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kap26 [50]
3 years ago
5

A package of aluminum foil is 63.2 yd long, 11 in. wide, and 0.00035 in. thick. If aluminum has a density of 2.70 g/cm³, what is

the mass, in grams, of the foil?
Chemistry
1 answer:
mina [271]3 years ago
3 0

Answer:

Mass of aluminium foil = 387.57_g

Explanation:

density of aluminium foil 2.70 g/cm³,

1_yd = 91.44_cm

1_in = 2.54_cm

Length of aluminum foil = 63.2_yd = 5779.008_cm

Width of aluminium foil = 11 in = 27.94_cm

Thickness of aluminium foil = 0.00035_in. = 0.000889_cm

Volume of aluminium foil = length × width × thickness = 5779.008_cm × 27.94_cm × 0.000889_cm = 143.54_cm^3

Mass of aluminium foil = (volume of aluminium foil) × (density of aluminium foil) = 143.54_cm³ × 2.70 g/cm³ = 387.57_g

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An oxygen atom has a mass of 2.66 x 10^-23 g and a glass of water has a mass of 0.050kg. Use this information to answer the ques
fomenos
<h2>ANSWER OF EACH PART ARE GIVEN BELOW</h2>

Explanation:

A)

We know, each mole contains N_A= 6.023 \times 10^{23} atoms.

It is given that mass of one oxygen atom is m= 2.66\times 10^{-23}\ g.

Therefore, mass of one mole of oxygen, M=m\times N_A.

Putting value of n and N_A,

M=2.66\times 10^{-23}\times 6.023\times 10^{23} \ gm\\M=16.0\ gm

B)

Given,

Mass of water in glass=0.050 kg = 50 gm.

From above part mass of one mole of oxygen atoms = 16.0 gm.

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3 years ago
If a 1.00 mL sample of the reaction mixture for the equilibrium constant experiment required 32.40 mL of 0.258 M NaOH to titrate
andrey2020 [161]

Answer:

The concentration of acetic acid is 8.36 M

Explanation:

Step 1: Data given

Volume of acetic acid = 1.00 mL = 0.001 L

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Step 3: Calculate the concentration of the acetic acid

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⇒with b = the coefficient of NaOH = 1

⇒with Ca = the concentration of CH3COOH = TO BE DETERMINED

⇒with Va = the volume of CH3COOH = 1.00 mL = 0.001L

⇒with a = the coefficient of CH3COOH = 1

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⇒with Vb = the volume of NaOH = 32.40 mL = 0.03240 L

Ca * 0.001 L = 0.258 * 0.03240

Ca = 8.36 M

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Answer:

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