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2 years ago

During an endothermic phase change, what happens to the potential energy and the kinetic energy?

2 answers:

7 0

The correct answer should be that Potential energy increases, and kinetic energy increases, since they both increase as the temperature changes.

m_a_m_a [10]2 years ago

7 0

Answer:

D.) Potential energy increases, and kinetic energy stays the same

Explanation:

edge 2020

hope this helps!!

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First person with the right answer gets brainliest thanks (btw the numbers on the right are the answers choose the right one)

kap26 [50]

Molar mass of FE2O3=2(55.85)+3(16)=159.7

2.56g*1mol/159.7*2mol/1mol*55.85g/1mol=1.79g

6 0

3 years ago

How many moles of oxygen are formed when 58.6 g of KNO3 decomposes according to the following reaction? 4 KNO3(s) → 2 K2O(s) + 2

Leya [2.2K]

Answer:

0.725 mol

Explanation:

Moles are calculated as the given mass divided by the molecular mass.

i.e. ,

moles = ( mass / molecular mass )

since,

mass of KNO₃ = 58.6 g ( given )

Molecular mass of KNO₃ = 101 g / mol

Therefore,

moles of KNO₃ = 58.6 g / 101 g / mol

moles of KNO₃ = 0.58 mol

From the balanced reaction ,

4 KNO₃ (s) ---> 2K₂O (s) + 2N₂ (g) + 5O₂ (g)

By the decomposition of 4 mol of KNO₃ , 5 mol of O₂ are formed ,

hence, unitary method is used as,

1 mol of KNO₃ gives 5 / 4 mol O₂

Therefore,

0.58 mol of KNO₃ , gives , 5 / 4 * 0.58 mol of O₂

Solving,

0.58 mol of KNO₃ , gives , 0.725 mol of O₂

Therefore,

58.6g of KNO₃ gives 0.725 mol of O₂.

3 0

2 years ago

A)The lower the greenhouse gases, the higher the ocean temperature.

HACTEHA [7]

The higher the greenhouse gases, the higher the ocean temperature

"The ocean absorbs most of the excess heat from greenhouse gas emissions, leading to rising ocean temperatures"

hope this helps :)

6 0

2 years ago

Read 2 more answers

A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.

irga5000 [103]

Answer:

1. pH = 1.23.

2. $H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

$H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+$

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the pKa is:

$pKa=-log(Ka)=-log(5.90x10^{-2})=1.23$

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

$pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23$

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

$n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol$

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

$H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Which is also shown in net ionic notation.

Best regards!

4 0

3 years ago

What is hexane?

Mamont248 [21]

Answer:

C. A hydrocarbon molecule containing six carbon atoms and only

single bonds

Explanation:

hope it helps

8 0

3 years ago