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german
2 years ago
9

A sample of metal has a mass of 24.64 g, and a volume of 5.91 mL. What is the density of this metal?

Chemistry
1 answer:
dlinn [17]2 years ago
3 0

Answer:

\boxed {\boxed {\sf 4.17 \ g/mL}}

Explanation:

We are asked to find the density of a metal. Density is the mass per unit volume. It is calculated by dividing the mass by the volume.

\rho= \frac{m}{v}

The mass of the metal sample is 24.64 grams and the volume is 5.91 milliliters. We can substitute these values into the formula.

  • m= 24.64 g
  • v= 5.91 mL

\rho=\frac{24.64 \ g }{5.81 \ mL}

Divide.

\rho= 4.169204738 \ g/mL

The mass measurement has 4 significant figures the volume measurement has 3 significant figures. Our answer for density must match the least number of significant figures, which is 3.

For the number we calculated, that is the hundredth place. The 9 in the thousandth place tells us to round the 6 up to a 7.

\rho \approx 4.17 \ g/mL

The density of this metal is approximately <u>4.17 grams per milliliter.</u>

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xenn [34]
<h3>Answer:</h3>

3.01 × 10²³ atoms C

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

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right
<h3>Explanation:</h3>

<u>Step 1: Define</u>

6.00 g C

<u>Step 2: Identify Conversions</u>

Avogadro's Number

Molar Mass of C - 12.01 g/mol

<u>Step 3: Convert</u>

<u />6.00 \ g \ C(\frac{1 \ mol \ C}{12.01 \ g \ C} )(\frac{6.022 \cdot 10^{23} \ atoms \ C}{1 \ mol \ C} ) = 3.00849 × 10²³ atoms C

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

3.00849 × 10²³ atoms C ≈ 3.01 × 10²³ atoms C

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A line that is diagonal and going up
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Ability to react chemically
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<h3>Reactivity is the ability of matter to react chemically with other substances. Flammability is the ability of matter to burn.</h3>
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A reaction produces .815 moles of H2O. How many molecules produces?
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Multiply 0.815 to this value.







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Gold is alloyed (mixed) with other metals to increase its hardness in making jewelry.
KiRa [710]

Answer:

A) 54.04%

B) 13-karat

Explanation:

A) From the problem we have

<em>1)</em> Mg + Ms = 9.40 g

<em>2)</em> Vg + Vs = 0.675 cm³

Where M stands for mass, V stands for volume, and g and s stand for gold and silver respectively.

We can rewrite the first equation using the density values:

<em>3)</em> Vg * 19.3 g/cm³ + Vs * 10.5 g/cm³ = 9.40

So now we have<em> a system of two equations</em> (2 and 3) <em>with two unknowns</em>:

We <u>express Vg in terms of Vs</u>:

  • Vg + Vs = 0.675 cm³
  • Vg = 0.675 - Vs

We <u>replace the value of Vg in equation 3</u>:

  • Vg * 19.3 + Vs * 10.5 = 9.40
  • (0.675-Vs) * 19.3 + Vs * 10.5 = 9.40
  • 13.0275 - 19.3Vs + 10.5Vs = 9.40
  • -8.8 Vs + 13.0275 = 9.40
  • <u>Vs = 0.412 cm³</u>

Now we <u>calculate Vg</u>:

  • Vg + Vs = 0.675 cm³
  • Vg + 0.412 cm³ = 0.675 cm³
  • Vg = 0.263 cm³

We <u>calculate Mg from Vg</u>:

  • 0.263 cm³ * 19.3 g/cm³ = 5.08 g

We calculate the mass percentage of gold:

  • 5.08 / 9.40 * 100% = 54.04%

B)

We multiply 24 by the percentage fraction:

  • 24 * 54.04/100 = 12.97-karat ≅ 13-karat
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