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nikklg [1K]
3 years ago
13

What is the reason that some elements are formed inside stars and nowhere else

Chemistry
2 answers:
elena-14-01-66 [18.8K]3 years ago
6 0

I think it would be answer B or D  . I gave you two answer because I dont want to be wrong with one

nikitadnepr [17]3 years ago
4 0
Your answer is D none of the above
You might be interested in
How would you prepare 2-methyl-2-propanol via a grignard with dimethyl carbonate as your carbonyl source? show all reagents?
Andreas93 [3]

Answer:

React it with CH₃MgBr and work up the product with saturated ammonium chloride solution

Explanation:

Grignard reagents convert esters into tertiary alcohols.

The general equation is

\text{RCOOR}' \xrightarrow[\text{2. H}^{+}]{\text{1. R$^{\prime \prime}$MgBr}}\text{RR$_{2}^{\prime \prime}$C-OH}

The Grignard reagent in this synthesis is methylmagnesium bromide. You prepare it by reacting a solution methyl bromide in anhydrous ether with magnesium and a few crystals of iodine.

The reaction consumes 3 mol of CH₃MgBr per mole of dimethyl carbonate, and everything happens in the same pot.

Acid workup of the product usually involves the addition of a saturated aqueous solution of ammonium chloride and extraction with a low-boiling organic solvent.

The mechanism involves:

Step 1. Nucleophilic attack and loss of leaving group

(a) The Grignard reagent attacks the carbonyl of dimethyl carbonate, followed by (b) the loss of a methoxide leaving group.

Step 2. Nucleophilic attack and loss of leaving group

(a) A second mole of the Grignard reagent attacks the carbonyl of methyl acetate, followed by (b) the loss of a methoxide leaving group.

Step 3. Nucleophilic attack and protonation of the adduct.  

(a) A third mole of the Grignard reagent attacks the carbonyl of acetone, followed by (b) protonation of the alkoxide to form 2-methylpropan-2-ol.

3 0
3 years ago
Identify the following as physical (P) or chemical (C) changes.
Stels [109]

holacomo es tu pregunta nola entiendo

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4 0
3 years ago
Read 2 more answers
How many grams of carbon dioxide are created from the complete
ololo11 [35]

Answer:

<u>167.2 g</u>

Explanation:

Known

VC4H10 = 21.3

T = 0.00 C (convert to Kelvin: 273 K)

P = 1.00 atm

Unknown

m = ?g

1. <u>Write the balanced chemical equation</u>

1 C4H10 + 1O2  ----->  4 CO2 + 5 H2O

2. <u>Find the volume ratio of Carbon Dioxide to Butane </u>

1 C4H10   4 CO2  = 4 volumes CO2 / 1 volume C4H10

3.<u> Multiply by the known volume of n (butane)</u>

21.3 L C4H10  x  4 volumes CO2 / 1 volume C4H10 = 85.2 L C4H10

4. <u>Use ideal gas law</u>

PV = nRT   solve for n  ----> n = PV/RT

n= (1.00 atm) (85.2 L) / (0.0821 L atm/mol K) (273)  = 3.80 mol CO2

5.<u> Find molar mass of CO2</u>

1 C x 12 + 2 O x 16 = 44.00

6. <u>Multiply the ideal gas law solution (3.80) by molar mass CO2 (44.00)</u>

3.80 mol CO2 x 44.00 g CO2

= 167.2 g CO2  

7 0
3 years ago
Directions: Convert the following word equations into formula equations then balance
UkoKoshka [18]

..............

......

8 0
2 years ago
UCI Chemistry researchers, Prof. F. Sherwood Rowland and Dr. Mario Molina werefirst to discovered in 1973 that chlorofluorocarbo
Olin [163]

Answer:

15.27895 x 10⁶kg of chlorine radical is added to the atmosphere in a year due to 100 million MVACs                                                                                              

Explanation:

Chlorofluorocarbons (CF₂Cl₂) from refrigerants produce chlorine radicals according to the following equation

         CF₂Cl₂ → CF2Cl  + Cl ⁻ .........(1)

From equation 1, one mole of CF₂Cl₂ produces one mole of Chlorine radical

From the question,

The emission rate of CF₂Cl₂ is 59.5mg/hour/MVAC

In one day the emission rate would be 59.5 x 24hours

                                                                  = 1428mg/day

In one year, the emission rate would be 1428mg/day x 365days

                                                                 =  521220mg/year

                                                                   = 521.220g/year/MVAC

Therefore the emission rate for 100 million MVAC using CF₂CL₂ in a year is

                                                                    = 52122 x 10⁶g/year/MVAC

                                                                    = 52122 x 10³kg/year/MVAC

The molar mass of CF₂CL₂                       = 120.913g/mol

No of moles  of CF₂CL₂                              = mass/ molar mass

                                                                     = 52122 x 10⁶g / 120.913g/mol

                                                                      = 431 x 10⁶ moles of CF₂Cl₂

From equation 1,  since one mole of CF₂Cl₂ produces one mole of Chlorine radical, it implies that

431 x 10⁶ moles of CF₂Cl₂ would produce 431 x 10⁶ moles of chlorine radical,

Therefore, to find the mass of chlorine radical produced, we use the formula

No of moles of chlorine radical  = mass/ molar mass

431 x10⁶ moles = mass of chlorine radical /molar mass of chlorine radical

431 x 10⁶ moles = mass/ 35.45g/mol

mass of chlorine =  431 x 10⁶ moles x 35.45 g/mol

                            =    15278.95 x 10⁶ g

In Kg, the mass    =   15,278.95 x 10³kg of cholrine radical

                             =   15.27895 x 10⁶ Kg of chlorine radical

                                                                                                 

6 0
3 years ago
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