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CaHeK987 [17]
3 years ago
15

Which statement describes the composition of potassium chlorate, KClO3?

Chemistry
2 answers:
Aliun [14]3 years ago
5 0
The statement that best describes the composition of potassium chlorate, KCIO3 is "<span> The proportion by mass of elements combined in potassium chlorate is fixed."</span>
Lisa [10]3 years ago
4 0

Answer : The correct option is, (1) The proportion by mass of elements combined in potassium chlorate is fixed.

Explanation :

The composition of any compound is studied on the basis of law of fixed proportion. This law stats that in a compound the elements are present in fixed proportion by their mass.

In the given options, the option (1) statement describes the composition of potassium chlorate.

While the other options (2), (3) and (4) statements does not describe the composition of potassium chlorate.

Hence, the correct option is (1).

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Given the following balanced chemical reaction:
Gnom [1K]

Answer:

Explanation:

according to balance chemical equation

  3 A2 moles produced 2 moles of A3B

so 12 moles A2 will produced moles of A3B= 12*2/3=24/3= 8

therefore 12 moles of A2 produced 8 moles of A3B

6 0
3 years ago
Gastric juice is made up of substances secreted from parietal cells, chief cells, and mucous-secreting cells. The cells secrete
neonofarm [45]

Answer:

The amount of energy required to transport hydrogen ions from a cell into the stomach is 37.26KJ/mol.

Explanation:

The free change for the process can be written in terms of its equilibrium constant as:

ΔG° = -RTInK_(eq)

where:

R= universal gas constant

T= temperature

K_eq= equilibrum constant for the process

Similarly, free energy change and cell potentia; are related to each other as follows;

ΔG= -nFE°

from above;

F = faraday's constant

n = number of electrons exchanged in the process; and  

E = standard cell potential

∴ The amount of energy required for transport of hydrogen ions from a cell into stomach lumen can be calculated as:

ΔG° = -RTInK_(eq)

where;

[texK_eq[/tex]=\frac{[H^+]_(cell)}{[H^+(stomach lumen)]}

For transport of ions to an internal pH of 7.4, the transport taking place can be given as:

H^+_{inside} ⇒ H^+_{outside}

Equilibrum constant for the transport is given as:

K_{eq}=\frac{[H^+]_{outside}}{[H^+]_{inside}}

=\frac{[H^+]_{cell}}{[H^+]_{stomach lumen}}

[H^+]_{cell}= 10⁻⁷⁴

=3.98 * 10⁻⁸M

[H^+]_{stomach lumen} = 10⁻²¹

=7.94 * 10⁻³M

Hence;

K_{eq}=\frac{[H^+]_{cell}}{[H^+]_{stomachlumen}}

=\frac{3.98*10^{-8}}{7.94*10{-3}}

= 5.012 × 10⁻⁶

Furthermore, free energy change for this reaction is related to the equilibrium concentration given as:

ΔG° = -RTInK_(eq)

If temperature T= 37° C ; in kelvin

=37° C + 273.15K

=310.15K; and

R-= 8.314 j/mol/k

substituting the values into the equation we have;

ΔG₁ = -(8.314J/mol/K)(310.15)TIn(5.0126*10^{-6})

= 31467.93Jmol⁻¹

≅ 31.47KJmol⁻¹

If the potential difference across the cell membrane= 60.0mV.

Energy required to cross the cell membrane will be:

ΔG₂ = -nFE°_{membrane}

ΔG₂ = -(1 mol)(96.5KJ/mol/V)(60*10^{-3})

= 5.79KJ

Therefore, for one mole of electron transfer across the membrane; the energy required is 5.79KJmol⁻¹

Now, we  can calculate the total amount of energyy required to transport H⁺ ions across the membrane:

Δ G_{total} = G_{1}+G_{2}

= (31.47+5.79) KJmol⁻¹

= 37.26KJmol⁻¹

We can therefore conclude that;

   The amount of energy required to transport ions from cell to stomach lumen is 37.26KJmol⁻¹

5 0
3 years ago
What is the smallest substance? <br> 1 .mixture<br> 2 .compound<br> 3 .element
Andrew [12]
Number 3 it’s element
8 0
3 years ago
Read 2 more answers
An important step in science is supporting a theory or idea without data. The questions we ask determine the type of data we col
san4es73 [151]

Answer:

What will happen to Uk if you double the mass?

Explanation:

Uk = 0.5 * m * v²

You see that both m and v are variable, which means that both m and v can be any number. Regardless of the numbers you put in for m or v, the formula to calculate the kinetic energy (Uk) remains valid.

You could ask

1. What will happen to Uk if you double the mass?

2. What will happen to Uk if you double the velocity?

please see and understand(!) that the relationship between Uk an v² is indeed the velocity squared....

EXTRA

Uk = 0.5 * m * (v)²

Suppose the m = 3kg and velocity = 5 m/s

What is the Uk?

Well if you know the formula you can use your calculator to find out:

Uk = 0.5 * m * (v)²

Uk = 0.5 * 3 * (5)²

Uk = 0.5 * 3 * 25

Uk = 37.5 kgm/s²

Again you ask what will happen to Uk if you double the velocity?

At first it was 5 m/s and now it doubles, which means it now has that value *2

The new velocity is 5 *2 = 10 m/s

Uk = 0.5 * m * (v)²

Uk = 0.5 * 3 * (10)²

Uk = 0.5 * 3 * 100

Uk = 150 kgm/s²

150 = 4 * 37.5

So now you see that if you double your velocity, the Uk will be 2² = 4 times as big !

3 0
3 years ago
When the equation below is balanced, what is the coefficient for oxygen?<br> C3H8+O2=CO2+H2O
Nataly [62]
Se for no balanceamento e 5. 
5 0
3 years ago
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