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3 years ago

_ Na3PO4 + _ HCl NaCl + _ H3PO4

Chemistry

1 answer:

tatiyna3 years ago

8 0

Answer:

Na₃PO₄ + 3HCl —> 3NaCl + H₃PO₄

The coefficients are: 1, 3, 3, 1

Explanation:

_Na₃PO₄ + __HCl —> __NaCl + _H₃PO₄

The above equation can be balance as follow:

Na₃PO₄ + HCl —> NaCl + H₃PO₄

There are 3 atoms of Na on the left side and 1 atom on the right side. It can be balance by writing 3 before NaCl as shown below:

Na₃PO₄ + HCl —> 3NaCl + H₃PO₄

There are 3 atoms of Cl on the right side and 1 atom on the left side. It can be balance by writing 3 before HCl as shown below:

Na₃PO₄ + 3HCl —> 3NaCl + H₃PO₄

Now, the equation is balanced.

The coefficients are: 1, 3, 3, 1

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which atomic particle determines the chemical behavior of an atom the nucleus, neutron, electron, proton or none of these

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The chemical behavior of an atoms is determine by the formation or destruction of chemical bonds. The chemical bonds are the result of the interaction of the electrons of the atoms. Chemical properties of the atoms are given by how attached are the shell electrons attached to the nucleus and how they interact with other atoms. Chemical changes are the result of exchange valence electrons of the atoms. So, the answer is the atomic particle that determines the chemical behavior of an atom is the electron, because it is the particle that is active in chemical bonding.

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4 years ago

When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2

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Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is $\Delta \ T_f = T_{solvent}- T_{solution}$

$\Delta \ T_f = 2.9 ^0 \ C$

Using the depression in freezing point, the molar depression constant of the solvent $K_f = \dfrac{\Delta T_f}{m}$

$K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}$

$K_f = 4.82 ^0 C / m}$

The freezing point of the solution $\Delta T_f = T_{solvent} - T_{solution}$

$\Delta T_f = 7.3^ 0 \ C$

The molality of the solution is:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

Molar depression constant of solvent X, $K_f = 4.82 ^0 \ C/m$

Hence, using the elevation in boiling point;

the Vant'Hoff factor $i = \dfrac{\Delta T_f}{k_f \times m}$

$i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}$

$\mathbf {i = 1.51 }$

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3 years ago

A sample of methane gas having a volume of 2.80 L at 25 degree C and 1.65 atm was mixed with a sample of oxygen gas having a vol

Svetradugi [14.3K]

Answer:

Volume of carbon dioxide = 2.47 L

Explanation:

Given that:

For methane gas:-

Temperature = 25.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (25.0 + 273.15) K = 298.15 K

V = 2.80 L

Pressure = 1.65 atm

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L atm/ K mol

Applying the equation as:

1.65 atm × 2.80 L = n ×0.0821 L atm/ K mol × 298.15 K

⇒n for methane gas = 0.1887 mol

For oxygen gas:-

Temperature = 31 °C

T = (31 + 273.15) K = 304.15 K

V = 35.0 L

Pressure = 1.25 atm

Using ideal gas equation as:

$PV=nRT$

Applying the equation as:

1.25 atm × 35.0 L = n ×0.0821 L atm/ K mol × 304.15 K

⇒n for oxygen gas = 1.7521 mol

According to the reaction:-

$CH_4+2O_2\rightarrow CO_2+2H_2O$

1 mole of methane gas reacts with 2 moles of oxygen gas

Also,

0.1887 mole of methane gas reacts with 2*0.1887 moles of oxygen gas

Moles of oxygen gas = 0.3774 moles

Available moles of oxygen gas = 1.7521 moles

Limiting reagent is the one which is present in small amount. Thus, methane gas is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of methane gas on reaction forms 1 mole of carbon dioxide

Thus,

0.1887 mole of methane gas on reaction forms 0.1887 mole of carbon dioxide

Moles of carbon dioxide = 0.1887 moles

Given that:- Pressure = 2.50 atm

Temperature = 125 °C

T = (125 + 273.15) K = 398.15 K

Using ideal gas equation as:

$PV=nRT$

Applying the equation as:

2.50 atm × V = 0.1887 moles ×0.0821 L atm/ K mol × 398.15 K

Volume of carbon dioxide = 2.47 L

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Answer:

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