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3 years ago

How many atoms of Fe are in 45.00 grams of Fe?

Chemistry

1 answer:

Licemer1 [7]3 years ago

6 0

Answer:

4.852*10^23

Explanation:

Use dimensional analysis.

First turn grams to moles using the molar mass of Fe (you can't convert to atoms from grams).

Then turn moles into atoms using Avogadro's constant.

45g Fe/55.85 g/mol Fe*6.022*10^23 atoms Fe

You might be interested in

A graduated cylinder contains 18.0ml of water. what is the new water level after 35.6g of silver metal with a density of 10.5 g/

AlekseyPX

Answer:

The answer to your question is

1.- Volume = 3.4 ml

2.- Volume = 0.61 ml

3.- Mass = 2872.8 pounds

Explanation:

Problem 1

Volume = 18 ml

mass = 35.6 g

density = 10.5 g/ml

Process

1.- Calculate the volume of silver

Formula

$density = \frac{mass}{volume}$

solve for volume

$volume = \frac{mass}{density}$

Substitution

$volume = \frac{35.6}{10.5}$

volume = 3.4 ml

2.- Problem 2

Total volume = ?

Volume = 18 + 3.4

Volume = 21.4 ml

Data

mass = 8.3 g

density = 13.6 g(ml

volume = ?

Formula

$density = \frac{mass}{volume}$

Solve for volume

$volume = \frac{mass}{density}$

Substitution

$volume = \frac{8.3}{13.6}$

Result

volume = 0.61 ml

3.- Problem 3

Data

volume = 345 gal

density = 1 g/ml

mass = ?

Formula

$density = \frac{mass}{volume}$

Solve for mass

mass = density x volume

Covert gal to ml

1 gal --------------- 3785 ml

345 gal ------------- x

x = (345 x 3785) / 1

x = 1305825 ml

Substitution

mass = 1 x 1305825

mass = 1305825 g

Convert g to pounds

1 g ------------------- 0.0022 pounds

1305825 g ---------------- x

x = (1305825 x 0.0022)

x = 2872.8 pounds

5 0

3 years ago

Hii pls help me to balance the equation thanksss

tiny-mole [99]

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7 0

2 years ago

Read 2 more answers

Bob measured out 1.60 grams of sodium. He calculates that 1.60 g of

saul85 [17]

Answer:

84.8%

Explanation:

Step 1: Given data

Bob measured out 1.60 g of Na. He forms NaCl according to the following equation.

Na + 1/2 Cl₂ ⇒ NaCl

According to this equation, he calculates that 1.60 g of sodium should produce 4.07 g of NaCl, which is the theoretical yield. However, he carries out the experiment and only makes 3.45 g of NaCl, which is the real yield.

Step 2: Calculate the percent yield.

We will use the following expression.

%yield = real yield / theoretical yield × 100%

%yield = 3.45 g / 4.07 g × 100% = 84.8%

6 0

3 years ago

(2 pts) The solubility of InF3 is 4.0 x 10-2 g/100 mL. a) What is the Ksp? Include the chemical equation and Ksp expression. MW

Aleonysh [2.5K]

Answer:

a) Ksp = 7.9x10⁻¹⁰

b) Solubility is 6.31x10⁻⁶M

Explanation:

a) InF₃ in water produce:

InF₃ ⇄ In⁺³ + 3F⁻

And Ksp is defined as:

Ksp = [In⁺³] [F⁻]³

4.0x10⁻²g / 100mL of InF₃ are:

4.0x10⁻²g / 100mL ₓ (1mol / 172g) ₓ (100mL / 0.1L) = 2.3x10⁻³M InF₃. Thus:

[In⁺³] = 2.3x10⁻³M InF₃ × (1 mol In⁺³ / mol InF₃) = 2.3x10⁻³M In⁺³

[F⁻] = 2.3x10⁻³M InF₃ × (3 mol F⁻ / mol InF₃) = 7.0x10⁻³M F⁻

Replacing these values in Ksp formula:

Ksp = [2.3x10⁻³M In⁺³] × [7.0x10⁻³M F⁻]³ = 7.9x10⁻¹⁰

b) 0.05 moles of F⁻ produce solubility of InF₃ decrease to:

7.9x10⁻¹⁰ = [x] [0.05 + 3x]³

Where x are moles of In⁺³ produced from solid InF₃ and 3x are moles of F⁻ produced from the same source. That means x is solubility in mol / L

Solving from x:

x = -0.018 → False solution, there is no negative concentrations.

x = 6.31x10⁻⁶M → Right answer.

Thus, solubility is 6.31x10⁻⁶M

3 0

3 years ago

In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe

const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = $10^{-9} m$ .

Hence, 0.283 nm = $0.283 \times 10^{-9} m$

Formula for coulombic energy is as follows.

$U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}$

where, e = $1.6 \times 10^{-19}$ C

$\epsilon_{o}$ = $8.85 \times 10^{-12}$

$U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}$

= $1.423 \times 10^{-18} J$

As 1 eV = $1.6 \times 10^{-19} J$

So, 1 J = $\frac{1 eV}{1.6 \times 10^{-19}}$

Hence, U = $\frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}$

= 8.9 eV

Also, 1 J = $\frac{10^{-3} kJ}{6.022 \times 10^{23}mol}$

= $1.67 \times 10^{-27}$ kJ/mol

Therefore, U = $1.423 \times 10^{-18} J \times 1.67 \times 10^{-27}$ kJ/mol

= $2.37 \times 10^{-45} kJ/mol$

7 0

3 years ago