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olga2289 [7]
3 years ago
7

3. A cylinder of compressed gas has a pressure of 4.882 atm on one day. The next

Chemistry
1 answer:
Debora [2.8K]3 years ago
7 0

Answer:

PART A

Explanation:

3. A cylinder of compressed gas has a pressure of 4.882 atm on one day. The next

day, the same cylinder of gas has a pressure of 4.690 atm, and its temperature is

8°C. What was the temperature on the previous day in °C? Ans: 20°C.

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7.66 Write balanced equations for the following reactions: (a) potassium oxide with water, (b) diphosphorus trioxide with water,
White raven [17]

Explanation:

(a) potassium oxide with water

K_2O(s)+H_2O(l)\rightarrow 2KOH(aq)

According to reaction,1 mole of potassium  oxide reacts with 1 mole of water to give 1 mole of potassium hydroxide.

(b) diphosphorus trioxide with water

P_2O_3(s)+3H_2O(l)\rightarrow 2H_3PO_3(aq)

According to reaction,1 mole of diphosphorus trioxide reacts with 2 moles of water  to give 2 moles of phosphorus acid.

(c) chromium(III) oxide with dilute hydrochloric acid,

Cr_2O_3(s)+6HCl(aq)\rightarrow 2CrCl_3(aq)+3H_2O(l)

According to reaction,1 mole of chromium(III) oxide reacts with 6 moles of hydrochloric acid to give 2 moles of chromium(III) chloride and 3 moles of water.

(d) selenium dioxide with aqueous potassium hydroxide

SeO_2 (s)+2KOH (aq)\rightarrow K_2SeO_3(aq)+H_2O(l)

According to reaction,1 mole of selenium dioxide reacts with 2 moles of potassium hydroxide to give 1 mole of potassium selenite and 1 mole of water.

6 0
3 years ago
First, I would balance the chemical equation.
andrezito [222]
......................ok
5 0
3 years ago
Read 2 more answers
The distance between the centers of the two oxygen atoms in an oxygen molecule is 1.21 x 10-8 cm. What is this distance in inche
Sergio [31]

Answer:

d=4.75\times 10^{-9}\ \text{inches}

Explanation:

Given that,

The distance between the centers of the two oxygen atoms in an oxygen molecule is 1.21\times 10^{-8}\ cm.

We need to convert this distance in inches.

We know that,

1 cm = 0.393 inches

We can solve it as follows :

1.21\times 10^{-8}\ cm=0.393\times 1.21\times 10^{-8}\\\\=4.75\times 10^{-9}\ \text{inches}

So, the distance between the centers of the two oxygen atoms is 4.75\times 10^{-9}\ \text{inches}.

5 0
2 years ago
Express the product of 9.0 x 10^-4m and 8.1 x 10^4m using the correct number of significant digits
Assoli18 [71]

Answer:

7.3 × 10¹ m²

Explanation:

The language "product" means that we must multiply the numbers given to get our answer.

9.0 × 10⁻⁴ m × 8.1 × 10⁴ m = 7.3 × 10¹ m²

The significant figures rule for multiplication is that we must us the amount of significant figures in the number with the least significant figures, in this question this is 2 significant digits from both 9.0 × 10⁻⁴ m and 8.1 × 10⁴ m.

7 0
3 years ago
Ethanol melts at -114 degree C. The enthalpy of fusion
Brut [27]

Answer: The heat required is 6.88 kJ.

Explanation:

The conversions involved in this process are :

(1):ethanol(s)(-135^0C)\rightarrow ethanol(s)(-114^0C)\\\\(2):ethanol(s)(-114^0C)\rightarrow ethanol(l)(-114^0C)\\\\(3):ethanol(l)(-114^0C)\rightarrow ethanol(l)(-50^0C)

Now we have to calculate the enthalpy change.

\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]

where,

\Delta H = enthalpy change = ?

m = mass of ethanol = 25.0 g

c_{p,s} = specific heat of solid ethanol= 0.97 J/gK

c_{p,l} = specific heat of liquid ethanol = 2.31 J/gK

n = number of moles of ethanol = \frac{\text{Mass of ethanol}}{\text{Molar mass of ethanol}}=\frac{25.0g}{46g/mole}=0.543mole

\Delta H_{fusion} = enthalpy change for fusion = 5.02 KJ/mole = 5020 J/mole

T_{final}-T_{initial}=\Delta T = change in temperature

The value of change in temperature always same in Kelvin and degree Celsius.

Now put all the given values in the above expression, we get

\Delta H=[25.0 g\times 0.97J/gK\times (-114-(-135)K]+0.534mole\times 5020J/mole+[25.0g\times 2.31J/gK\times (-50-(-114))K]

\Delta H=6885.93J=6.88kJ     (1 KJ = 1000 J)

Therefore, the heat required is 6.88 kJ

3 0
2 years ago
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