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4 years ago

MgNH4PO4*6H2O, loses all water of hydration at what temperature?

2 answers:

8 0

Ammonium magnesium phosphate hexa-hydrate usually looses its water of crystallization in a gradual manner. At the temperature above 100 degree Celsius the compound would have lost all its water of crystallization.<span />

iragen [17]4 years ago

4 0

Answer:

T = 100 C (212 F)

Explanation:

'Water of hydration' is the terminology used to refer to the water molecules present in the crystals of a given substance that can be removed by heating the substance without altering its composition.

MgNH4PO4.6H2O is essentially magnesium ammonium phosphate which has 6 molecules of water of hydration. These can be removed stepwise by gradually increasing the temperature. At a final temperature of 100 C or 212 F the compound will lose all water of hydration.

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Consider the reaction to produce methanolCO(g) + 2H2 (g) <-----> CH3OHAn equilibrium mixture in a 2.00-L vessel is found t

MariettaO [177]

Answer : The value of $K_c$ of the reaction is 10.5 and the reaction is product favored.

Explanation : Given,

Moles of $CH_3OH$ at equilibrium = 0.0406 mole

Moles of $CO$ at equilibrium = 0.170 mole

Moles of $H_2$ at equilibrium = 0.302 mole

Volume of solution = 2.00 L

First we have to calculate the concentration of $CH_3OH,CO\text{ and }H_2$ at equilibrium.

$\text{Concentration of }CH_3OH=\frac{\text{Moles of }CH_3OH}{\text{Volume of solution}}=\frac{0.0406mole}{2.00L}=0.0203M$

$\text{Concentration of }CO=\frac{\text{Moles of }CO}{\text{Volume of solution}}=\frac{0.170mole}{2.00L}=0.085M$

$\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.302mole}{2.00L}=0.151M$

Now we have to calculate the value of equilibrium constant.

The balanced equilibrium reaction is,

$CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)$

The expression of equilibrium constant $K_c$ for the reaction will be:

$K_c=\frac{[CH_3OH]}{[CO][H_2]^2}$

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

$K_c=\frac{(0.0203)}{(0.085)\times (0.151)^2}$

$K_c=10.5$

Therefore, the value of $K_c$ of the reaction is, 10.5

There are 3 conditions:

When $K_{c}>1$ ; the reaction is product favored.

When $K_{c}$ ; the reaction is reactant favored.

When $K_{c}=1$ ; the reaction is in equilibrium.

As the value of $K_{c}>1$ . So, the reaction is product favored.

7 0

3 years ago

given a density of 350 grams/cm3 and a mass of 25 grams. What is the volume using the equation involving density?

Eduardwww [97]

0.07cm³

Explanation:

Given parameters:

Density of the body = 350g/cm³

Mass of the body = 25g

Unknown:

Volume of the body = ?

Solution:

Density is the amount of substance contained in a volume of body. It is expressed as:

Density = $\frac{mass}{volume}$

Since the unknown is volume, we make it the subject of the expression;

Volume = $\frac{mass}{density}$

Volume = $\frac{25}{350}$ = 0.07cm³

learn more:

Density brainly.com/question/2690299

#learnwithBrainly

6 0

4 years ago

How many grams of glucose are in 11.5 moles?

qaws [65]

There are 2071.4662 grams of glucose in 11.5 moles.

Per 1 mole there are 180.15588 grams of glucose. 180.5588 x 11.5 =2076.4262

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4 years ago

You measure the mass of a model car to be 230 grams.The actual mass is 218 grams .what is your percent error

LenKa [72]

Your percent is 5.50458716

8 0

4 years ago

Draw the products of the complete hydrolysis of an acetal. Draw all products of the reaction.

jolli1 [7]

Answer: the product is ketone or aldehyde

Explanation:

The first step is the conversion of acetal to hemiacetal in the presence of H3O+/ ROH, and then the final conversion of hemiacetal to ketone/aldehyde using

H3O+/ ROH...

Attached is the structural conversion

8 0

4 years ago