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blondinia [14]
2 years ago
14

Is mass conserved when 10 g of sodium hydroxide undergoes a chemical change during an interaction with 57 g of copper sulfate? U

se complete sentences to support your answer by explaining how this can be demonstrated.
Chemistry
1 answer:
Olin [163]2 years ago
7 0
<span>Yes, mass is conserved. In any chemical reaction mass is neither created, nor destroyed. That is a basic principle that remained unaltered and unchallenged until the discovery of nuclear power. And even then, the basic principle of conservation of mass was simply altered to include the conservation of energy as well. If you with to demonstrate this conservation by the reaction of sodium hydroxide with copper sulfate, you can do so by measuring the mass of all reactants and reaction vessels both before and after the reaction. Just put the beaker with sodium hydroxide solution along with the beaker of copper sulfate solution on the balance scale and measure their total mass. Then pour one solution into the other, and place both the full beaker with the reactants and product as well as the empty beaker back on the balance scale and measure their combined mass. If you perform the measurements accurately, the combined masses both before and after the reaction will be identical.</span>
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Three compounds containing potassium and oxygen are compared. Analysis shows that for each 1.00 g of O, the compounds have 1.22
kicyunya [14]
<span>The law of proportion states that elements combine in whole number ratios. The gram readings for K are multiples of each other, both in grams and moles.
Let us compare the ratios:
</span>2.44 grams/1.22 grams = 2
<span>4.89 grams/2.44 grams = 2</span>

<span>Therefore, Potassium always combines with Oxygen in a ratio of 2  is to 1.</span>
4 0
2 years ago
At high temperature, 2.00 mol of HBr was placed in a 4.00 L container where it decomposed in the reaction: 2HBr(g) H2(g) Br2(g)
viktelen [127]

Answer: K_c for this reaction at this temperature is 0.029

Explanation:

Moles of  HBr = 2.00 mole

Volume of solution = 4.00 L

Initial concentration of HBr=\frac{moles}{Volume}=\frac{2.00}{4.00L}=0.500M

The given balanced equilibrium reaction is,

                            2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial conc.              0.500 M              0  M        0 M  

At eqm. conc.            (0.500-2x) M   (x) M   (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[H_2\times [Br_2]}{[HBr]^2}

Equilibrium concentration of [Br_2] = x =  0.0955 M

Now put all the given values in this expression, we get :

K_c=\frac{0.0955\times 0.0955}{0.500-2\times 0.0955}

K_c=0.029

Thus K_c for this reaction at this temperature is 0.029

7 0
2 years ago
Consider the equation: 2NO2(g) N2O4(g). Using ONLY the information given by the equation which of the following changes would in
Reika [66]
I believe the correct answer is the first option. To increase the molar concentration of the product N2O4, you should increase the pressure of the system. You cannot determine the effect of changing the temperature since we cannot tell whether it is an endothermic or an exothermic reaction. Also, decreasing the number of NO2 would not increase the product rather it would shift the equilibrium to the left forming more reactants. The only parameter we can change would be the pressure. And, since NO2 takes up more space than the product increasing the pressure would allow the reactant to collide more forming the product.
7 0
2 years ago
Read 2 more answers
Based on the activity series provided, which reactants will form products?
allsm [11]

Answer: CuI₂ + Br₂

Explanation:

1) The activity series F > Cl > Br > I means that F is the most active and I is the least active of those four elements (the halogens, group 17 in the periodic table).

The activity is a measure of how eager is an element to react compared to other elements in the series in a single replacement reaction.

2) Choice 1: CuI₂ + Br₂

Since the activity of Br is higher than that of I, Br will react with CuI₂, displacing I, which will be left alone, as per this chemical equation:

CuI₂ + Br₂ → CuBr₂ + I₂

Being I less active than Br, it cannot displace Br in CuBr₂.

3) Choice 2: Cl₂ + AlF₃

Being Cl less active than F, the former will not displace the latter, and the reaction will not proceed.

4) Choice 3: Br₂ + NaCl

Again, being Br less active than Cl, the former will not displace the latter, and the reaction will not proceed.

5) Choice 4: CuF₂ + I₂

Once more, being I less active than F, the former will not displace the latter, and the reaction will not proceed.

6 0
2 years ago
Read 2 more answers
Use the PhET simulation to identify what happens to the concentrations of a dilute Drink mix mixture or a pure Drink mix solutio
elena-s [515]

Hello user, you did not add a picture of an attachment to enable me help you solve this problem. Please check the attachment I added i believe it is the question that needs to be solved.

Explanation:

1. In the first column

<u>INCREASE CONCENTRATION</u>: these are the items that needs to be placed in this bin

a. Add drink mix solid to a diluted mixture of drink mix in pure water

b. Add drink mix solution to a diluted mixture of drink mix in pure water

c. Add drink mix solid to pure drink mix solution

d Evaporate water from a diluted mixture of drink mix in pure water

e Evaporate water from pure drink mix solution

2. The items below are what should be placed in this bin,<u>DECREASE CONCENTRATION:</u>

a. Add water to a diluted mixture of drink mix in pure water

b. Add water to pure drink mix solution

3. These items here should be placed here. <u>DOES NOT AFFECT CONCENTRATION:</u>

a. Add pure drink mix solution to pure drink mix solution

b. Drain the pure drink mix solution

c. Drain the diluted mixture

7 0
2 years ago
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