All categories

3 years ago

The molar mass of copper is 63.55 g/mol. How many moles of copper were reacted?

2 answers:

8 0

The moles of any substance are equal to the substance's mass divided by its molar mass. Therefore, in order to calculate the moles of copper, you would divide the reacted mass by 63.55

KengaRu [80]3 years ago

5 0

Answer:

Explanation:

You might be interested in

25 g of a compound is added to 500 mL of water if the freezing point of the resulting solution is

vlabodo [156]

Answer:

a) 119 g/mol

Explanation:

-We apply the formula for freezing point depression to obtain the molality of the solution:

$\bigtriangleup T_f=K_fm, \ \ K_f=1.36\textdegree C/m\\\\\therefore m=\frac{\bigtriangleup T_f}{K_f}\\\\=\frac{0.57\textdegree C}{1.36\textdegree C}\\\\=0.4191\ mol/Kg\\\\$

#We use the molality above to calculate the molar mass:

$m=\frac{0.4191\ mol}{1\ Kg}=\frac{25\ g}{0.5\ Kg}\\\\\therefore 1 \ mol=\frac{25\ g}{0.5\ Kg}\times\frac{1\ Kg}{ 0.4191}\\\\=119.3033\approx 119\ g/mol$

Hence, the molar mass of the compound is 119 g/mol

5 0

3 years ago

Plz answer question! Due Today!!! Question is in picture! Will make BRAINLIEST for first correct answer!!! No links plz!

iris [78.8K]

Answer:

I'd Go. B. weigh everything, let the reaction happen, then weigh everything again.

3 0

3 years ago

Read 2 more answers

A student experimentally obtained the density of osmium the densest element as 22.57g/cm3 the density of osmium is reported to b

Airida [17]

Answer:

The answer to your question is: % error = 0.4

Explanation:

Data

real value = 22.48%

estimated value = 22.57 %

Formula

% error = |real value - estimated value|/real value x 100

%error = |22.48 - 22.57|/22.48 x 100

% error = |-0.09|/22.48 x 100

%error = 0.09/22.48 x 100

% error = 0.004 x 100

% error = 0.4

3 0

3 years ago

Hiii pls help me to write out the ionic equation

emmasim [6.3K]

Answer:

STEP I

This is the balanced equation for the given reaction:-

$2KOH_{(aq)} + H_2SO_4{}_{(aq)} \rightarrow K_2SO_4{}_{(aq)} + 2H_2O_{(l)}$

STEP II

The compounds marked with (aq) are soluble ionic compounds. They must be

broken into their respective ions.

see, in the equation KOH, H2SO4, and K2SO4 are marked with (aq).

On breaking them into their respective ions :-

2KOH -> 2K+ + 2OH-
H2SO4 -> 2H+ + (SO4)2-
K2SO4 -> 2K+ + (SO4)2-

STEP III

Rewriting these in the form of equation

$\underline{\pmb{2K^+} }+ 2OH^- + 2H^+ + \pmb{\underline{{SO_4{}^{2-}}} \: \rightarrow \: \underline{\pmb{2K^+}}} + \underline{\pmb{SO_4{}^{2-}}} + 2H_2O$

STEP IV

Canceling spectator ions, the ions that appear the same on either side of the equation

(note: in the above step the ions in bold have gotten canceled.)

$\boxed{ \mathfrak{ \red{ 2OH^-{}_{(aq)} + 2H^+{(aq.)} \rightarrow H_2O{}_{(l)}}}}$

This is the net ionic equation.

____________________________

$\\$

$\mathfrak{\underline{\green{ Why\: KOH \:has\: been\: taken\: as\: aqueous ?}}}$

KOH has been taken as aqueous because the question informs us that we have a solution of KOH. by solution it means that KOH has been dissolved in water before use.

[Alkali metal hydroxides are the only halides soluble in water ]

4 0

3 years ago

Can anyone fill in the answers?

Arte-miy333 [17]

Hi, P.S. The atomic number will always be the same as the number of protons...

6 0

2 years ago