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Jet001 [13]
3 years ago
12

According to the website of the National Aeronautics and Space Administration (NASA), the average temperature of the universe is

2.7 K. Convert this temperature to degrees Celsius.
Chemistry
1 answer:
charle [14.2K]3 years ago
3 0

Answer : The temperature in degree Celsius is, -270.45^oC

Explanation :

The conversion used for the temperature from Kelvin to degree Celsius is:

^oC=K-273.15

where,

K = temperature in Kelvin

^oC = temperature in centigrade

As we are given the temperature in Kelvin is, 2.7

Now we have to determine the temperature in Kelvin.

^oC=K-273.15

^oC=2.7-273.15

^oC=-270.45

Therefore, the temperature in degree Celsius is, -270.45^oC

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Salt water is considered to be a solution
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Answer:

the answer is unsaturated

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A saturated solution contains more solute per volume of solvent than an unsaturated solution. The solute has dissolved until no more can, leaving undissolved matter in the solution. ... In a supersaturated solution, there is more dissolved solute than in a saturated solution.

7 0
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How many moles of gas X are present if the gas has a volume of 2dm³ at room temperature and pressure? Give your answer to 2 deci
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Answer:

Approximately 0.08\; \rm mol, assuming that this gas is an ideal gas.

Explanation:

Look up the standard room temperature and pressure:25\; \rm ^{\circ}C and P = 101.325 \; \rm kPa.

The question states that the volume of this gas is V = 2\; \rm dm^{3}.

Convert the unit of all three measures to standard units:

\begin{aligned} T &= 25\; \rm ^{\circ}C \\ &= (25 + 273.15)\; \rm K \\ &= 293.15\; \rm K\end{aligned}.

\begin{aligned}P &= 101.325\; \rm kPa \\ &= 101.325 \; \rm kPa \times \frac{10^{3}\; \rm Pa}{1\; \rm kPa} \\ &= 1.01325 \times 10^{5}\; \rm Pa\end{aligned}.

\begin{aligned}V &= 2\; \rm dm^{3} \\ &= 2 \; \rm dm^{3} \times \frac{1\; \rm m^{3}}{10^{3}\; \rm dm^{3}} \\ &= 2 \times 10^{-3}\; \rm m^{3}\end{aligned}.

Look up the ideal gas constant in the corresponding units: R \approx 8.31\; \rm m^{3}\cdot Pa \cdot mol^{-1} \cdot K^{-1}.

Let n denote the number of moles of this gas in that V = 2\; \rm dm^{3}. By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:

P \cdot V = n \cdot R \cdot T.

Rearrange this equation and solve for n:

\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &\approx \frac{1.01325 \times 10^{5}\; {\rm Pa} \times 2 \times 10^{-3}\; {\rm m^{3}}}{8.31 \; {\rm m^{3} \cdot Pa \cdot mol^{-1} \cdot K^{-1}} \times 293.15\; {\rm K}} \\ &\approx 0.08\; \rm mol\end{aligned}.

In other words, there is approximately 2\; \rm mol of this gas in that V = 2\; \rm dm^{3}.

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A double displacement reaction takes place when strontium bromide and potassium sulfate are mixed and a precipitate of strontium sulfate is formed:

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