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antoniya [11.8K]
3 years ago
9

How many moles of FeCI3 are present in 345.0 g FeCI3

Chemistry
1 answer:
Sphinxa [80]3 years ago
6 0

Answer:

\boxed {\boxed {\sf About \ 2.127 \ moles \ of \ FeCl_3}}

Explanation:

To convert from moles to grams, the molar mass must be used.

1. Find Molar Mass

The compound is iron (III) chloride: FeCl₃

First, find the molar masses of the individual elements in the compound: iron (Fe) and chlorine (Cl).

  • Fe:  55.84 g/mol
  • Cl:  35.45 g/mol

There are 3 atoms of chlorine, denoted by the subscript after Cl. Multiply the molar mass of chlorine by 3 and add iron's molar mass.

  • FeCl₃: 3(35.45 g/mol)+(55.84 g/mol)=162.19 g/mol

This number tells us the grams of FeCl₃ in 1 mole.

2. Calculate Moles

Use the number as a ratio.

\frac{162.19 \ g \ FeCl_3}{1 \ mol \ FeCl_3}

Multiply by the given number of grams, 345.0.

345.0 \ g \ FeCl_3 *\frac{162.19 \ g \ FeCl_3}{1 \ mol \ FeCl_3}

Flip the fraction so the grams of FeCl₃ will cancel.

345.0 \ g \ FeCl_3 *\frac{1 \ mol \ FeCl_3}{162.19 \ g \ FeCl_3}

345.0 *\frac{1 \ mol \ FeCl_3}{162.19 }

\frac{345.0 \ mol \ FeCl_3}{162.19 }

Divide.

2.12713484 \ mol \ FeCl_3

3. Round

The original measurement of grams, 345.0, has 4 significant figures. We must round our answer to 4 sig figs.

For the answer we calculated, that is the thousandth place.

The 1 in the ten thousandth place tells us to leave the 7 in the thousandth place.

\approx 2.127 \ mol \ FeCl_3

There are about <u>2.127 mole</u>s of iron (III) chloride in 345.0 grams.

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Assume that a daily diet of 2000 calories (i.e. 8.37 x 106 J) is converted completely to body heat.
slava [35]

Answer:

(a) the mass of the water is 3704 g

(b) the mass of the water is 199, 285.7 g

Explanation:

Given;

Quantity of heat, H= 8.37 x 10⁶ J

Part (a) mass of water (as sweat) need to evaporate to cool that person off

Latent heat of vaporization of water, Lvap. = 2.26 x 10⁶ J/kg

H = m x Lvap.

m = \frac{H}{L._{vap}} =\frac{8.37 * 10^6.J}{2.26*10^6\ \frac{J}{kg}} = 3.704 \ kg

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Part (b) quantity of water raised from 25.0 °C to 35.0 °C by 8.37 x 10⁶ J

specific heat capacity of water, C, 4200 J/kg.°C

H = mcΔθ

where;

Δθ is the change in temperature = 35 - 25 = 10°C

m =\frac{H}{c* \delta \theta} = \frac{8.37 *10^6}{4200*10} = 199.2857 kg

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5 0
3 years ago
Please help me
Wittaler [7]

Answer:

pH = 6.999

The solution is acidic.

Explanation:

HBr is a strong acid, a very strong one.

In water, this acid is totally dissociated.

HBr + H₂O  →  H₃O⁺  +  Br⁻

We can think pH, as - log 7.75×10⁻¹² but this is 11.1

acid pH can't never be higher than 7.

We apply the charge balance:

[H⁺] = [Br⁻] + [OH⁻]

All the protons come from the bromide and the OH⁻ that come from water.

We can also think [OH⁻] = Kw / [H⁺] so:

[H⁺] = [Br⁻] + Kw / [H⁺]

Now, our unknown is [H⁺]

[H⁺] =  7.75×10⁻¹² + 1×10⁻¹⁴ / [H⁺]

[H⁺] = (7.75×10⁻¹² [H⁺] + 1×10⁻¹⁴) /  [H⁺]

This is quadratic equation:  [H⁺]² - 7.75×10⁻¹² [H⁺] - 1×10⁻¹⁴

a = 1 ; b = - 7.75×10⁻¹² ; c = -1×10⁻¹⁴

(-b +- √(b² - 4ac) / (2a)

[H⁺] = 1.000038751×10⁻⁷

- log [H⁺] = pH → 6.999

A very strong acid as HBr, in this case, it is so diluted that its pH is almost neutral.

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