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Naddik [55]
2 years ago
12

HELP! Ethanol has a density of 0.8 g/cm3. a. What is the mass of 225 cm3 of ethanol? b. What is the volume of 75.0 g of ethanol?

Chemistry
1 answer:
IRINA_888 [86]2 years ago
5 0

Answer:

The correct answers are:

a) 180 g

b) 93.7 cm³

Explanation:

The density of a substance is the mass of the substance per unit of volume. So, it is calculated as follows:

density= mass/volume

From the data provided in the problem:

density = 0.8 g/cm³

a) Given: volume= 225 cm³

mass= density x volume = 0.8 g/cm³ x 225 cm³ = 180 g

b) Given: mass= 75.0 g

volume = mass/density = 75.0 g/(0.8 g/cm³)= 93.75 cm³≅ 93.7 cm³

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zavuch27 [327]
Mass =  mass/molar  mass  of   ch3ch2nh2
the   molar  mass  of CH3CH2NH2 =  12  +(1x3)+12+(1 x2)+14+(1x2) =45  g/mol

moles  is therefore  =  1.50g /45g/mol =   0.033 moles
6 0
2 years ago
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A gold wire has a diameter of 1.00 mm. What length of this wire contains exactly 1.00 mol of gold? (density of Au = 17.0 g/cm3)
allochka39001 [22]

Answer:

The answer to your question is 7160 cm

Explanation:

Data

diameter = 1 mm

length = ?

amount of gold = 1 mol

density = 17 g/cm³

Process

1.- Get the atomic mass of gold

Atomic mass = 197 g

then, 197g ------------ 1 mol

2.- Calculate the volume of this wire

density = mass/volume

volume = mass/density

volume = 197/17

volume = 5.7 cm³

3.- Calculate the length of the wire

Volume = πr²h

solve for h

h = volume /πr²

radius = 0.05 cm

substitution

h = 5.7/(3.14 x 0.05²)

h = 5.7 / 0.0025

h = 7159.2 cm ≈ 7160 cm

8 0
2 years ago
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Boron - 10 emits alpha particles and cesium - 137 emits beta particles. Write balanced nuclear
anzhelika [568]

Answer:

B10  5N +5P= Li6 3N +3P

Cs 137 82N+55P = I 133  80N + 53P

Explanation:

8 0
2 years ago
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What volume in<br><br> L<br><br> of a 0.724 M Nal solution contains 0.405 mol of Nal ?
Amanda [17]

Answer:

0.5594\ \text{L}

Explanation:

Mol of NaI = 0.405 mol

Molarity of solution = 0.724 M

Molarity is given by

M=\dfrac{\text{mol}}{\text{Volume of solution in }NaI}\\\Rightarrow \text{Volume of solution in }NaI=\dfrac{0.405}{0.724}\\\Rightarrow \text{Volume of solution in }NaI=0.5594\ \text{L}

The required volume is 0.5594\ \text{L}.

5 0
2 years ago
Consider the reaction: 2 Al + 3Br2 → 2 AlBr3 Suppose a reaction vessel initially contains 5.0 mole Al and 6.0 mole Br2. What is
timama [110]

Answer:

4 moles of AlBr_3 and 1 mole of Al will be present in the reaction vessel

Explanation:

The reaction given in the question is

2Al +  3 Br_2 ⇒ 2AlBr_3

According to the stoichiometric coefficients of the reaction, 2 moles of Al requires 3 moles of Br_2 so in this reaction, Br_2 is a limiting reagent. So we will consider that Al is in excess.

Now,

Since 3 moles of Br_2 requires 2 moles of Al

So, for 6 moles of Br_2 the moles of Al required = \frac{2}{3} \times 6 = 4 moles.

Moles of Al remaining after the completion of reaction = 5 - 4 = 1 mole.

Again,

Since 3 moles of Br_2 produces 2 moles of AlBr_3

So, moles of AlBr_3 produced by 6 moles of Br_2 = \frac{2}{3} \times 6 = 4 moles.

Therefore, after the completion of reaction, 4 moles of AlBr_3 and 1 mole of Al will be present in the reaction vessel.

5 0
2 years ago
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