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3 years ago

What is the term for a solid that forms when two liquids are mixed together

Chemistry

2 answers:

polet [3.4K]3 years ago

7 0

C.) Precipitate because it is the term for a solid that forms when two liquids are mixed together.

MrRissso [65]3 years ago

5 0

The correct answer would be C. a precipitate
Hope this helped!

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Can someone please please help....what will happen to these glands if it go on strike?

sammy [17]

Answer:

World War 3

Explanation:

People will go crazy wife or go to h*** in a become World War 3

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3 years ago

If you use 1 mole of NaOH, how much NaAl(OH)4 is produced

abruzzese [7]

Answer:
81.97 g of NaAl(OH)₄

Solution:
The reaction for the preparation of Sodium Aluminate from Aluminium Metal and NaOH is as follow,

2 NaOH + 2 Al + 6 H₂O → 2 NaAl(OH)₄ + 3 H₂

According to this equation,

2 Moles of NaOH produces = 163.94 g (2 mole) of NaAl(OH)₄
So,
1 Mole of NaOH will produce = X g of NaAl(OH)₄

Solving for X,
X = (1 mol × 163.94 g) ÷ 2 mol

X = 81.97 g of NaAl(OH)₄

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4 years ago

Alkenes can be hydrated via the addition of borane to yield alcohols with non-Markovnikov regiochemistry.

galina1969 [7]

Answer:

The given statement is true.

Explanation:

Initially, the addition of borane to the alkene takes place in the form of a concerted reaction owing to the dissociation of the bond and subsequent formation, which occurs at a similar time. After that the Anti Markovnikov supplementation of boron takes place. The addition of this atom takes place with the less substituted carbon of the alkene that then substitutes the molecule of hydrogen on the more substituted carbon.

Then through the donation of a pair of electrons from the hydrogen peroxide ion, the process of oxidation takes place resulting in the formation of trialkylborane. After this realignment of an R group with its pair of bonding, electrons take place with adjacent oxygen resulting in the withdrawal of a hydroxide ion. Eventually, the trialkylborate reacts with the aqueous NaOH to generate alcohol and sodium borate as the side product.

8 0

3 years ago

The normal boiling point of a liquid is 282 °C. At what temperature (in °C) would the vapor pressure be 0.350 atm? (∆Hvap = 28.5

victus00 [196]

Answer:

The temperature at which the vapor pressure would be 0.350 atm is 201.37°C

Explanation:

The relationship between variables in equilibrium between phases of one component system e.g liquid and vapor, solid and vapor , solid and liquid can be obtained from a thermodynamic relationship called Clapeyron equation.

Clausius- Clapeyron Equation can be put in a more convenient form applicable to vaporization and sublimation equilibria in which one of the two phases is gaseous.

The equation for Clausius- Clapeyron Equation can be expressed as:

$\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }$

where ;

$P_1$ is the vapor pressure at temperature 1

$P_ 2$ is the vapor pressure at temperature 2

∆Hvap = enthalpy of vaporization

R = universal gas constant

Given that:

$P_1$ = 1 atm

$P_ 2$ = 0.350 atm

∆Hvap = 28.5 kJ/mol = 28.5 × 10³ J/mol

$T_1$ = 282 °C = (282 + 273) K = 555 K

R = 8.314 J/mol/k

Substituting the above values into the Clausius - Clapeyron equation, we have:

$\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }$

$\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{T_2 - 555}{555T_2} \end {pmatrix} }$

$\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }$

$- 1.0498= 3427.953 \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }$

$\dfrac{- 1.0498}{3427.953}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }$

$- 3.06246906 \times 10^{-4}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }$

$\dfrac{1}{T_2} = \begin {pmatrix} \dfrac{1}{555}+ (3.06246906 \times 10^{-4} ) \end {pmatrix} }$

$\dfrac{1}{T_2} = 0.002108048708$

$T_2 = \dfrac{1}{0.002108048708}$

$\mathbf{T_2 }$ = 474.37 K

To °C ; we have $\mathbf{T_2 }$ = (474.37 - 273)°C

$\mathbf{T_2 }$ = 201.37 °C

Thus, the temperature at which the vapor pressure would be 0.350 atm is 201.37 °C

3 0

3 years ago

Lewis Structure of HOClO

rosijanka [135]

At first you need to draw the lewis dots around each atom that you have.
Next thing that you should do is to connect the dots of HOClO and do not forget that each dot connect with one lone electron.
Then draw valence of each electron and connect them with a line.

8 0

3 years ago