Ask question

Ask question

All categories

2 years ago

12

Of protons, neutrons and electrons which of these can never change an atom during an ordinary chemical or physical change. Expla

in why

1 answer:

hjlf2 years ago

6 0

Answer:

protons

Explanation:

electron number changes when the atom reacts with another atom to gain a full octet

neutron number changes when it goes through radioactive decay

but proton number never changes

You might be interested in

Given equilibrium partial pressures of PNO2= 0.247 atm, PNO = 0.0022atm, and PO2 = 0.0011 atm calculate the equilibrium constant

maxonik [38]

Answer 1:
Equilibrium constant (K) mathematically expressed as the ratio of the concentration of products to concentration of reactant. In case of gaseous system, partial pressure is used, instead to concentration.

In present case, following reaction is involved:

2NO2 ↔ 2NO + O2

Here, K = $\frac{[PNO]^2[O2]}{[PNO2]^2}$

Given: At equilibrium, <span>PNO2= 0.247 atm, PNO = 0.0022atm, and PO2 = 0.0011 atm
</span>
Hence, K = $\frac{[0.0022]^2[0.0011]}{[0.247]^2}$
= 8.727 X 10^-8

Thus, equilibrium constant of reaction = 8.727 X 10^-8
.......................................................................................................................

Answer 2:
Given: <span>PNO2= 0.192 atm, PNO = 0.021 atm, and PO2 = 0.037 atm.

Therefore, Reaction quotient = </span> $\frac{[PNO]^2[O2]}{[PNO2]^2}$
= $\frac{[0.021]^2[0.037]}{[0.192]^2}$
= 4.426 X 10^-4.

Here, Reaction quotient > Equilibrium constant.

Hence, <span>the reaction need to go to reverse direction to reattain equilibrium </span>

5 0

2 years ago

Read 2 more answers

Unit Conversion Help Thank you

AlekseyPX

Answer : 1721.72 g/qt are in 18.2 g/cL

Explanation :

As we are given: 18.2 g/cL

Now we have to convert 18.2 g/cL to g/qt.

Conversions used are:

(1) 1 L = 100 cL

(2) 1 L = 1000 mL

(3) 1 qt = 946 qt

The conversion expression will be:

$\frac{18.2g}{1cL}\times \frac{100cL}{1L}\times \frac{1L}{1000mL}\times \frac{946mL}{1qt}$

$=1721.72\text{ g/qt}$

Therefore, 1721.72 g/qt are in 18.2 g/cL

5 0

1 year ago

What is the change in enthalpy in kilojoules when 3.24 g of CH3OH is completely reacted according to the following reaction 2 CH

vodka [1.7K]

Answer:

12.78 kJ

Explanation:

The correct balanced reaction would be

$2CH_3OH\rightarrow 2CH_4+O_2\Delta H=252.8\ \text{kJ}$

Mass of methanol = $3.24\ \text{g}$

Moles of methanol can be obtained by dividing the mass of methanol with its molar mass $(32.04\ \text{g/mol})$

$\dfrac{3.24}{32.04}=0.10112\ \text{moles}$

Enthalpy change for the number of moles is given by

$\dfrac{\text{Number of moles of methanol in the reaction}}{\text{Enthalpy change in the reaction}}=\dfrac{\text{Number of moles in 3.24 g of methanol}}{\text{Enthaply in change in the mass of methanol}}$

$\\\Rightarrow\dfrac{2}{252.8}=\dfrac{0.10112}{\Delta H}\\\Rightarrow \Delta H=\dfrac{0.10112\times 252.8}{2}\\\Rightarrow \Delta H=12.781568\approx 12.78\ \text{kJ}$

The change in enthalpy is 12.78 kJ.

5 0

2 years ago

Notice that "SO4" appears in two different places in this chemical equation. SO42− is a polyatomic ion called "sulfate." What nu

masya89 [10]

Answer : The number placed in front of $CuSO_4$ should be, three (3).

Explanation :

Balanced chemical reaction : It is defined as the number of atoms of individual elements present on reactant side must be equal to the product side.

The given unbalanced chemical reaction is,

$CaSO_4+AlCl_3\rightarrow CaCl_2+Al_2(SO_4)_3$

This chemical reaction is an unbalanced reaction because in this reaction, the number of atoms of chloride and sulfate ion are not balanced.

In order to balanced the chemical reaction, the coefficient 3 is put before the $CuSO_4$ , the coefficient 2 is put before the $AlCl_3$ and the coefficient 3 is put before the $CaCl_2$ .

Thus, the balanced chemical reaction will be,

$3CaSO_4+2AlCl_3\rightarrow 3CaCl_2+Al_2(SO_4)_3$

Therefore, the number placed in front of $CuSO_4$ should be, three (3).

5 0

2 years ago

A box has a volume of 45m3 and is filled with air held at 25∘C and 3.65atm. What will be the pressure (in atmospheres) if the sa

Marina CMI [18]

Answer:

Given:

Initial pressure: $3.65\; \rm atm$ .
Volume was reduced from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ .
Temperature was raised from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ .

New pressure: approximately $3.4\times 10\; \rm atm$ ( $34\; \rm atm$ .) (Assuming that the gas is an ideal gas.)

Explanation:

Both the volume and the temperature of this gas has changed. Consider the two changes in two separate steps:

Reduce the volume of the gas from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ . Calculate the new pressure, $P_1$ .
Raise the temperature of the gas from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ . Calculate the final pressure, $P_2$ .

By Boyle's Law, the pressure of an ideal gas is inversely proportional to the volume of this gas (assuming constant temperature and that no gas particles escaped or was added.)

For this gas, $V_0 = 45\; \rm m^{3}$ while $V_1 = 5.0\; \rm m^{3}$ .

Let $P_0$ denote the pressure of this gas before the volume change ( $P_0 = 3.65\; \rm atm$ .) Let $P_1$ denote the pressure of this gas after the volume change (but before changing the temperature.) Apply Boyle's Law to find the ratio between $P_1\!$ and $P_0\!$ :

$\displaystyle \frac{P_1}{P_0} = \frac{V_0}{V_1} = \frac{45\; \rm m^{3}}{5.0\; \rm m^{3}} = 9.0$ .

In other words, because the final volume is $(1/9)$ of the initial volume, the final pressure is $9$ times the initial pressure. Therefore:

$\displaystyle P_1 = 9.0\times P_0 = 32.85\; \rm atm$ .

On the other hand, by Amonton's Law, the pressure of an ideal gas is directly proportional to the temperature (in degrees Kelvins) of this gas (assuming constant volume and that no gas particle escaped or was added.)

Convert the unit of the temperature of this gas to degrees Kelvins:

$T_1 = (25 + 273.15)\; \rm K = 298.15\; \rm K$ .

$T_2 = (35 + 273.15)\; \rm K = 308.15\; \rm K$ .

Let $P_1$ denote the pressure of this gas before this temperature change ( $P_1 = 32.85\; \rm atm$ .) Let $P_2$ denote the pressure of this gas after the temperature change. The volume of this gas is kept constant at $V_2 = V_1 = 5.0\; \rm m^{3}$ .

Apply Amonton's Law to find the ratio between $P_2$ and $P_1$ :

$\displaystyle \frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{308.16\; \rm K}{298.15\; \rm K}$ .

Calculate $P_2$ , the final pressure of this gas:

$\begin{aligned} P_2 &= \frac{308.15\; \rm K}{298.15\; \rm K} \times P_1 \\ &= \frac{308.15\; \rm K}{298.15\; \rm K} \times 32.85\; \rm atm \approx 3.4 \times 10\; \rm atm\end{aligned}$ .

In other words, the pressure of this gas after the volume and the temperature changes would be approximately $3.4\times 10\; \rm atm$ .

8 0

1 year ago