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3 years ago

25.0 grams of Iron (III) oxide reacts with 30.0 grams of Silver sulfate in a double replacement reaction. Find the

Chemistry

1 answer:

VladimirAG [237]3 years ago

3 0

Answer:

iron

Explanation:

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If the acceleration is 600 m/s/s and the mass is 300 kg, what force is acting on the object?

vampirchik [111]

Explanation:

Acceleration (a) = 600 m/s²

Mass (m) = 300 kg

Force (F) = ?

We know

F = m * a

= 300 * 600

=180000 Newton

The force acting on the object is 18000 Newton.

5 0

3 years ago

Read 2 more answers

How many moles of copper are in 1.51 x 1024 Cu atoms?

kkurt [141]

<h3>Answer:</h3>

2.51 mol Cu

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<h3>Explanation:</h3>

Step 1: Define

1.51 × 10²⁴ atoms Cu

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

Set up: $\displaystyle 1.51 \cdot 10^{24} \ atoms \ Cu(\frac{1 \ mol \ Cu}{6.022 \cdot 10^{23} \ atoms \ Cu})$
Multiply/Divide: $\displaystyle 2.50747 \ mol \ Cu$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

2.50747 mol Cu ≈ 2.51 mol Cu

8 0

3 years ago

Calculate the energy required to heat 406.0mg of cyclohexane from 33.5°C to 38.9°C . Assume the specific heat capacity of cycloh

N76 [4]

Answer:

Q = 4.056 J

Explanation:

Q = mCΔT

∴ m = 406.0 mg = 0.406 g

∴ C = 1.85 J/g.K

∴ T1 = 33.5°C ≅ 306.5 K

∴ T2 = 38.9°C = 311.9 K

⇒ ΔT = 311.9 - 306.5 = 5.4 K

⇒ Q = (0.406 g)(1.85 J/gK)(5.4 K)

⇒ Q = 4.056 J

6 0

3 years ago

Silver (Ag) has two stable isotopes: 107Ag, 106.90 amu, and 109Ag, 108.90 amu. If the average atomic mass of silver is 107.87 am

inysia [295]

Answer:

The abundance of 107Ag is 51.5%.

The abundance of 109Ag is 48.5%.

Explanation:

The average atomic mass of silver can be expressed as:

107.87 = 106.90 * A1 + 108.90 * A2

Where A1 is the abundance of 107Ag and A2 of 109Ag.

Assuming those two isotopes are the only one stables, we can use the equation:

A1 + A2 = 1.0

So now we have a system of two equations with two unknowns, and what's left is algebra.

First we use the second equation to express A1 in terms of A2:

A1 = 1.0 - A2

We replace A1 in the first equation:

107.87 = 106.90 * A1 + 108.90 * A2

107.87 = 106.90 * (1.0-A2) + 108.90 * A2

107.87 = 106.90 - 106.90*A2 + 108.90*A2

107.87 = 106.90 + 2*A2

2*A2 = 0.97

A2 = 0.485

So the abundance of 109Ag is (0.485*100%) 48.5%.

We use the value of A2 to calculate A1 in the second equation:

A1 + A2 = 1.0

A1 + 0.485 = 1.0

A1 = 0.515

So the abundance of 107Ag is 51.5%.

3 0

3 years ago

Explain why potassium forms ions more easily than sodium

Evgesh-ka [11]

So potassium is more reactive than lithium because the outer electron of a potassium atom is further from its nucleus than the outer electron of a lithium atom.

7 0

3 years ago