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Andreas93 [3]
3 years ago
14

Help meeee! 3874 dkg = _____ dg

Chemistry
1 answer:
musickatia [10]3 years ago
6 0

Answer:

the answer is 387400 but I might be wrong

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What volume (in mL) of a 0.113 M NaCl solution contains 0.087 mol NaCl?
Helen [10]
Hello I a m s o m o n e I n t h e w o r l d
5 0
4 years ago
The diagram below shows a cell placed in a solution.
Sholpan [36]

it will expand as water moves into it.


3 0
3 years ago
What are the list of catalysts that can be used for hydrogenation of nitrobenzene???​
Alex Ar [27]

Answer:

The multiring aromatic hydrocarbons in the coal liquid were hydrogenated to give saturated molecules that contained only one aromatic ring. Of the several organic bases investigated, potassium bis(trimethylsilyl)amide was found to be the most effective catalyst.

Explanation:

5 0
3 years ago
He heat of solution of kcl is 17. 2 kj/mol and the lattice energy of kcl(s) is 701. 2 kj/mol. calculate the total heat of hydrat
Alexandra [31]

The total heat of hydration of 1. 00 mol of gas phase K^{+} ions and Cl^{-}ions is - 684 kJ/mol.

Calculation ,

Given data ,

Heat of solution =  17. 2 kJ/mol

lattice energy of KCl(s) = 701. 2 kJ/mol.

heat of hydration = ?

The KCl is formed byK^{+} ions and Cl^{-}– ions

ΔH_{solution} = U° + ΔH_{hydration}

ΔH_{hydration} = ΔH_{solution} - U° =  17. 2 kJ/mol - 701. 2 kJ/mol = - 684 kJ/mol

Hence, heat of hydration of 1. 00 mol of gas phase K^{+} ions and Cl^{-} ions is - 684 kJ/mol.

To learn more about  hydration

brainly.com/question/15724859

#SPJ4

7 0
2 years ago
in order to make tea, 32,000 J of energy were added to 100.0g of water. what was the temperature Chang of the water? ​
mina [271]

Answer:

ΔT  = 76.5 °C

Explanation:

Given data:

Amount of water = 100.0 g

Energy needed = 32000 J

Change in temperature = ?

Solution,

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Now we will put the values in formula.

Q = m.c. ΔT

ΔT  = Q / m.c

ΔT  = 32000 j/ 100.0 g × 4.184 j/g. °C

ΔT  =  32000 j / 418.4 j /°C

ΔT  = 76.5 °C

5 0
4 years ago
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