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kakasveta [241]
3 years ago
9

The table below describes the temperature of different colors of stars.

Chemistry
1 answer:
nydimaria [60]3 years ago
3 0

Answer:

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Explanation:

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Find the mass in kilograms of the liquid air that is required to produce 600L of oxygen. In normal condition, 1L of liquid air h
IgorC [24]

Mass in kilograms of liquid air required = 0.78 kg

<u>Given that </u>

1 Litre of liquid air contains 1.3 grams of oxygen ( air )

<u />

<u> Determine the a</u><u>mount of liquid air</u><u> in Kg</u>

volume of air given = 600 L

mass of liquid air required = x

1 litre = 1.3 grams

600 L =  x

∴ x ( mass of liquid air ) = 1.3 * 600

                                       = 780 g  = 0.78 kg

Hence we can conclude that Mass in kilograms of liquid air required = 0.78 kg

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2 years ago
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Determine the volume in mL of 0.242 M NaOH(aq) needed to reach the equivalence (stoichiometric) point in the titration of 46.79
Alla [95]

The volume (in mL) of 0.242 M NaOH solution needed for the titration reaction is 39.44 mL

<h3>Balanced equation </h3>

CH₃CH₂COOH + NaOH —> CH₃CH₂COONa + H₂O

From the balanced equation above,

  • The mole ratio of the acid, CH₃CH₂COOH (nA) = 1
  • The mole ratio of the base, NaOH (nB) = 1

<h3>How to determine the volume of NaOH</h3>
  • Volume of acid, CH₃CH₂COOH (Va) = 46.79 mL
  • Molarity of acid, CH₃CH₂COOH (Ma) = 0.204 M
  • Molarity of base, NaOH (Mb) = 0.242 M
  • Volume of base, KOH (Vb) =?

MaVa / MbVb = nA / nB

(0.204 × 46.79) / (0.242 × Vb) = 1

Cross multiply

0.242 × Vb = 0.204 × 46.79

Divide both side by 0.242

Vb = (0.204 × 46.79) / 0.242

Vb = 39.44 mL

Thus, the volume of NaOH needed for the reaction is 39.44 mL

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Explanation:

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3 years ago
Suppose that you add 26.7 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f Kf of 5.12 oC/m. With the a
kiruha [24]

From the calculation, the molar mass of the solution is 141 g/mol.

<h3>What is the molar mass?</h3>

We know that;

ΔT = K m i

K = the freezing constant

m = molality of the solution

i = the Van't Hoft factor

The molality of the solution is obtained from;

m = ΔT/K i

m = 3.89/5.12 * 1

m = 0.76 m

Now;

0.76 =  26.7 /MM/0.250

0.76 = 26.7 /0.250MM

0.76 * 0.250MM =  26.7

MM= 26.7/0.76 * 0.250

MM = 141 g/mol

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there are 2 copper atoms on the left side of the arrow, but none on the other side. there are also 7 oxygen atoms on the left hand side, but only 5 on the right hand side of the arrow.
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