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ElenaW [278]
3 years ago
14

Select all that apply. Select all the items that are public goods or services.

Chemistry
2 answers:
IRINA_888 [86]3 years ago
7 0
Education millitary and police force roads is not a public good or servicev he is wrong
Gekata [30.6K]3 years ago
4 0
All of them are technically a public good or a service because for a public goods it would be all and service it would be all so.... yah
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What are the products of the following equation?
Artyom0805 [142]
The products are on the right side of the equation. For this one it would be 2AlPO4 + 3CaSO4
5 0
3 years ago
How many moles of chlorine gas would occupy a volume of 35.5 L at a pressure of 100.0 kPa and a temperature of 100.0 degrees Cel
o-na [289]

1.137448506 mol moles of chlorine gas would occupy a volume of 35.5 L at a pressure of 100.0 kPa and a temperature of 100.0 degrees Celsius.

<h3>What is an ideal gas equation?</h3>

The ideal gas equation, pV = nRT, is an equation used to calculate either the pressure, volume, temperature or number of moles of a gas. The terms are: p = pressure, in pascals (Pa). V = volume, in m^3.

We apply the formula of the ideal gases, we clear n (number of moles); we use the ideal gas constant R = 0.082 l atm / K mol:

PV= nRT

Given data:

P=100.0 kPa =0.986923 atm

T=100 degree celcius= 100 + 273 =373 K

V=35.5 L

Substituting the values in the equation.

n= \frac{\;0,98 \;atm \;X \;35,5 \;L }{\;0,082\;atm / \;K mol \;X \;373 K}

n= 1.137448506 mol

Hence, 1.137448506 mol moles of chlorine gas would occupy a volume of 35.5 L at a pressure of 100.0 kPa and a temperature of 100.0 degrees Celsius.

Learn more about ideal gas here:

brainly.com/question/16552394

#SPJ1

7 0
2 years ago
Two solutions namely, 500 ml of 0.50 m hcl and 500 ml of 0.50 m naoh at the same temperature of 21.6 are mixed in a constant-pre
weeeeeb [17]

24.6 ℃

<h3>Explanation</h3>

Hydrochloric acid and sodium hydroxide reacts by the following equation:

\text{HCl} \; (aq) + \text{NaOH} \; (aq) \to \text{NaCl} \; (aq) + \text{H}_2\text{O} \; (aq)

which is equivalent to

\text{H}^{+} \; (aq) + \text{OH}^{-} \; (aq) \to \text{H}_2\text{O}\; (l)

The question states that the second equation has an enthalpy, or "heat", of neutralization of -56.2 \; \text{kJ}. Thus the combination of every mole of hydrogen ions and hydroxide ions in solution would produce 56.2 \; \text{kJ} or 56.2 \times 10^{3}\; \text{J} of energy.

500 milliliter of a 0.50 mol per liter "M" solution contains 0.25 moles of the solute. There are thus 0.25 moles of hydrogen ions and hydroxide ions in the two 0.500 milliliter solutions, respectively. They would combine to release 0.25 \times 56.2 \times 10^{3} = 1.405 \times 10^{4} \; \text{J} of energy.

Both the solution and the calorimeter absorb energy released in this neutralization reaction. Their temperature change is dependent on the heat capacity <em>C</em> of the two objects, combined.

The question has given the heat capacity of the calorimeter directly.

The heat capacity (the one without mass in the unit) of water is to be calculated from its mass and <em>specific</em> heat.

The calorimeter contains 1.00 liters or 1.00 \times 10^{3} \; \text{ml} of the 1.0 gram per milliliter solution. Accordingly, it would have a mass of 1.00 \times 10^{3} \; \text{g}.

The solution has a specific heat of 4.184 \; \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1}. The solution thus have a heat capacity of 4.184 \times 1.00 \times 10^{3} = 4.184 \times 10^{3} \; \text{J} \cdot\text{K}^{-1}. Note that one degree Kelvins K is equivalent to one degree celsius ℃ in temperature change measurements.

The calorimeter-solution system thus has a heat capacity of 4.634 \times 10^{3} \; \text{J} \cdot \text{K}^{-1}, meaning that its temperature would rise by 1 degree celsius on the absorption of 4.634 × 10³ joules of energy. 1.405 \times 10^{4} \; \text{J} are available from the reaction. Thus, the temperature of the system shall have risen by 3.03 degrees celsius to 24.6 degrees celsius by the end of the reaction.

4 0
3 years ago
8. When a 2.5 mol of sugar (C12H22O11) are added to a certain amount of water the boiling point is raised by 1 Celsius degree. I
Julli [10]

Hey there!:

8) ΔTb = i*Kb*m  

 m is molality

 Since same number of mol is added to same amount of water in both cases

m will be same for both

is 1 for glucose since it is covalent compound

is 4 of Al(NO3)3 as it breaks into 1 Al₃⁺ and 3 NO₃⁻

So,  ΔTb will be 4 times in aluminum nitrate case

So, boiling point will change by 4ºC


9) use Q = m*  L

L =  heat of vaporization so:

T1=T2=100ºC

5.40 * 1000 => 5400  cal/g

Q =   5400 / 540

Q = 10 grams


Hope that thlps!

5 0
3 years ago
Use the following equation to answer the questions below.
sertanlavr [38]

Answer:

Explanation:

Here is ur answer what you wanna do first is calculate the first and second numbers and letters this should be ur answer = -394 + 2 x (-242) - (-75) + 2 x 0 = -803 kJ/mol

4 0
2 years ago
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