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nikklg [1K]
3 years ago
11

You decide to establish a new temperature scale on which the melting point of ammonia (-77.75 ∘c) is 0∘a, and the boiling point

of ammonia (-33.35∘c) is 100 ∘a. part a what would be the boiling point of water in ∘a?
Chemistry
2 answers:
Hoochie [10]3 years ago
7 0
First we need to know that the boiling point of water in C is 100 and we just need to solve for x in the equation:

-33.75-(-77.75) / 100 = 100-(-77.75) / x
44.4/100 = 177.75 / x
 x = 177.75*100/44.4 = 400.33

The boiling point of water in ∘a would be 400.33∘a.

AveGali [126]3 years ago
3 0

<u>Answer:</u> The boiling point of water is 400.34^oA

<u>Explanation:</u>

To convert the units of temperature, we use the equation:

\frac{\text{X-LFP}}{\text{UFP-LFP}}=\frac{\text{Y-LFP}}{\text{UFP-LFP}}

where,

X = temperature in ^oA

Y = temperature in ^oC

LFP = lower fixed point

UFP = upper fixed point

LFP of ^oA is 0°A

UFP of ^oA is 100°A

LFP of ^oC is -77.75°A

UFP of ^oC is -33.35°A

Normal boiling point of water = 100^oC

Putting values in above equation, we get:

\frac{T(^oA)-0^oA}{100^oA-0^oA}=\frac{100^oC-(-77.75^oC)}{-33.35^oC-(-77.75^oC)}\\\\\frac{T(^oA)}{100^oA}=\frac{177.75^oC}{44.4^oC}\\\\T(^oA)=400.34^oA

Hence, the boiling point of water is 400.34^oA

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Xg --------------------- 6300 mL
X = 0,000179×6300
X = 1,1277g ≈ 1,1g

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7 0
3 years ago
A carbon compound with a covalently bonded chlorine or bromine is called _____. an amide an halocarbon an alcohol an aldehyde
german
Halogens are elements that can be found in group 7 of the periodic table. They have 7 electrons in their outer shell and thus can form only a single covalent bond with other elements. Examples of halogens include chlorine, bromine and fluorine. A carbon compound that is covalently bonded with chlorine or bromine is called a halocarbon.
7 0
3 years ago
How many kilojoules of energy would be required to heat a 225g block of aluminum from 23.0 C to 73.5 C?
gulaghasi [49]

Answer:

\boxed {\boxed {\sf 10.2 \ kJ}}

Explanation:

We are asked to find how many kilojoules of energy would be required to heat a block of aluminum.

We will use the following formula to calculate heat energy.

q=mc \Delta T

The mass (m) of the aluminum block is 225 grams and the specific heat (c) is 0.897 Joules per gram degree Celsius. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature.

  • ΔT = final temperature - inital temperature

The aluminum block was heated from 23.0 °C to 73.5 °C.

  • ΔT= 73.5 °C - 23.0 °C = 50.5 °C

Now we know all three variables and can substitute them into the formula.

  • m= 225 g
  • c= 0.897 J/g° C
  • ΔT= 50.5 °C

q= (225 \ g )(0.897 \ J/g \textdegree C)(50.5 \textdegree C)

Multiply the first two numbers. The units of grams cancel.

q= (225 \ g  * 0.897 \ J/g \textdegree C)(50.5 \textdegree C)

q= (225   * 0.897 \ J / \textdegree C)(50.5 \textdegree C)

q= (201.825\ J / \textdegree C)(50.5 \textdegree C)

Multiply again. This time, the units of degrees Celsius cancel.

q= 201.825 \ J * 50.5

q= 10192.1625 \ J

The answer asks for the energy in kilojoules, so we must convert our answer. Remember that 1 kilojoule contains 1000 joules.

\frac { 1  \ kJ}{ 1000 \ J}

Multiply by the answer we found in Joules.

10192.1625 \ J * \frac{ 1 \ kJ}{ 1000 \ J}

10192.1625  * \frac{ 1 \ kJ}{ 1000 }

\frac {10192. 1625}{1000} \ kJ

10.1921625 \ kJ

The original values of mass, temperature, and specific heat all have 3 significant figures, so our answer must have the same. For the number we found, that is the tneths place. The 9 in the hundredth place tells us to round the 1 up to a 2.

10.2 \ kJ

Approximately <u>10.2 kilojoules</u> of energy would be required.

3 0
3 years ago
Calculate the number of molecules in a deep breath of air whose volume is 2.15 L at body temperature, 37 ∘C, and a pressure of 7
Zinaida [17]
PV = nRT 
R = 0.0821 L * atm / mol * K 
(ideal gas constant) 

First, convert 735 torr to atm. Divide by 760. 
(1 atm = 760 torr) 

735 torr * 1 atm / 760 torr = 0.967 atm 
Then, convert 37 C to Kelvin. Just add 273. 
37 C = 310K 

n = PV / RT 
= (0.967)(2.07) / (0.0821)(310) 
= 0.0786 mol 

<span>0.0786 mol * 6.02 * 10^23 molecules / 1 mol = 4.73 * 10^22 molecules </span>
7 0
3 years ago
Read 2 more answers
1-propanol (P1° = 20.9 Torr at 25 °C) and 2-propanol (P2° = 45.2 Torr at 25 °C) form ideal solutions in all proportions. Let x1
jekas [21]

Answer:

y1 = 0.3162

y2 = 0.6838

Explanation:

ok let us begin,

first we would be defining the parameters;

at 25°C;

1-propanol P1° = 20.90 Torr

2-propanol P2° = 45.2 Torr

From Raoults law:

P(1-propanol) = P⁰ × X(1-propanol)

P(1-propanol) = 20.9 torr × 0.45 = 9.405

P(1-propanol) = 9.405 torr

Also P(2-propanol) = P⁰ × X(2-propanol)

P(2-propanol) = 45.2 torr × 0.45

P(2-propanol) = 20.34 torr

but the total pressure = sum of individual pressures

total pressure = 9.405 + 20.34

total pressure = 29.745 torr

given that y1 and y2 represent the mole fraction of each in the vapor phase

y1 = P1 / total pressure

y1 = 9.405/29.745

y1 = 0.3162

Since y1 + y2 = 1

y2 = 1 - y1

∴ y2 = 1 -  0.3162

y2 = 0.6838

cheers, i hope this helps.

7 0
3 years ago
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