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VashaNatasha [74]
3 years ago
6

What is the precipitate that will form from the reaction of barium chloride dihydrate with sodium sulfate decahydrate?

Chemistry
2 answers:
iragen [17]3 years ago
4 0
 <span>immediate milky appearance 
after 5-10 minutes a precipitate of BaSO4 will settle out of solution the the solution will be clear.
</span>but mostly a white precipitate

MakcuM [25]3 years ago
4 0

The formation of an insoluble salt on mixing of two solutions which contains soluble salts is refer to a precipitation reaction. The insoluble salt formed in the reaction is said to be a precipitate and can be separated by filtration or any other separation method.

The formula for barium chloride dihydrate is BaCl_2.2H_2O.

The formula for sodium sulfate decahydrate is Na_2SO_4.10H_2O.

The hydrates on the above compounds represents that they are in the aqueous phase. So, the reaction is:

BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + NaCl(aq)

The above equation is not balanced as the number of atoms of elements on the reactant side is not equal to the number of atoms of elements on the product side. So, in order to balance the reaction, NaCl on the reactant side is multiplied by 2. Thus, the balanced chemical equation is:

BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaCl(aq)

Hence, the precipitate that will form from the reaction of barium chloride dihydrate with sodium sulfate decahydrate is barium sulfate, BaSO_4.

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Someone please help me I got a few minutes
HACTEHA [7]

Answer:

30 feet /second

Explanation:

60 feet/ 2 sec = 30 feet/sec

6 0
3 years ago
Read 2 more answers
A hypothetical covalent molecule, x–y, has a dipole moment of 1.44 d and a bond length of 163 pm. calculate the partial charge o
Aneli [31]
As we know,
                                     1 D  =  3.34 × 10⁻³⁰ C.m
So,
                                     1.44 D  =  ?
Solving for 1.44 D,
                                     =  (3.34 × 10⁻³⁰ C.m × 1.44 D) ÷ 1 D
                    
                         1.44 D  =  4.80 × 10⁻³⁰ C.m

Dipole Moment 
is given as,
 
                         Dipole Moment  =  q  ×  r    
Solving for q,
                         q  =  Dipole Moment / r    ------ (1)
Where,
                         Dipole Moment  =  4.80 × 10⁻³⁰ C.m

                         r  =  163 pm  =  1.63 × 10⁻¹⁰ m

Putting values in eq. 1,

                            q  =  4.80 × 10⁻³⁰ C.m / 1.63 × 10⁻¹⁰ m

                            q  =  2.94 × 10⁻²⁰ C

As,
                            1.602 × 10⁻¹⁹ C  =  1 e⁻
So,
                             2.94 × 10⁻²⁰ C  =  X e⁻

Solving for X,

                            X  =  (2.94 × 10⁻²⁰ C × 1 e⁻) ÷ 1.602 × 10⁻¹⁹ C

                                = 0.183 e⁻

Result:
           
So one element is containing + 0.183 e⁻ while the other element is containing - 0.183 e⁻.
4 0
3 years ago
If the half life of iridium-182 is 15 years, how much of a 3 gram sample is left after 2 half-lives?
padilas [110]

Answer:

D. 0.75 grams

Explanation:

The data given on the iridium 182 are;

The half life of the iridium 182, t_{(1/2)} = 15 years

The mass of the sample of iridium, N₀ = 3 grams

The amount left, N(t) after two half lives is given as follows;

N(t) = N_0 \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{1/2}} }

For two half lives, t = 2 × t_{(1/2)}

∴ t = 2 × 15 = 30

\dfrac{t}{t_{(1/2)}} = \dfrac{30}{15} = 2

\therefore N(t) = 3 \times\left (\dfrac{1}{2} \right )^2 = 0.75

∴ The amount left, N(t) = 0.75 grams

4 0
2 years ago
How is n1 in the Rydberg equation related to the quantum number n in the Bohr model?How is n1 in the Rydberg equation related to
andre [41]

Ansddwdccwer:

ExplanationBecause:

7 0
3 years ago
A 1.800-g sample of solid phenol (C6H5OH(s)) was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/?C. The temp
vichka [17]

Answer:

The balanced chemical equation:

C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)

Heat of combustion per gram of phenol is 32.454 kJ/g

Heat of combustion per gram of phenol is 3,050 kJ/mol

Explanation:

C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)

Heat capacity of calorimeter = C = 11.66 kJ/°C

Initial temperature of the calorimeter = T_1= 21.36^oC

Final temperature of the calorimeter = T_2= 26.37^oC

Heat absorbed by calorimeter = Q

Q=C\times \Delta T

Heat released during reaction = Q'

Q' = -Q ( law of conservation of energy)

Energy released on combustion of 1.800 grams of phenol = Q' = -(58.4166 kJ)

Heat of combustion per gram of phenol:

\frac{Q'}{1.800 g}=\frac{-58.4166 kJ}{1.800 g}=32.454 kJ/g

Molar mass of phenol = 94 g/mol

Heat of combustion per gram of phenol:

\frac{Q'}{\frac{1.800 g}{94 g/mol}}=\frac{-58.4166 kJ\times 94 g/mol}{1.800 g}=3,050 kJ/mol

3 0
2 years ago
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