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3 years ago

How many atom of cd are in 3.00g of cadmium

2 answers:

8 0

The molar mass of Cadmium is 112.41 grams, so first you need to convert your given value to moles. To do that divide 3 grams by the molar mass of 112.41 grams. That gives you an answer of .0267 moles of cadmium. Then to convert that value to atoms, you must use the constant of 6.022X10^23 atoms in one mole. Multiply the moles by that number, which will give you a final answer of (1.61X10^22) atoms in 3 grams of cadmium.

Hope this helps!

Crazy boy [7]3 years ago

5 0

I wish i was smart enough

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How many moles of ethanol are produced starting with 500.g glucose?

Monica [59]

<h3>Answer:</h3>

5.55 mol C₂H₅OH

<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> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Tables
Moles

<u>Stoichiometry</u>

Using Dimensional Analysis
Analyzing Reactions RxN

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

[Given] 500. g C₆H₁₂O₆ (Glucose)

[Solve] moles C₂H₅OH (Ethanol)

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

[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH

[PT] Molar mass of C - 12.01 g/mol

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol

<u>Step 3: Stoichiometry</u>

[DA] Set up conversion: $\displaystyle 500 \ g \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{180.18 \ g \ C_6H_{12}O_6})(\frac{2 \ mol \ C_2H_5OH}{1 \ mol \ C_6H_{12}O_6})$
[DA} Multiply/Divide [Cancel out units]: $\displaystyle 5.55001 \ mol \ C_2H_5OH$

<u>Step 4: Check</u>

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

5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH

8 0

3 years ago

Explain how to produce a longitudinal wave on a spring that has large energy. There are two answers; one involves wavelength and

melamori03 [73]

Compressions are regions of high pressure due to particles being close together
rarefactions are regions of low pressure due to particles being spread further apart
Longitudinal waves are often demonstrated by pushing and pulling a stretched slinky spring

4 0

2 years ago

_____ The primary function of molecular membranes is the transport of ions and molecules into and out of cells. The movement of

ale4655 [162]

<h2>Active Transport </h2>

Explanation:

A cell uses energy for the movement of substances against a concentration gradient or an electrochemical gradient that maintains a balanced concentration of ions and molecules in living cells.
This transport of substances is known as Active transport mechanisms which are the basic function of the cell membranes.
The energy used is in the form of ATP.
Active transport occurs when there is a difference in the concentration of particles on both sides.

5 0

3 years ago

Express the following in scientific notation 0.000365

d1i1m1o1n [39]

3.65 X 10^-4

8 0

3 years ago

If 4.65 LL of CO2CO2 gas at 22 ∘C∘C at 793 mmHg mmHg is used, what is the final volume, in liters, of the gas at 35 ∘C∘C and a p

otez555 [7]

Answer:

About 7.9 L.

Explanation:

We can utilize the ideal gas law. Recall that:

$\displaystyle PV = nRT$

Because the amount of carbon dioxide does not change, we can rearrange to formula to:
$\displaystyle \frac{PV}{T}= nR$

Because the right-hand side stays constant, we have that:
$\displaystyle \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} = nR$

Hence substitute initial values and known final values:
$\displaystyle \begin{aligned} \frac{(793\text{ mm Hg})(4.65 \text{ L})}{(22 \text{ $^\circ$C})} & = \frac{(743 \text{ mm Hg})V_2}{(35\text{ $^\circ$C})} \\ \\ V_2 & = 7.9\text{ L}\end{aligned}$

Therefore, the final volume is about 7.9 L.

8 0

2 years ago