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Dima020 [189]
3 years ago
7

14. הכניסו בקבוק מים במסה של 100 גר' לתוך מקפיא, מה מסתו של הקרח לאחר שהמים קפאו ?

Chemistry
1 answer:
Andrei [34K]3 years ago
8 0

Answer:

whattt, can u write it in English so i can help you

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How does thermal energy transfer when a room is heated by a furnace?
Anna007 [38]
The answer is B,This is an example of convection, where fluid will circulate.This is because when part of air is warmed up, they expand and become less dense then rise,Therefore, the colder air contracts and become denser and sinks, in order to take in the place of the warm air. This phrase circulates and therefore the it can transfer heat a lot more faster.
 
5 0
3 years ago
Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.85 M of reagent
elixir [45]

Answer : The initial rate for a reaction will be 3.8\times 10^{-4}Ms^{-1}

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The chemical equation will be:

A+B+C\rightarrow P

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b[C]^c

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

6.1\times 10^{-5}=k(0.2)^a(0.2)^b(0.2)^c ....(1)

Expression for rate law for second observation:

1.8\times 10^{-4}=k(0.2)^a(0.2)^b(0.6)^c ....(2)

Expression for rate law for third observation:

2.4\times 10^{-4}=k(0.4)^a(0.2)^b(0.2)^c ....(3)

Expression for rate law for fourth observation:

2.4\times 10^{-4}=k(0.4)^a(0.4)^b(0.2)^c ....(4)

Dividing 1 from 2, we get:

\frac{1.8\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.2)^a(0.2)^b(0.6)^c}{k(0.2)^a(0.2)^b(0.2)^c}\\\\3=3^c\\c=1

Dividing 1 from 3, we get:

\frac{2.4\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.4)^a(0.2)^b(0.2)^c}{k(0.2)^a(0.2)^b(0.2)^c}\\\\4=2^a\\a=2

Dividing 3 from 4, we get:

\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.4)^a(0.4)^b(0.2)^c}{k(0.4)^a(0.2)^b(0.2)^c}\\\\1=2^b\\b=0

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^0[C]^1

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

6.1\times 10^{-5}=k(0.2)^2(0.2)^0(0.2)^1

k=7.6\times 10^{-3}M^{-2}s^{-1}

Now we have to calculate the initial rate for a reaction that starts with 0.85 M of reagent A and 0.70 M of reagents B and C.

\text{Rate}=k[A]^2[B]^0[C]^1

\text{Rate}=(7.6\times 10^{-3})\times (0.85)^2(0.70)^0(0.70)^1

\text{Rate}=3.8\times 10^{-3}Ms^{-1}

Therefore, the initial rate for a reaction will be 3.8\times 10^{-3}Ms^{-1}

6 0
3 years ago
What is an orbital shell
Lyrx [107]
<span>orbital shell is the</span><span> the circular paths around the nucleus of an atom along which the electrons traverse.</span>
4 0
3 years ago
Copper(II) carbonate undergoes thermal decomposition. One of the pro
Llana [10]

Answer: The green copper (II) carbonate (CuCO_3) changes to black copper oxide (CuO)

CuCO_3(s)\stackrel{\Delta }\rightarrow CaO(s)+CO_2(g)

Explanation:

Decomposition is defined as the chemical reaction in which a single compound gives two or more simple substances. It requires energy to break the bonds between reactants, thus is an endothermic process.

Thermal decomposition uses heat for decomposition.

The chemical equation for thermal decomposition of copper (II) carbonate is:

CuCO_3(s)\stackrel{\Delta }\rightarrow CaO(s)+CO_2(g)

The green copper (II) carbonate (CuCO_3) changes to black copper oxide (CuO)

8 0
3 years ago
A 34.53 ml sample of a solution of sulfuric acid, h2s04, reacts with 27.86 ml of 0.08964 m naoh solution. calculate the molarity
Gnoma [55]
The balanced equation between NaOH and H₂SO₄ is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of moles of NaOH moles reacted = molarity of NaOH x volume
number of NaOH moles = 0.08964 mol/L x 27.86 x 10⁻³ L = 2.497 x 10⁻³ mol
according to molar ratio of 2:1
2 mol of NaOH reacts with 1 mol of H₂SO₄
therefore 2.497 x 10⁻³ mol of NaOH reacts with - 1/2 x 2.497 x 10⁻³ mol of H₂SO₄
number of moles of H₂SO₄ reacted - 1.249 x 10⁻³ mol 
Number of H₂SO₄ moles in 34.53 mL - 1.249 x 10⁻³ mol 
number of H₂SO₄ moles in 1000 mL - 1.249 x 10⁻³ mol / 34.53 x 10⁻³ L = 0.03617 mol 
molarity of H₂SO₄ is 0.03617 M
6 0
3 years ago
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