A piece of sodium metal can be described as

John Dalton believed which of the following about atoms?

Snowcat [4.5K]

Answer : Option B) All atoms of a single substance are identical.

Explanation : The scientist John Dalton proposed the atomic theory which had the postulates as follows.

i) All matter/substances consists of indivisible particles known as atoms.

ii) Atoms of the same element/substance are similar in mass,shape and size, but differ from the atoms of other elements.

iii) Atoms obey the law of conservation of energy which says atoms cannot be created or destroyed.

iv) Atoms of different elements may combine with each other in a fixed, simple, whole number ratios to form any compound atoms.

v) Atoms of same element can combine in ratio with more than one to form two or more compounds.

vi) The atom is considered to be the smallest unit of matter that can take part in a chemical reaction

8 0

2 years ago

A sample of hydrated magnesium sulfate (MgSO4)

yaroslaw [1]

Answer:

MgSO4.7H2O

Explanation:

Let the formula for the hydrated magnesium sulphate be MgSO4.xH2O

Mass of the hydrated salt (MgSO4.xH2O) = 12.845g

Mass of anhydrous salt (MgSO4) = 6.273g

Mass of water molecule(xH2O) = Mass of the hydrated salt — Mass of anhydrous salt = 12.845 — 6.273 = 6.572g

Now,we can obtain the number of mole of water molecule present in the hydrated salt as follows:

Molar Mass of hydrated salt (MgSO4.xH2O) = 24 + 32 + (16x4) + x(2 + 16) = 24 + 32 + 64 + x(18) = 120 + 18x

Mass of xH2O/ Molar Mass of MgSO4.xH2O = Mass of water / mass of hydrated salt

18x/120 + 18x = 6.572/12.845

Cross multiply to express in linear form

18x x 12.845 = 6.572(120 + 18x)

231.21x = 788.64 + 118.296x

Collect like terms

231.21x — 118.296x = 788.64

112.914x = 788.64

Divide both side by 112.914

x = 788.64 /112.914

x = 7

Therefore the formula for the hydrated salt (MgSO4.xH2O) is MgSO4.7H2O

6 0

2 years ago

Is color change a physical or chemical change?

bagirrra123 [75]

It's a chemical change because it's changing the object's property.

3 0

2 years ago

Consider this reaction at equilibrium at a total pressure P1: 2SO2(g) + O2(g) â†’ 2SO3(g) Suppose the volume of this system is c

oksian1 [2.3K]

Answer:

The new equilibrium total pressure will be increased to one-half to initial total pressure.

Explanation:

From the information given :

The equation of the reaction can be represented as;

$2SO_{2(g)}+O_{2(g)} \to2SO_{3(g)}$

From above equation:

2 moles of sulphur dioxide reacts with 1 mole of oxygen (i.e 2 moles +1 mole =3 moles ) to give 2 moles of sulphur trioxide

So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.

So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.

Let the total pressure at the initial equilibrium be $P_1$

and the total pressure at the final equilibrium be $P_2$

According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.

Thus;

P ∝ 1/V

P = K/V

PV = K

where K = constant

So;

PV = constant

Hence;

$P_1V_1 = P_2V_2$

From the foregoing; since the volume is decreased to one- half to initial Volume; then ,

$V_2 = \dfrac{V_1}{\dfrac{3}{2}} ----- (1)$

also;

Thus ;

$P_1V_1 = P_2( \dfrac{V_1}{\frac{3}{2}})$

$P_1V_1 = P_2 * 2 \dfrac{V_1}{3}$

$3 P_1 V_1 = 2 P_2 V_1$

Dividing both sides by $V_1$

$3P_1 = 2P_2$

$P_2 =P_1 \dfrac{3}{2} ----- (2)$

From ;

$P_1V_1 = P_2V_2$

$P_2 V_2 = P_1 * \dfrac{3}{2}* \dfrac{V_1}{\frac{3}{2}}$

$P_2 V_2 = P_1 * \dfrac{3}{2}* \dfrac{2 }{3}}*V_1$

$P_2 V_2 = P_1 V_1$

Thus; The new equilibrium total pressure will be increased to one-half to initial total pressure.

7 0

3 years ago

Consider the following reaction where Kc = 1.80×10-2 at 698 K:

Klio2033 [76]

Answer:

The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.

Explanation:

The reaction quotient Qc is a measure of the relative amount of products and reagents present in a reaction at any given time, which is calculated in a reaction that may not yet have reached equilibrium.

For the reversible reaction aA + bB⇔ cC + dD, where a, b, c and d are the stoichiometric coefficients of the balanced equation, Qc is calculated by:

$Qc=\frac{[C]^{c}*[D]^{d} } {[A]^{a}*[B]^{b}}$

In this case:

$Qc=\frac{[H_{2} ]*[I_{2} ] } {[HI]^{2}}$

Since molarity is the concentration of a solution expressed in the number of moles dissolved per liter of solution, you have:

$[H_{2} ]=\frac{2.09*10^{-2} moles}{1 Liter}$ =2.09*10⁻² $\frac{moles}{liter}$

$[I_{2} ]=\frac{4.14*10^{-2} moles}{1 Liter}$ =4.14*10⁻² $\frac{moles}{liter}$

$[I_{2} ]=\frac{0.280 moles}{1 Liter}$ = 0.280 $\frac{moles}{liter}$

So,

$Qc=\frac{2.09*10^{-2} *4.14*10^{-2} } {0.280^{2} }$

Qc= 0.011

Comparing Qc with Kc allows to find out the status and evolution of the system:

If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.

If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

Being Qc=0.011 and Kc=1.80⁻²=0.018, then Qc<Kc. <u><em>The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.</em></u>

8 0

3 years ago