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2 years ago

Question 1:

Chemistry

1 answer:

Gwar [14]2 years ago

8 0

Answer:

1.a

2.c

hope helps

Explanation:

mark

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Are normal salts acidic, basic or neutral salt?

RSB [31]

Neutral salt is the answer

Hope this helps!

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3 years ago

What is the mass-action expression Qc for the following chemical reaction:Zn(s)+2Ag+(aq) â‡Œ Zn+2(aq)+2Ag(s) a. [Zn-2]/[Ag-]b. [

sattari [20]

Answer : The expression for reaction quotient will be :

$Q_c=\frac{[Zn^{2+}]}{[Ag^{+}]^2}$

Explanation :

Reaction quotient (Qc) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given redox reaction is :

$2Ag^{+}(aq)+Zn(s)\rightarrow 2Ag(s)+Zn^{2+}(aq)$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

The expression for reaction quotient will be :

$Q_c=\frac{[Zn^{2+}]}{[Ag^{+}]^2}$

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2 years ago

AN object has a volume<br> of 77ml and a mass of<br> 429. What is the density<br> of this object?

pantera1 [17]

Answer:

4.254320865115cm or simplied 4.25cm

Explanation:

6 0

3 years ago

A 603.0 g/hr stream of liquid methyl alcohol, also called methanol, (CH3OH) at 8.10 atm and 26.0°C was held at constant pressure

mafiozo [28]

Answer:

$Q=\frac{1385270 J}{hr}$

Explanation:

Methanol's molecular weight: $M=32 g/mol$

Methanol's heat of vaporization: $\Delta H_{vap}=69.69 kJ/mol$

Ideal gas heat capacity (Cp):

$Cp=\frac{5}{2}*R$ where R is the gas constant

$Cp=\frac{5}{2}*8.314 \frac{J}{mol*C}=20.78\frac{J}{mol*C}$

The heat needed to vaporize and bring the gs to 210°C is:

$Q=\frac{603 g/hr}{32 g/mol}*(69690\frac{J}{mol}+20.78\frac{J}{mol*C}*(210-26)C)$

$Q=\frac{1385270 J}{hr}$

4 0

2 years ago

Calculate the change in ph when 3.00 ml of 0.100 m hcl(aq) is added to 100.0 ml of a buffer solution that is 0.100 m in nh3(aq)

nikklg [1K]

The change in pH is calculated by:

pOH = Protein kinase B + log [NH4+]/ [NH3]

Protein kinase B of ammonia = 4.74

initial potential of oxygen hydroxide= 4.74 + log 0.100/0.100 = 4.74
pH = 14 - 4.74=9.26

moles NH4+ = moles NH3 = 0.100 L x 0.100 M = 0.0100
moles H+ added = 3.00 x 10^-3 L x 0.100 M=0.000300

NH3 + H+ = NH4+
moles NH3 = 0.0100 - 0.000300=0.00970
moles NH4+ = 0.0100 + 0.000300=0.0103

pOH = 4.74 + log 0.0103/ 0.00970= 4.77
oH = 14 - 4.77 = 9.23

the change is = 9.26 - 9.23 =0.03

7 0

3 years ago