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Vladimir79 [104]
2 years ago
11

In the reaction K2CrO4 (aq) + PbCl2 (aq) 2KCl (aq) + PbCrO4(s), how many grams of PbCrO4 will precipitate out from the reaction

between 25.0 milliliters of 3.0 M K2CrO4 in an excess of PbCl2?
Chemistry
1 answer:
tankabanditka [31]2 years ago
6 0
The answer is 24 grams                       
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What metalloid has commonly been used as an insecticide due<br> to its effectiveness as a poison.
Taya2010 [7]

Answer:

Arsenic.

Explanation:

Hello there!

In this case, since insecticides are substances that act as poisons to get rid of insects in order to prevent their presence and/or reproduction in houses, companies, crops and others, a substance that has been widely used is the metalloid arsenic due to its direct affection of the insect's body (movement, performance, cellular functions).

In addition, high levels of arsenic in food could cause arsenic poisoning in humans as well, that is why such practice must be properly performed and by using the correct security protocol.

Best regards!

5 0
3 years ago
Which of the following is a completely elastic collision? A. a ball rebounds against a wall, reversing its direction, but at onl
Dmitry_Shevchenko [17]

Answer:

Answer E.

For a collision to be completely elastic, there must be NO LOSS in kinetic energy.

We can go through each answer choice:

A. Since the ball rebounds at half the initial speed, there is a loss in kinetic energy. This is NOT an elastic collision.

B. A collision involving sticking is an example of a perfectly INELASTIC collision. This is NOT an elastic collision.

C. A reduced speed indicates that there is a loss of kinetic energy. This is NOT elastic.

D. The balls traveling at half the speed after the collision indicates a loss of kinetic energy, making this collision NOT elastic.

E. This collision indicates an exchange of velocities, characteristic of an elastic collision. We can prove this:

Let:

m = mass of each ball

v = velocity

We have the initial kinetic energy as:

KE = \frac{1}{2}mv^2 + 0 = \frac{1}{2}mv^2KE=21mv2+0=21mv2

And the final as:

KE = 0 + \frac{1}{2}mv^2 = \frac{1}{2}mv^2KE=0+21mv2=21mv2

5 0
2 years ago
Read 2 more answers
How many grams of hydrogen in 1.85moles
olga2289 [7]
2H2(g) + O2(g) → 2H2O(1) 0 260 g 0.2068 0.180 g 2008 When 45.0 g of CH4 reacts with excess O2, the actual yield of CO2 is 118 g. What is the percent yield? CHA(g) + 2O2(g) - CO2(g) + 2H2O(g) 73.6% 67.9% 95.2% 86.4% For the reaction: 2503(g) + 790 kcal - 25(s) + 3O2(g), how many kcal are needed to form 1.5 moles O2(g)? 790 kcal 395 kcal 2370 kcal 411 kcal When 3 moles of Ny are mixed with 5 moles of H2 the limiting reactant is N2(g) + 3H2(g) - 2NH3(g) H2 NH3 ОООО H20 O N₂
5 0
2 years ago
Please help me with this chemistry problem
Margarita [4]

Answer:

50 g of K₂CO₃ are needed

Explanation:

How many grams of K₂CO₃ are needed to make 500 g of a 10% m/m solution?

We analyse data:

500 g is the mass of the solution we want

10% m/m is a sort of concentration,  in this case means that 10 g of solute (K₂CO₃) are contained in 100 g of solution

Therefore we can solve this, by a rule of three:

In 100 g of solution we have 10 g of K₂CO₃

In 500 g of solution we may have, (500 . 10) / 100 = 50 g of K₂CO₃

6 0
3 years ago
For the reaction C2H4(g) + H2O(g) --&gt; CH3CH2OH(g)
Dominik [7]

Answer : The value of equilibrium constant for this reaction at 262.0 K is 3.35\times 10^{2}

Explanation :

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

where,

\Delta G^o = standard Gibbs free energy  = ?

\Delta H^o = standard enthalpy = -45.6 kJ = -45600 J

\Delta S^o = standard entropy = -125.7 J/K

T = temperature of reaction = 262.0 K

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

\Delta G^o=(-45600J)-(262.0K\times -125.7J/K)

\Delta G^o=-12666.6J=-12.7kJ

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln k

where,

\Delta G^o = standard Gibbs free energy  = -12666.6 J

R = gas constant  = 8.314 J/K.mol

T = temperature  = 262.0 K

K = equilibrium constant = ?

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

-12666.6J=-(8.314J/K.mol)\times (262.0K)\times \ln k

k=3.35\times 10^{2}

Therefore, the value of equilibrium constant for this reaction at 262.0 K is 3.35\times 10^{2}

3 0
3 years ago
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