.

___P4 + ___O2 ⟶ ___P2O3

Explanation:

Serhud [2]2 years ago
3 0

According to the conversation of mass, mass cannot be created or destroyed. This means whatever is done to one side, must be done to the other.

There are 4 Phosphorus atoms on the left, there must be 4 on the right. To do this, you must multiply the P2O3 by 2 to get 4 Phosphorus atoms and 6 Oxygen atoms. Now to balance the Oxygen atoms, you must multiply the oxygen atoms on the left by 3.

1 P4 + 3 O2 —-> 2 P2O3

Lastly, this equation type is synthesis (combination) because two reactants are becoming a single product.

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The K w for water at 0 ∘ C is 0.12 × 10 − 14 M 2 . Calculate the pH of a neutral aqueous solution at 0 ∘ C . p H = Is a pH = 7.2
labwork [276]

Answer:

pH → 7.46

Explanation:

We begin with the autoionization of water. This equilibrium reaction is:

2H₂O  ⇄   H₃O⁺  +  OH⁻            Kw = 1×10⁻¹⁴         at 25°C

Kw = [H₃O⁺] . [OH⁻]

We do not consider [H₂O] in the expression for the constant.

[H₃O⁺] = [OH⁻] = √1×10⁻¹⁴   →  1×10⁻⁷ M

Kw depends on the temperature

0.12×10⁻¹⁴ = [H₃O⁺] . [OH⁻]  → [H₃O⁺] = [OH⁻]         at 0°C

√0.12×10⁻¹⁴ = [H₃O⁺] → 3.46×10⁻⁸ M

- log [H₃O⁺] = pH

pH = - log 3.46×10⁻⁸ → 7.46

5 0
2 years ago
Many drugs decompose in blood by a first-order process. two tablets of aspirin supply 0.60 g of the active compound. after 30 mi
Firdavs [7]
<span>The half-life of a first-order reaction is determined as follows: 

</span>t½<span>=ln2/k

From the equation, we can calculate the </span><span>first-order rate constant:

</span>k = (ln(2)) / t½ = 0.693 / 90 = 7.7 × 10⁻³

When we know the value of k we can then calculate concentration with the equation:

A₀ = 2 g/100 mL 

t = 2.5 h = 150min

A = A₀ × e^(-kt) =2 × e^(-7.7 × 10⁻³ × 150) = 0.63 g / 100ml

   = 6.3 × 10⁻⁴ mg / 100ml


3 0
2 years ago
Read 2 more answers
In May 2016, William Trubridge broke the world record in free diving (diving underwater without the use of supplemental oxygen)
Maru [420]

Answer:

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

Explanation:

Using Boyle's law  

{P_1}\times {V_1}={P_2}\times {V_2}

Given ,  

V₁ = 3.6 L  

V₂ = ?

P₁ = 1.0 atm

P₂ = 13.3 atm (From correct source)

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{1.0\ atm}\times {3.6\ L}={13.3\ atm}\times {V_2}

{V_2}=\frac{{1.0}\times {3.6}}{13.3}\ L

{V_2}=0.27\ L

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

7 0
2 years ago
Water flows over Niagara Falss at the average rate of 2,400,000 kg/s, and the average height of the falls is about 50 m. Knowing
liberstina [14]

Answer:

1. 176 × 10^12 W ; 78400000000

Explanation:

Given the following :

Fall rate = 2,400,000kg/s

Average height of fall = 50m

Gravitational Potential of falling water = mgh = mass × acceleration due to gravity × height =

How many 15 W LED light bulbs could it power?

Recall : power = workdone / time

Workdone = gravitational potential energy

Mass of water = density * volume

Density of water = 1 * 10^3kg/m^3

Rate of fow = volume / time = 2400000

Hence,

Power = 1000 * 2,400,000 * 9.8 * 50

Power = 1176000000000

Power = 1. 176 × 10^12 W

How many 15 W LED light bulbs could it power?

1176000000000 / 15 = 78400000000

= 78400000000 15 W bulbs

4 0
2 years ago
The pH of a 0.30 M solution of a weak acid is 2.67. What is the Ka for this acid?
vova2212 [387]

Answer:

Ka  → 1.5×10⁻⁵

Option E. None of the above

Explanation:

We propose the reaction of equlibrium

Weak ac.H +  H₂O ⇄  Weak ac⁻  +  H₃O⁺

Initially we have 0.30 moles of acid in 1 L

In equilibrium we would have:

Weak ac.H +  H₂O ⇄  Weak ac⁻  +  H₃O⁺

0.30 - x                               x               x

We have the pH, where we can obtanined the x, the [H₃O⁺] in the equilibrium.

pH = - log [H₃O⁺] → [H₃O⁺] = 10^⁻(pH)

[H₃O⁺] = 10⁻²'⁶⁷ = 2.14×10⁻³

So let's determine the concentration of the acid, in the equilibrium

0.30 - 2.14×10⁻³ = 0.29786 → [Weak ac.H]

2.14×10⁻³ →  [H₃O⁺] = Conjugate base (Weak ac.⁻)

Let's make the expression for Ka

Ka = [Weak ac.⁻] .  [H₃O⁺]  / [Weak ac.H]

Ka = x² / 0.30 - x

Ka = (2.14×10⁻³)² / 0.29786 → 1.5×10⁻⁵

6 0
2 years ago
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