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3 years ago

9

electric current is passed through water, it decomposes inti hydrogen and oxygen. write full balanced chemical equation for the

reaction

1 answer:

Tems11 [23]3 years ago

6 0

Answer:

2H2O ------------------] 2H2 + O2

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In descriptive scientific investigations, scientists often make observations to understand the interacting parts of a complex (b

wlad13 [49]

Answer:

Complex System

Explanation:

Given that, a descriptive scientific investigation is one of the three main types of investigation which formulates and quantify the natural phenomenon. This natural phenomenon oftentimes involves Complex System, such as microscopic organisms, thereby, scientists often make observations to understand the interacting parts of this COMPLEX SYSTEM

Hence, the right answer is a COMPLEX SYSTEM

8 0

3 years ago

A sample of 4.0 L of nitrogen, at 1.2 atmospheres, is transferred to a 12 L container.

Zielflug [23.3K]

Answer:

The pressure will be 0.4 atm.

Explanation:

The gas laws are a set of chemical and physical laws that allow determining the behavior of gases in a closed system. The parameters evaluated in these laws are pressure, volume, temperature and moles.

As the volume increases, the gas particles (atoms or molecules) take longer to reach the walls of the container and therefore collide with them less times per unit of time. This means that the pressure will be lower because it represents the frequency of collisions of the gas against the walls. In this way pressure and volume are related, determining Boyle's law which says:

"The volume occupied by a certain gaseous mass at constant temperature is inversely proportional to pressure"

Boyle's law is expressed mathematically as:

P*V= k

If you initially have the gas at a volume V1 and press P1, when the conditions change to a volume V2 and pressure P2, the following is satisfied:

P1*V1= P2*V2

In this case:

P1= 1.2 atm
V1= 4 L
P2= ?
V2= 12 L

Replacing:

1.2 atm* 4 L= P2* 12 L

Solving:

$P2=\frac{1.2 atm*4 L}{12 L}$

P2= 0.4 atm

<u><em>The pressure will be 0.4 atm.</em></u>

7 0

3 years ago

Blank c2h4 + blank o2 → blank co2 + blank h2o how many moles of o2 are in the chemical equation when balanced using the lowest w

umka21 [38]

The reaction is<span>
C</span>₂H₄ + O₂ → CO₂ + H₂O<span>

To balance the equation, both side have same number of elements. Here,</span>

In left hand side has in right hand side has

4 H atoms 2 H atoms

2 C atoms 1 C atom

<span> 2 O atoms 3 O atoms

First, we have to balance number of C atoms and number of H atoms in both side.
To balance C atoms, '2' should be added before CO</span>₂ and to balance H atoms, '2' should be added before H₂<span>O.
Then number of oxygen atoms is </span>2 x 2 + 2 = 6 in right hand side. So, 3 should be added before O₂<span> in left hand side.
After balancing the equation should be,</span>

C₂H₄<span> + 3O</span>₂<span> → 2CO</span>₂<span> + 2H</span>₂O

8 0

3 years ago

Can someone help Plz

Marrrta [24]

Answer: A. 1.60 liters

Explanation:

$2Na_2O_2+2CO_2\rightarrow 2Na_2CO_3+O_2$

As can be seen from the balanced chemical equation, 2 moles of $CO_2$ produce 1 mole of $O_2$ .

According to Avogadro's law, every 1 mole of the gas occupies 22.4 liters at STP.

Thus $2$ moles of $CO_2$ occupies $22.4\times 2=44.8L$ at STP and produce 1 mole of $O_2$ i.e. 22.4 L

44.8 L of $CO_2$ will produce 22.4 L of $O_2$

Thus $3.20L$ of $CO_2$ will produce $\frac{22.4}{44.8}\times 3.20=1.60L$

6 0

3 years ago

Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka is given

tekilochka [14]

Answer:

a) = 0.704%

b) = 1.30%

c) = 2.60%

Explanation:

Given that:

$K_a$ = $1.34*10^{-5$

For Part A; where Concentration of A = 0.270 M

Percentage Ionization(∝) $\alpha = \sqrt{\frac{K_a}{C} }$

$\alpha = \sqrt{\frac{1.34*10^{-5}}{0.270} }$

$\alpha = \sqrt{4.9629*10^{-5}}$

$\alpha = 7.044*10^{-3$

percentage% (∝) = $7.044*10^{-3}*100$

= 0.704%

For Part B; where Concentration of B = $7.84*10^{-2$ M

$\alpha = \sqrt{\frac{1.34*10^{-5}}{7.84*10^{-2}} }$

$\alpha = \sqrt{1.709*10^{-4} }$

$\alpha = 0.0130\\$

percentage% (∝) = 0.0130 × 100%

= 1.30%

For Part C; where Concentration of C= $1.92*10^{-2} M$

$\alpha = \sqrt{\frac{1.34*10^{-5}}{1.97*10^{-2}} }$

$\alpha = \sqrt{6.802*10^{-4}}$

$\alpha =0.02608$

percentage% (∝) = 0.02608 × 100%

= 2.60%

7 0

3 years ago