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aleksklad [387]
3 years ago
9

Manganese(IV) sulfate O Mn(SO4)2 O MnSO4 O MgSO4 DONE

Chemistry
2 answers:
oee [108]3 years ago
7 0
Mn(SO4)2 is right opinion
LiRa [457]3 years ago
7 0

Answer:

A

Explanation:

You might be interested in
Three substances A, B, and C melt at 00C,500C and -1500C
Drupady [299]
Answer:

That information is better presented and analyzed in a table.

This table shows you all the information and the answers:


Substance         melting point   boiling point    room temperature    conclusion
                                    °C                  °C                      °C                    (state)


A                                  0                    100                    25                    liquid

B                                  50                  200                    25                    solid
C                               -150                   10                     25                    gas

Explanation:

1) Substance A at 25° is above the melting point and below the boiling point, then it is liquid (just like water)


2) Substance B at 25°C is below the melting point, so it is solid.

3) Substance C at 25°C is above the boiling point, so it is gas.
7 0
3 years ago
How many grams of C3H8 are in a 7 L tank at 293 K and 5.45 atm?
Sophie [7]
1.59 moles















































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8 0
3 years ago
why do we heat the test tube in a glass beaker during a chemistry experiment and not directly over a flame?
bearhunter [10]
It is not directly over a flame because it depends on the substance you might not want to heat it too much.you never know what could happen
7 0
4 years ago
When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2
MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is \Delta \ T_f = T_{solvent}- T_{solution}

\Delta \ T_f = 2.9 ^0 \ C

Using the depression in freezing point, the molar depression constant of the solvent K_f = \dfrac{\Delta T_f}{m}

K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}

K_f = 4.82 ^0 C / m}

The freezing point of the solution \Delta T_f = T_{solvent} - T_{solution}

\Delta T_f = 7.3^ 0 \ C

The molality of the solution is:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

Molar depression constant of solvent X, K_f = 4.82 ^0 \ C/m

Hence, using the elevation in boiling point;

the Vant'Hoff factor i = \dfrac{\Delta T_f}{k_f \times m}

i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}

\mathbf {i = 1.51 }

3 0
3 years ago
PLEASE HELP!!!
Darina [25.2K]
None of the above I believe it is what can form a tornado??
3 0
3 years ago
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