Answer
_2 HNO₃ + 1 Mg(OH)₂ → 1 Mg(NO₃)₂ + 2 H₂O
Explanation
Given:
______HNO3 + Mg(OH)2 ------>
Solution:
Note that the reaction between an acid and a base will give salt and water only.
Hence the complete reaction of the given equation is:
___HNO₃ + Mg(OH)₂ → Mg(NO₃)₂ + H₂O
To get the balanced equation for the acid-base reaction, 2 moles of HNO₃ will react with 1 mole of Mg(OH)₂ to produced 1 mole of Mg(NO₃)₂ and 2 moles of H₂O.
Therefore, the complete and balanced equation for the given acid-base reaction is:
_2 HNO₃ + 1 Mg(OH)₂ → 1 Mg(NO₃)₂ + 2 H₂O
Answer:
The maximum pressure is 612.2 Pa
Explanation:
The pressure of the ice (P1) = 624 Pa
The temperature of the ice = 273.16 K
The maximum temperature the specimen = - 5 oC
= -5 + 273 = 268K
The maximum Pressure the freeze drying can be will be (P2) = ?
Using Pressure law, which shows the relationship between pressure and temperature.
P1 / T1 = P2 / T2
P2 T1 = P1 T2
P2 = P1 T2 / T1
P2 = 624 × 268 / 273.16
P2 = 612.2 Pa
The maximum pressure at which drying can be carried out is 612.2 Pa
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Explanation:
When a metal replaces another metal in solution, we say such a reaction has undergone a single displacement reaction.
In such a reaction, metal higher up in the activity series replaces another one due to their position.
To known the metal or metals that will replace the given copper, we need to reference the activity series of metals.
Every metal higher than copper in the series will displace copper from the solution.
So, there metals are: potassium, sodium, lithium, barium, strontium etc.
Answer: im in 8th they got me doin this
Explanation: i need help quick
Answer : The correct answer for change in freezing point = 1.69 ° C
Freezing point depression :
It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .
SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .
It can be expressed as :
ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m
Where : ΔTf = change in freezing point (°C)
i = Von't Hoff factor
kf =molal freezing point depression constant of solvent.
m = molality of solute (m or
)
Given : kf = 1.86 
m = 0.907
)
Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1
Plugging value in expression :
ΔTf = 1* 1.86
* 0.907
)
ΔTf = 1.69 ° C
Hence change in freezing point = 1.69 °C