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3 years ago

How many moles are in 23.9 grams of titanium

Chemistry

2 answers:

harina [27]3 years ago

8 0

Answer:

$\boxed {\boxed {\sf About \ 0.499 \ grams \ of \ Titanium}}$

Explanation:

To convert from grams to moles, we must use the molar mass, which can be found on the Periodic Table.

Titanium (Ti): 47.87 g/mol

Use this number as a fraction.

$\frac{47.87 \ g \ Ti}{ 1 \ mol \ Ti}$

Multiply by the given number of grams: 23.9

$23.9 \ g\ Ti *\frac{47.87 \ g \ Ti}{ 1 \ mol \ Ti}$

Flip the fraction so the grams of titanium cancel out.

$23.9 \ g\ Ti *\frac{ 1 \ mol \ Ti}{47.87 \ g \ Ti}$

$23.9 *\frac{ 1 \ mol \ Ti}{47.87 }$

$\frac{ 23.9 \ mol \ Ti}{47.87 }$

Divide.

$0.499268853 \ mol \ Ti$

The original measurement had 3 significant figures, so our answer must have the same. For the number we calculated, that is is the thousandth place. The 2 in the ten thousandth place tells us to keep the 9.

$0.499 \ mol \ Ti$

There are <u>0.499 moles of titanium</u> in 23.9 grams.

Stels [109]3 years ago

6 0

<h3>Answer:</h3>

0.499 mol Ti

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

23.9 g Ti (Titanium)

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

[PT] Molar Mass of Ti - 47.88 g/mol

<u>Step 3: Convert</u>

[Dimensional Analysis] Set up: $\displaystyle 23.9 \ g \ Ti(\frac{1 \ mol \ Ti}{47.88 \ g \ Ti})$
[Dimensional Analysis] Multiply/Divide [Cancel out units]: $\displaystyle 0.499165 \ mol \ Ti$

<u>Step 4: Check</u>

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

0.499165 mol Ti ≈ 0.499 mol Ti

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vivado [14]

Answer:

2 eV

Explanation:

The energy of a photon of light is given by the formula

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

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$\lambda$ is the wavelength of the photon

In this problem we have:

$h=6.63\cdot 10^{-34} Js$

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Therefore, the energy in Joules is

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3 years ago

The decomposition of potassium chlorate into kcl & o2 is used as a source of oxygen in the laboratory. How many moles of pot

fredd [130]

Answer:

10moles of kcl

Explanation:

2 K C l O3 → 2 K C l + 3 O 2

Notice that you have a 2 : 3 mole ratio between potassium chlorate and oxygen gas, which means that, regardless of how many moles of the former react, you'll always produce 2/3 times more moles of the latter.

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4 years ago

A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som

jek_recluse [69]

The question is incomplete, the complete question is:

A certain substance X has a normal freezing point of $-6.4^oC$ and a molal freezing point depression constant $K_f=3.96^oC.kg/mol$ . A solution is prepared by dissolving some glycine in 950. g of X. This solution freezes at $-13.6^oC$ . Calculate the mass of urea that was dissolved. Round your answer to 2 significant digits.

<u>Answer:</u> The mass of glycine that can be dissolved is $1.3\times 10^2g$

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

$\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m$

$\text{Freezing point of pure solvent}=\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

where,

Freezing point of pure solvent = $-6.4^oC$

Freezing point of solution = $-13.6^oC$

i = Vant Hoff factor = 1 (for non-electrolytes)

$K_f$ = freezing point depression constant = $3.96^oC/m$

$m_{solute}$ = Given mass of solute (glycine) = ?

$M_{solute}$ = Molar mass of solute (glycine) = 75.07 g/mol

$w_{solvent}$ = Mass of solvent = 950. g

Putting values in equation 1, we get:

$-6.4-(-13.6)=1\times 3.96\times \frac{m_{solute}\times 1000}{75.07\times 950}\\\\m_{solute}=\frac{7.2\times 75.07\times 950}{1\times 3.96\times 1000}\\\\m_{solute}=129.66g=1.3\times 10^2g$

Hence, the mass of glycine that can be dissolved is $1.3\times 10^2g$

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3 years ago