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LUCKY_DIMON [66]
3 years ago
13

*WILL GIVE BRAINLIEST IF SOMEONE ANSWERS THIS CORRECTLY ASAP!!!!*

Chemistry
2 answers:
o-na [289]3 years ago
6 0

Answer:

A.Chemical change, because a substance in a state different from the state of the original substances is formed

Explanation:

kondaur [170]3 years ago
4 0
I would go with the first answer
You might be interested in
When the pressure and number of particles of a gas are constant, which of the following is also constant
kvv77 [185]
The temperature is also constant
4 0
3 years ago
What is the oxidation state for a Mn atom?<br> (1) 0 (2) +7 (3) +3 (4) +4
ss7ja [257]

Answer : The correct option is, (1) 0

Explanation :

Oxidation state or oxidation number : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with sign (+ and -) first and then the magnitude.

When the atoms are present in their elemental state then the oxidation number will be zero.

As per question, 'Mn' atom is a free element that means it is present in their elemental state then the oxidation state will be zero.

Hence, the correct option is, (1) 0

5 0
4 years ago
Name:_____________________________________________________ Date:___________ Period:_________ 3/23 - 3/27 Assignment 1: Gas Law P
crimeas [40]

Answer:

The answers are;

1. 8.2 liters

2. 1214.84 ml

3. 318.027 K

4. 4.00 l.

Explanation:

1. Boyle's law states that the volume of  given mass of gas is inversely proportional to its pressure at constant temperature

that is

P₁·V₁ = P₂·V₂

Where:

P₁ = Initial pressure = 40.0 mm Hg

V₁ = Initial volume = 12.3 liters

P₂ = Final pressure = 60.0 mm Hg

V₂ = Final volume = Required

From P₁·V₁ = P₂·V₂, V₂ is given by

V_2=\frac{P_1\cdot V_1}{P_2} = \frac{40.0 mm Hg\cdot 12.3 l}{60.0 mm Hg} =  8.2 l

The volume reduces to V₂ = 8.2 liters

2. Here Charles law states that

\frac{T_1}{V_1} =\frac{T_2}{V_2}

T₁ = Initial temperature = 27.0 °C = 300.15 K

V₁ = Initial volume = 900.0 mL

T₂ = Final temperature = 132.0 °C = 405.15 K

V₂ = Final volume = Required

Therefore  V_2 =\frac{T_2\cdot V_1}{T_1} = \frac{405.15 K\times 900.0 mL}{300.15 K} = 1214.84 ml

V₂ = 1214.84 ml

3.  Gay-Lussac's Law states that

\frac{T_1}{P_1} =\frac{T_2}{P_2}

Where:

P₁ = Initial pressure = 15.0 atmospheres

T₁ = Initial temperature = 25.0 °C = 298.15 K

P₂ = Final pressure = 16.0 atmospheres

T₂ = Final temperature = Required

∴ T_2 = \frac{T_1\times P_2}{P_1}

=  \frac{298.15 K\times 16.0atm}{15.0atm} = 318.027 K

T₂ = 318.027 K

4. Avogadro's law states that,

Equal volume of all gases at the same temperature and pressure contain equal number of molecules.

Therefore if 5.00 moles of gas occupies 2.00 l volume, then

1 moles will occupy 2.00/5 l volume and

10 moles will occupy 2.00/5 × 10 or 4.00 l volume.

6 0
4 years ago
A 0.25-mol sample of a weak acid with an unknown Pka was combined with 10.0-mL of 3.00 M KOH, and the resulting solution was dil
Masteriza [31]

Answer : The value of pK_a of the weak acid is, 4.72

Explanation :

First we have to calculate the moles of KOH.

\text{Moles of }KOH=\text{Concentration of }KOH\times \text{Volume of solution}

\text{Moles of }KOH=3.00M\times 10.0mL=30mmol=0.03mol

Now we have to calculate the value of pK_a of the weak acid.

The equilibrium chemical reaction is:

                          HA+KOH\rightleftharpoons HK+H_2O

Initial moles     0.25     0.03        0

At eqm.    (0.25-0.03)   0.03      0.03

                     = 0.22

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{[HK]}{[HA]}

Now put all the given values in this expression, we get:

3.85=pK_a+\log (\frac{0.03}{0.22})

pK_a=4.72

Therefore, the value of pK_a of the weak acid is, 4.72

7 0
3 years ago
Is the haber process for the industrial synthesis of ammonia spontaneous or nonspontaneous under standard conditions at 25 ∘c? n
Ostrovityanka [42]
The  industrial  synthesis  of  ammonia   is  spontaneous.

 to  to  prove this  one  calculate  the  gibbs  free  energy(delta G)

=   delta  G  =  Delta  H  -T delta  s

delta H = -92.2  kj  in  joules  -92.2  x1000 =-92200j
delta  s  =  -199j/k
T=  25+273=  298k

delta  G  is  therefore=   -92200-298(-199)=-32898j
since  delta G  is  negative the  reaction  is  therefore  spontaneous
6 0
4 years ago
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