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prisoha [69]
3 years ago
15

A factory at a low elevation ships bags of popcorn to cities at high elevations. Why does the factory manager need to monitor th

e amount of air in each bag
Chemistry
2 answers:
motikmotik3 years ago
8 0

Explanation:

Since factory is at low elevation, therefore, there will be more pressure. Similarly at high elevations there will be low pressure.

So, when bags of popcorn move from low elevation to high elevation the factory manager need to monitor the amount of air in each bag because if there are too many molecules in the bag then it will pop as it goes from high to low pressure areas.



german3 years ago
7 0

A factory at a low elevation ships bags of popcorn to cities at high elevations. The factory manager need to monitor the amount of air in each bag because when the bags of popcorn are delivered into high elevation cities, it will shrink due to the high pressure exerted by the surroundings towards the bag. As the elevation increases, the pressure surrounding it also increases.

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Answer:

Below

Explanation:

Balanced form;

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1.Benzene + Dioxygen = Carbon Dioxide + Water

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4.This means that the carbon dioxide and limewater react to produce calcium carbonate and water.

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5 0
3 years ago
A solution is made by dissolving 10.7 g of magnesium sulfate, MgSO4, in enough water to make exactly 100 mL of solution. Calcula
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Answer:

[MgSO₄] = 890 mM/L

Explanation:

In order to determine molarity we need to determine the moles of solute that are in 1L of solution.

Solute: MgSO₄ (10.7 g)

Solvent: water

Solution: 100 mL as volume. (100 mL . 1L / 1000mL) = 0.1L

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Identify the limiting reactant when 32. 0 g hydrogen is allowed to react with 16. 0 g oxygen
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Oxygen is limiting reactant

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2 H2  +   O2  ======> 2 H2 O

from this equation (and periodic table) you can see that

  4 gm of H combine with 32 gm O2  

     H / O  =  4/32 = 1/8

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3 0
1 year 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}.

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