How many moles are 5.55 x 102 atoms of Mg? M2) an> NU X 02.0

Chemistry

1 answer:

cupoosta [38]2 years ago

4 0

Explanation:

number of atoms = moles × avegadro number

so 5.55 × 10^2 = moles × 6.023 × 10^23

moles = 5.55 × 10^2 ÷ 6.023 × 10^23 = 9.214×10−22 moles

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C12H26O + SO3+NaOH ----> C12H25NaSO4+ H2O

Vlad [161]

Answer:

We can solve this by the method of which i solved your one question earlier

so again here molar mass of C12H25NaSO4 is 288.372 and number of moles for 11900 gm of C12H25NaSO4 will be = 11900/288.372

which is almost = 41.26 moles

so to get one mole of C12H25NaSO4 we need one mole of C12H26O

so for 41.26 moles of C12H25NaSO4 it will require 41 26 moles of C12H26O

so the mass of C12H26O = 41.26× its molar mass

C12H26O = 41.26×186.34

= 7688.38 gm!!

so the conclusion is If you need 11900 g of C12H25NaSO4 (Sodium Lauryl Sulfate) you need C12H26O 7688.38 gm !!

Again i d k wether it's right or wrong but i tried my best hope it helped you!!

4 0

3 years ago

Based on the chemical equation, use the drop-down menu to choose the coefficients that will balance the chemical equation:

Leni [432]

Answer: (1)CaSO4 -> (2)O2 + (1)CaS

Explanation: edge 2020 chem

5 0

2 years ago

Give the symbols for metalloids

Elina [12.6K]

Answer:

The metalloids; boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), polonium (Po) and astatine (At) are the elements found along the step like line between metals and non-metals of the periodic table.

Elements: Germanium; Boron; Arsenic

Explanation:

6 0

3 years ago

What is my theoretical yield (in moles) of Potassium Bromide (KBr) if I start with 40 grams of Iron (II) Bromide [FeBr2]? moles

Verizon [17]

The reaction will be: FeBr2 + K --> KBr + Fe
Balancing gives: FeBr2 + 2K --> 2KBr + Fe
The molar mass of FeBr2 is 55.85 + 2*79.9 = 215.65 g/mol.
We divide 40 g / 215.65 g/mol = 0.185 mol FeBr2
Based on stoichiometry:
(0.185 mol FeBr2)(2 mol KBr/1 mol FeBr2) = 0.370 mol KBr

4 0

3 years ago

The half life for the decay of carbon-14 is 5.73 x 10 years. Suppose the activity due to the radioactive decay of the carbon-14

Elena-2011 [213]

Answer:

Age of the atifact is $4.2\times 10^{2}$ years

Explanation:

For first -order radioactive decay- $A_{t}=A_{0}(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$
$A_{t}$ represents activity of radioactive nuclide after t time, $A_{0}$ represents initial activity of radioactive nuclide and $t_{\frac{1}{2}}$ represents half-life
Here, $A_{t}=19Bq$ , $A_{0}=20Bq$ and $t_{\frac{1}{2}}=5.73\times 10^{3}years$