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galina1969 [7]
3 years ago
5

Can someone help? Ill give brainliest!

Chemistry
1 answer:
igomit [66]3 years ago
6 0

Answer:

Yes, it becomes 2 of H20 which is water

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What is the formula for flow rate?
Natasha2012 [34]

Answer:Solve for X+ 2x=

Explanation:

6 0
3 years ago
How many hydrogen's are in 6HCI
wolverine [178]

Answer:

6 atoms

Explanation:

In 6 moles of HCL (hydrochloric acid), there are 6 hydrogen atoms. Think about it this way: there is one hydrogen atom in HCL, or one hydrogen atom in one mole of HCL. Since there are 6 moles of HCL instead of one, this means there are six times as many hydrogen atoms. 6 times 1 is 6.

4 0
3 years ago
How many milliliters of 0.200 M NH4OH are needed to react with 12.0 mL of 0.550 M FeCl3?
trapecia [35]

Answer:

9.9 ml of 0.200M NH₄OH(aq)

Explanation:

3NH₄OH(Iaq) + FeCl₃(aq) => NH₄Cl(aq) + Fe(OH)₃(s)

?ml of 0.200M NH₄OH(aq) reacts completely with  12ml of 0.550M FeCl₃(aq)

1 x Molarity NH₄OH x Volume Am-OH Solution(L) = 2 x Molarity FeCl₃ x Volume FeCl₃ Solution

1(0.200M)(Vol Am-OH Soln) = 3(0.550M)(0.012L)

=> Vol Am-OH Soln = 3(0.550M)(0.012L)/1(0.200M) = 0.0099 Liter = 9.9 milliliters  

5 0
3 years ago
C8H10N4O2 is the chemical formula for caffeine, there are 8 carbons. List the 3 remaining elements in caffeine and how many atom
Nataly [62]
Caffeine   that  is C8H10N402  is molecule which made   up  of   4 different elements. apart  from  carbon  the  other  element  includes

element        symbol
nitrogen          N
oxygen           O
hydrogen        H
The number  of  atom  present in caffeine molecule  are  as follows

Nitrogen=4  atoms
Oxygen=2  atoms
Hydrogen=10 atoms

4 0
3 years ago
A student dissolves 3.9g of aniline (C6H5NH2) in 200.mL of a solvent with a density of 1.05 g/mL . The student notices that the
Tresset [83]

Answer:

a. Molarity= M =2.1x10^{-1}M

b. Molality= m=2.0x10^{-1}m

Explanation:

Hello,

In this case, given the information about the aniline, whose molar mass is 93g/mol, one could assume the volume of the solution is just 200 mL (0.200 L) as no volume change is observed when mixing, therefore, the molarity results:

M=\frac{n_{solute}}{V_{solution}} =\frac{3.9g*\frac{1mol}{93g} }{0.2L} =2.1x10^{-1}M

Moreover, the molality:

m=\frac{n_{solute}}{m_{solvent}} =\frac{3.9g*\frac{1mol}{93g} }{0.2L*\frac{1.05kg}{1L} } =2.0x10^{-1}m

Best regards.

7 0
3 years ago
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