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julsineya [31]
3 years ago
14

A gas occupies 4.0 L of air at 1.0 atm pressure and 30.0°C. If the gas occupies 6.0 L of air at 37.0°C, what is the pressure of

the gas?
Chemistry
1 answer:
V125BC [204]3 years ago
3 0

Answer:

Gas pressure is measured in units such as torr, mmHg, atm, and Pa. ... You can see mass is small, Volume is large. - hense a small ... The pressure increases if volume decreases. ... For each of the following changes of the initial balloon, select ... A sample of nitrogen (N2) has a volume of 60.0 L at a pressure of 725 mmHg.

Explanation:

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Four elements are shown as they appear in the periodic table. which statement makes an accurate comparison of some of these elem
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Write the name of the following compound: H3PO4
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Which of the following is true when 1 M KCl is diluted to semi-molar KCl?
eimsori [14]

The solution before dilution and after dilution contains same number of moles, and water is added for dilution.

Option B

<h3><u>Explanation:</u></h3>

Suppose before dilution, the solution contains x moles of KCl in Y liter of water. Now as the concentration got halved, then the solution contains x moles of KCl in 2Y kiters of solution. So the number of moles of KCl in the solution remained constant.

Again, as the solution is diluted to half of the concentration, water must have been added with the solution to make it dilute.

4 0
3 years ago
Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 1.30 MJ o
Sunny_sXe [5.5K]

Answer:

97 000 g Na

Explanation:

The absortion (or liberation) of energy in form of heat is expressed by:

q=m*Cp*ΔT

The information we have:

q=1.30MJ= 1.30*10^6 J

ΔT = 10.0°C = 10.0 K (ΔT is the same in °C than in K)

Cp=30.8 J/(K mol Na)

If you notice, the Cp in the question is in relation with mol of Na. Before using the q equation, we can find the Cp in relation to the grams of Na.

To do so, we use the molar mass of Na= 22.99g/mol

Cp= \frac{30.8J}{K*mol Na}*\frac{1 mol Na}{22.99 g Na}=1.34\frac{J}{K*g Na}

Now, we are able to solve for m:

m=\frac{1.30*10^6 J}{1.34\frac{J}{K*g Na} *10.0K}=\frac{1.30*10^6J}{13.4\frac{J}{g Na} }  = 9.70*10^4 g Na=97 Kg Na=97 000 g Na

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3 years ago
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