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Assoli18 [71]
4 years ago
12

Engine size is measured by how much air the cylinders displace. A 1996 Pontiac V8 has an engine size of 316.6 cubic inches. What

is the size of this engine in cubic meters?
Chemistry
1 answer:
inna [77]4 years ago
3 0

The size of this engine in cubic meters is 0.0052 m^{3}.

<u>Explanation:</u>

As it is stated that engine size can be measured depending upon the amount of air the cylinders displace, it is also stated that the engine of model Pontiac V8 has a size of 316.6 cubic inches. The problem is to convert the size of engine in cubic meters. We know that 1 inch = 0.0254 m.

Then 1 inch^{3}=(0.0254)^{3} m^{3}

As here the size of engine is 316.6 cubic inches, then in metres it will be

    316.6 \text { inches }^{3}=316.6 \times(0.0254)^{3} \mathrm{m}^{3}=0.0052 \mathrm{m}^{3}

Thus, the size of the engine in cubic meters will be 0.0052 m^{3}.

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A chemist titrates 60.0 mL of a 0.1935 M benzoic acid (HC (H5CO2) solution with 0.2088 M KOH solution at 25 °C. Calculate the pH
erik [133]

Answer:

pH at the equivalence point is 8.6

Explanation:

A titulation between a weak acid and a strong base, gives a basic pH at the equivalence point. In the equivalence point, we need to know the volume of base we added, so:

mmoles acid = mmoles of base

60 mL . 0.1935M = 0.2088 M . volume

(60 mL . 0.1935M) /0.2088 M = 55.6 mL of KOH

The neutralization is:

HBz + KOH  ⇄  KBz  +  H₂O

In the equilibrum:

HBz + OH⁻   ⇄  Bz⁻  +  H₂O

mmoles of acid are: 11.61 and mmoles of base are: 11.61

So in the equilibrium we have, 11.61 mmoles of benzoate.

[Bz⁻] = 11.61 mmoles / (volume acid + volume base)

[Bz⁻] = 11.61 mmoles / 60 mL + 55.6 mL = 0.100 M

The conjugate strong base reacts:

  Bz⁻  +  H₂O  ⇄  HBz + OH⁻    Kb

0.1 - x                       x        x

(We don't have pKb, but we can calculate it from pKa)

14 - 4.2 = 9.80 → pKb  → 10⁻⁹'⁸ = 1.58×10⁻¹⁰ → Kb

Kb = [HBz] . [OH⁻] / [Bz⁻]

Kb = x² / (0.1 - x)

As Kb is so small, we can avoid the quadratic equation

Kb =  x² / 0.1 → Kb . 0.1 = x²

√ 1.58×10⁻¹¹ = [OH⁻] = 3.98 ×10⁻⁶ M

From this value, we calculate pOH and afterwards, pH (14 - pOH)

- log [OH⁻] =  pOH → - log 3.98 ×10⁻⁶  = 5.4

pH = 8.6

7 0
3 years ago
A reaction produces 0.829 moles of H2O. How many molecules are produced?
mr_godi [17]
You should get 5.05 x 10^23 moles.
8 0
4 years ago
Isaac newton eyes optics is it a theory or law?
kumpel [21]

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4 0
4 years ago
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Charcoal is primarily carbon. what mass of co2 is produced if you burn enough carbon (in the form of charcoal) to produce 4.80 k
Reika [66]
When carbon is burned in air carbon iv oxide gas is formed.
C (s) + O2 (g) = CO2(g)  ΔH = - 393.5 kj/mol
The enthalpy change of the  reaction is -393.5 j/mol which means that when one mole of carbon is completely burnt in air then 393.5 j of energy is evolved.
Thus, 1 mole = -393.5 j , then for 480 kj 
            = 480 × 1/393.5
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6 0
4 years ago
How many moles are in 175g<br> (Ca(NO3)2
zheka24 [161]

Answer:

appx. 1.07 moles

Explanation:

175g of molecule Ca(NO3)2

To find the # of moles, use stoichoimetry.

1. We need the molar mass of Ca(NO3)2.

Ca mass: 40.08 g

NO3 mass: N + 3(O) --> 14.01 + 3(16.00) --> 62.01

Molar mass = Ca + 2(NO3) --> 40.08 + 2(62.01) --> 164.1 g

2. write out the calculation

175g Ca(NO3)2 * (1 mole Ca(NO3)2)/(molar mass of Ca(NO3)2)

175 g Ca(NO3)2 * (1 mole Ca(NO3)2)/(164.1 g Ca(NO3)2)

The g units cancel out and we're left with moles.

Simply perform the calculation now: 175*1/164.1 ≅1.07 moles

6 0
3 years ago
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