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marishachu [46]
3 years ago
7

1. A gas is confined in a cylinder with a movable piston at one end. When the volume of the

Chemistry
1 answer:
Sedaia [141]3 years ago
7 0

Answer: 211 kPa

Explanation:

Given that,

Original pressure of gas (P1) = 125 kPa

Original volume of cylinder (V1) = 760 cm3

New pressure of gas (P2) = ?

New volume of cylinder (V2) = 450 cm3

Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law

P1V1= P2V2

125 kPa x 760 cm3= P2 x 450 cm3

95000 kPa•cm3 = 450 cm3•P2

Divide both sides by 450 cm3

95000 kPa•cm3/450 cm3 = 450 cm3•P2/450 cm3

211.1 kPa = P2

[Round the value of P2 to 3 significant numbers. Hence, P2 becomes 211 kPa]

Thus, the new pressure within the cylinder is 211 kPa

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Which of the five most abundant elements are metals?
Gala2k [10]

Answer:

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Hope it helps :)

Mark me as brainliest

6 0
3 years ago
Calculate δ h for the reaction:no (g) + o2 (g) ↔ no2 (g). given: 2o3(g) ↔ 3o2(g) δh=-426 kj o2(g) ↔ 2o(g) δh=+ 490 kj no(g) + o3
maks197457 [2]
To calculate the <span>δ h, we must balance first the reaction: 

NO + 0.5O2 -----> NO2

Then we write all the reactions,

2O3 -----> 3O2    </span><span>δ h = -426 kj        eq. (1)

O2 -----> 2O    </span><span>δ h = 490 kj             eq. (2)

NO + O3 -----> NO2 + O2    </span><span>δ h = -200 kj          eq. (3)


We divide eq. (1) by 2, we get

</span>O3 -----> 1.5O2    δ h = -213  kj             eq. (4)

Then, we subtract eq. (3) by eq. (4) 

NO + O3 ----->  NO2 + O2   δ h = -200 kj
-       (O3 -----> 1.5 O2         δ h = -213  kj)
NO -----> NO2 - 0.5O2        δ h = 13  kj               eq. (5)


eq. (2) divided by -2. (Note: Dividing or multiplying by negative number reverses the reaction)

O -----> 0.5O2  <span>δ h = -245  kj         eq. (6)
</span>
Add eq. (6) to eq. (5), we get

NO -----> NO2 - 0.5O2        δ h = 13  kj 
+  O -----> 0.5O2                 δ h = -245  kj
NO + O ----> NO2               δ h = -232 kj

<em>ANSWER:</em> <em>NO + O ----> NO2               δ h = -232 kj</em>


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3 years ago
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3 years ago
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How much sucrose (g) do you need to weight in order to prepare 19.16 g of a 13.1 % (weight percent) solution?
Nonamiya [84]
<span>2.51 grams
   You want to prepare 19.16 g of some solution which will have 13.1% of it's mass being sucrose. So we just need to perform some simple multiplication: 19.16g * 0.131 = 2.50996g
   Rounding to 3 significant figures gives 2.51 g.</span>
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