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puteri [66]
1 year ago
6

What is pure substancesubscribe this channel​

Chemistry
1 answer:
irina [24]1 year ago
8 0

A pure substance is a material made of only one type of particle.

What is pure substance?

A substance that has fixed in chemical composition throughout is called a pure substance such as water, air and nitrogen. The  pure substance doesn't have to be of a single element or compound. A pure substance consists only of one element for the or one compound. a mixture consists of two or more different substances, not chemically joined together.

Examples of pure substance include tin, sulfur, diamond, water, pure sugar (sucrose), table salt (sodium chloride) and the  baking soda (sodium bicarbonate). Crystals, in general, are pure substances.

Elements are the pure substances that cannot be broken down into simpler substances by any physical or chemical means we know  as they have only one kind of atom at the  entire composition. Thus when gold is tjr broken down it still remains gold and hence is considered a pure substance is and an element.

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Round off each number to the indicated number of significant figures (sf)
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At standard pressure, ammonia melts at 195 K and boils at 240 K. If a sample of ammonia at standard pressure is cooled from 200
Svetradugi [14.3K]

Answer:

I) the heat capacity of ammonia(s)

II) the heat capacity of ammonia(ℓ)

IV) the enthalpy of fusion of ammonia

Explanation:

Initially, ammonia at 200 K is liquid. To calculate the change of enthalpy from 200 K to 195 K (melting point) we need to know the heat capacity of ammonia(ℓ).

At 195, ammonia is in the transition from liquid to solid (solidification). To calculate the change of enthalpy in that process we need to know the enthalpy of solidification of ammonia, which has the same value but opposite sign to the enthalpy of fusion of ammonia.

From 195 K to 0 K, ammonia is solid. To calculate the change of enthalpy in that process we need to know the heat capacity of ammonia(s).

7 0
3 years ago
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
bezimeni [28]

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}.

6 0
2 years ago
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