All categories

3 years ago

How many ml of a 2.0 m nabr solution are needed to make 200.0 ml of 0.50 m nabr?

Chemistry

1 answer:

gogolik [260]3 years ago

4 0

To solve this we use the equation,

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

2.0 M x V1 = 0.50 M x 200 mL

V1 = 50 mL needed

You might be interested in

How many formula units are contained in 0.57 g Cao?

joja [24]

Then you will multiply the number of moles by 6.022×1023formula units/mol . To determine the molar mass of a compound, add the atomic weight on the periodic table in g/mol times each element's subscript. Since the formula unit CaO has no subscripts, they are understood to be 1

7 0

3 years ago

A cup of hot water cannot melt a large block of ice. However, the same block melts when dropped into a lake. This is because

masya89 [10]

Explanation:

When a large block of ice is placed in a hot cup of water then some of the ice will melt into the cup. As a result, when ice melts then after sometime temperature of cup will become equal to the temperature of ice.

Therefore, ice won't melt further as surface area of cup is small so, only some of the ice will melt into it.

Whereas when same block melts when dropped into a lake it completely melts because lake has large surface area and hence, ice will melt into it easily. This is also because there is different in temperature of both ice and lake.

So, ice will melt into it as temperature of ice will be less than the temperature of lake.

6 0

4 years ago

Read 2 more answers

Al2(SO4)3(s) + H2O(l) ---- Al2O3(s) + H2SO4 (aq) calculate enthalpy formation for this reaction. Balance the reaction. Calculate

katen-ka-za [31]

The given chemical equation is:

$Al_{2}(SO_{4})_{3}(aq)+H_{2}O(l)-->Al_{2}O_{3}(aq)+H_{2}SO_{4}(aq)$

On balancing the equation we get,

$Al_{2}(SO_{4})_{3}(aq)+3H_{2}O(l)-->Al_{2}O_{3}(aq)+3H_{2}SO_{4}(aq)$

Calculating enthalpy of formation of this reaction from the standard heats of formation of the products and reactants:

Δ $H_{reaction}^{0}=[H_{f}^{0}(Al_{2}O_{3}(s)) + (3*H_{f}^{0}(H_{2}SO_{4}(aq))] - [H_{f}^{0}(Al_{2}SO_{4}(aq)) + (3*H_{f}^{0}(H_{2}O(l))]$

=[(-1669.8kJ/mol)+ {3* (-909.27 kJ/mol)}]-[(-3442kJ/mol)+{3*(-285.8 kJ/mol)}]

=[(-4397.61kJ/mol)]-[(-4299.4kJ/mol)]

=-98.21kJ/mol

Total enthalpy change when 15 mol of $Al_{2}(SO_{4})_{3}$ reacts will be=

$15 mol Al_{2}(SO_{4})_{3}*\frac{-98.21kJ}{1 molAl_{2}(SO_{4})_{3}} =-1473.15kJ/mol$

4 0

3 years ago

What is the mass of the hydrogen atoms in 1 mole of water

Katen [24]

Answer:

2.0158 grams

Explantion:

We are to find the mass of the hydrogen atoms in 1 mole of water.

We know that the formula of water is: $H_2O$

We can see, from the above mentioned formula, that water has 2 hydrpgen atoms.

From the periodic table, we get to know that Hydrogen has an atomic mass of 1.00794 grams.

As there are 2 atoms of hydrogen in water so $2 \times 1.00794 = 2.0158$ grams is the answer

3 0

4 years ago

Read 2 more answers

A galvanic (voltaic) cell consists of an electrode composed of titanium in a 1.0 M titanium(II) ion solution and a second electr

Readme [11.4K]

Answer: The standard potential for this cell is +1.49 V at 25C.

Explanation:

$E^0_{[Sn^{2+}/Sn]}=-0.14V$

$E^0_{[Ti^{2+}/Ti]}=-1.63V$

As titanium has lower reduction potential, it will act as anode and tin will acts as cathode.

$Ti+Sn^{2+}\rightarrow Ti^{2+}+Sn$

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Ti^{2+}]}{[Sn^{2+}]}$

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = $25^oC=273+25=298K$

n = number of electrons in oxidation-reduction reaction = 2

$E^0=E^0_{cathode}- E^0_{anode}=-0.14-(-1.63)=1.49V$

Where both $E^0$ are standard reduction potentials.

$E^o_{cell}$ = standard electrode potential of the cell = 1.49 V

$E_{cell}$ = emf of the cell = ?

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

$E_{cell}=1.49-\frac{2.303\times (8.314)\times (298)}{1\times 96500}\log \frac{1}{1}$

$E_{cell}=1.49V$

6 0

4 years ago