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tangare [24]
3 years ago
11

What impact would adding twice as much Na2 CO3 than required for stoichiometric quantities have on the quantity of product produ

ced
Chemistry
1 answer:
IrinaK [193]3 years ago
5 0

Answer:

There will be no observed impact of adding twice as much Na2CO3 on the product

Explanation:

Stoichiometry gives the relationship between reactants and products in terms of mass, mole and volume.

If we consider the stoichiometry of the reaction, we will discover that the reaction occurs in a 1:1 ratio. This implies that use of twice the amount of Na2CO3 will only lead to an excess of Na2CO3 making the other reactant the limiting reactant. Once the other reactant is used up, the reaction quenches.

Hence, use of twice as much Na2CO3 has no impact on the quantity of product produced.

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Consider the reaction of peroxydisulfate ion (S2O2−8) with iodide ion (I−) in aqueous solution: S2O2−8(aq)+3I−(aq)→2SO2−4(aq)+I−
kolbaska11 [484]

Answer:

r = 3.61x10^{-6} M/s

Explanation:

The rate of disappearance (r) is given by the multiplication of the concentrations of the reagents, each one raised of the coefficient of the reaction.

r = k.[S2O2^{-8} ]^{x} x [I^{-} ]^{y}

K is the constant of the reaction, and doesn't depends on the concentrations. First, let's find the coefficients x and y. Let's use the first and the second experiments, and lets divide 1º by 2º :

\frac{r1}{r2} = \frac{0.018^{x} x0.036^{y} }{0.027^xx0.036^y}

\frac{2.6x10^{-6}}{3.9x10^{-6}} = (\frac{0.018}{0.027})^xx(\frac{0.036}{0.036})^y

0.67 = 0.67^x

x = 1

Now, to find the coefficient y let's do the same for the experiments 1 and 3:

\frac{r1}{r3} = \frac{0.018x0.036^y}{0.036x0.054^y}

\frac{2.6x10^{-6}}{7.8x10^{-6}} = (\frac{0.018}{0.036})x(\frac{0.036}{0.054})^y

0.33 = 0.5x 0.67^y

0.67 = 0.67^y

y = 1

Now, we need to calculate the constant k in whatever experiment. Using the first :

2.6x10^{-6} = kx0.018x0.036kx6.48x10^{-4} = 2.6x10^{-6}

k = 4.01x10^{-3} M^{-1}s^{-1}[/tex]

Using the data given,

r = 4.01x10^{-3}x1.8x10^{-2}x5.0x10^{-2}

r = 3.61x10^{-6} M/s

7 0
3 years ago
The four fundamental forces are arranged in the increasing order of their relative strength.
natali 33 [55]

Answer:

The correct answer is - both act only between non-atomic particles.

Explanation:

The decreasing order of their relative strength is - the strong force, electromagnetism, weak force, gravity.

so, A would be here = gravity

B would be  = weak force

C  would be = strong force

The weak and strong forces both are fundamental forces that act on sub-atomic particles only such as quarks.

3 0
3 years ago
How would a disease that damages the nucleus most directly affect the functioning of cells?
RSB [31]

Answer:

Hereditary information in the cell would be destroyed.

Explanation:

The nucleus can be defined as a membrane bound organelle that is found in eukaryotic cells. The main function of the nucleus is that it controls all activities that is related to the growth of the cell and also reproduction. The nucleus contains the cell hereditary information(DNA).

The nucleus is the most important organelle in the cell, It can sometimes be referred to as the brain of the cell. Therefore any health related condition that affects the nucleus would directly destroy all hereditary information that is stored in the cell.

3 0
3 years ago
How to make crystals using table salt?<br> Please help I"ll mark you as the BRAINLIEST!!!
Dafna11 [192]

Answer:

1.heat a pan of water with just a little bit of water,have a boil

2.chosse ure salt

3.stir in has much salt has u can than take the pan off the heat

4.pour the mix into a glass jar

5.tie a string to an objeet that can lay accross the top and put just the string in ure mix

Explanation oh and look at it everyday       hope that helps

8 0
3 years ago
how many kilograms of a 35% m/m sodium chlorate solution is needed to react completely with 0.29 l of a 22% m/v aluminum nitrate
Stolb23 [73]

Answer:- 0.273 kg

Solution:- A double replacement reaction takes place. The balanced equation is:

3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})

= 63.8 g aluminum nitrate

From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})

= 95.7gNaClO_3

sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate  and then the grams are converted to kg.

95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})

= 0.273 kg

So, 0.273 kg of 35% m/m sodium chlorate solution are required.

7 0
3 years ago
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