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3 years ago

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write the chemical formula of the following compounds sodium sulfate magnesium chloride aluminum nitrate ammonium sulfate

1 answer:

tekilochka [14]3 years ago

3 0

Explanation:

Magnesium chloride = MgCl₂

aluminum nitrate= Al(NO₃)₃

ammonium sulfate=( NH₄)₂SO₄

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What happens when phenol is treated with bromine water?

aleksley [76]

Answer:

Polyhalogen derivatives are given when Phenol is treated with bromine water, in which all the H-atoms present at the o- and p- positions are substituted by Bromine with respect to the -OH group.

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2 years ago

Draw the structural formula of the major product of the reaction of (S)-2,2,3-trimethyloxirane with MeOH, H . Use the wedge/hash

Katarina [22]

Answer:

(S)-3-methoxy-3-methylbutan-2-ol

Explanation:

In this case, we have an <u>epoxide opening in acid medium</u>. The first step then is the <u>protonation of the oxygen</u>. Then the epoxide is broken to generate the most <u>stable carbocation</u>. The nucleophile ( $CH_3OH$ ) will attack the carbocation generating a new bond. Finally, the oxygen is <u>deprotonated</u> to obtain an ether functional group and we will obtain the molecule <u>(S)-3-methoxy-3-methylbutan-2-ol</u>.

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8 0

3 years ago

State the number of protons contained in the atomic number 8<br>

Nat2105 [25]

Answer:

There are 8 protons

Explanation:

${ \bf{atomic \: number = proton \: number}}$

6 0

3 years ago

What is the volume of 0.1 mole of methane (CH4) ? (One mole of any gas occupies 22.4 L under certain conditions of temperature a

Veronika [31]

Answer:

Option A = 2.2 L

Explanation:

Given data:

volume of one mole of gas = 22.4 L

Volume of 0.1 mole of gas at same condition = ?

Solution:

It is known that one mole of gas at STP occupy 22.4 L volume. The standard temperature is 273.15 K and standard pressure is 1 atm.

For 0.1 mole of methane.

0.1/1 × 22.4 = 2.24 L

0.1 mole of methane occupy 2.24 L volume.

8 0

3 years ago

Read 2 more answers

A 825 g iron block is heated to 352 degrees C and is placed in an insulated container (of negligible heat capacity) containing 4

Stella [2.4K]

Answer : The final equilibrium temperature of the water and iron is, 537.12 K

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of iron = 560 J/(kg.K)

$c_1$ = specific heat of water = 4186 J/(kg.K)

$m_1$ = mass of iron = 825 g

$m_2$ = mass of water = 40 g

$T_f$ = final temperature of water and iron = ?

$T_1$ = initial temperature of iron = $352^oC=273+352=625K$

$T_2$ = initial temperature of water = $20^oC=273+20=293K$

Now put all the given values in the above formula, we get:

$(825\times 10^{-3}kg)\times 560J/(kg.K)\times (T_f-625K)=-(40\times 10^{-3}kg)\times 4186J/(kg.K)\times (T_f-293K)$

$T_f=537.12K$

Therefore, the final equilibrium temperature of the water and iron is, 537.12 K

8 0

3 years ago