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LenKa [72]
3 years ago
9

Which process is occurring at the center of all active stars? A) fusion B) fission C) flatulation D) hydrogen decay

Chemistry
1 answer:
dusya [7]3 years ago
8 0
The answer is A) fusion. Occurring at the center of all active stars including our sun, this nuclear reaction of combining two hydrogen atoms, and using an enormous amount of heat, forms a single more massive helium atom. This fusion of hydrogen atoms also produces the energy of the stars and creates the helium gases existing in the universe. 
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Laura jala un trineo una distancia de 2000 cm usando una cuerda, con una tensión constante de 80 N. La tarea requiere de 800 J d
Andre45 [30]

Respuesta:

60°

Explicación:

Paso 1: Información dada

  • Distancia recorrida (d): 2000 cm
  • Fuerza aplicada (F): 80 N
  • Trabajo realizado (W): 800 J

Paso 2: Convertir "d" a metros

Usaremos el factor de conversión 1 m = 100 cm.

2000 cm × 1 m/100 cm = 20 m

Paso 3: Calcular el ángulo entre la cuerda y el suelo (θ)

W = F × d × cosθ

800 J = 80 N × 20 m × cosθ

θ = 60°

5 0
3 years ago
What is the mass of one mole of S8?​
vredina [299]

Answer:

256.52 g/m

Explanation:

Each S has a molecular mass of 32.056 g/mol so multiply that by 8 and you get 256.52.

4 0
3 years ago
Now we need to find the amount of NF3 that can be formed by the complete reactions of each of the reactants. If all of the N2 wa
Alex777 [14]

The question is incomplete, the complete question is:

Nitrogen and fluorine react to form nitrogen fluoride according to the chemical equation:

N_2(g)+3F_2(g)\rightarrow 2NF_3(g)

A sample contains 19.3 g of N_2 is reacted with 19.3 g of F_2. Now we need to find the amount of NF_3 that can be formed by the complete reactions of each of the reactants.

If all of the N_2 was used up in the reaction, how many moles of NF_3 would be produced?

<u>Answer:</u> 1.378 moles of NF_3 are produced in the reaction.

<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       ......(1)

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

In the given chemical reaction, N_2 is considered as a limiting reagent because it limits the formation of the product and it was completely consumed in the reaction.

We are given:

Mass of N_2 = 19.3 g

Molar mass of N_2 = 28.02 g/mol

Putting values in equation 1:

\text{Moles of }N_2=\frac{19.3g}{28.02g/mol}=0.689mol

For the given chemical reaction:

N_2(g)+3F_2(g)\rightarrow 2NF_3(g)

By the stoichiometry of the reaction:

1 mole of N_2 produces 2 moles of NF_3

So, 0.689 moles of N_2 will produce = \frac{2}{1}\times 0.689=1.378mol of NF_3

Hence, 1.378 moles of NF_3 are produced in the reaction.

8 0
3 years ago
If I titrate an acid with 255mL of 3.4 M NaOH (base) and reach the equivalence point, how many moles of H+ ions were in the acid
FromTheMoon [43]

Answer:

Molar Concentration = \frac{Moles of Solute}{Litres of Solution}

                                     = \frac{3.4}{0.255}

                                    =  13.33

No. of H+ ions present = 13.33

pH value = - log[13.33]

               =  -1.12

Explanation:

The equivalence point, or stoichiometric point, of a substance response is the point at which synthetically identical amounts of reactants have been blended. As such, the moles of corrosive are equal to the moles of base, as per the condition (this doesn't really infer a 1:1 molar proportion of acid:base, simply that the proportion is equivalent to in the condition). It tends to be found by methods for a marker, for instance phenolphthalein or methyl orange. The endpoint (identified with, however not equivalent to the equivalence point) alludes to the point at which the marker changes shading in a colorimetric titration.

7 0
3 years ago
How many moles of phosphoric acid can be formed from 3000 g of diphosphorus pentoxide?
Semenov [28]

Answer:

42.2 moles of H3PO4

Explanation:

The equation of the reaction is:

P2O5(s) + 3 H2O(l) ⟶ 2 H3PO4.

First we must obtain the number of moles of P2O5 from

Number of moles of P2O5= reacting mass of P2O5/molar mass of P2O5

Molar mass of P2O5= 141.9445 g/mol

Number of moles= 3000g/141.9445 g/mol = 21.1 moles of P2O5

From the reaction equation;

1 mole of P2O5 yields 2 moles of H3PO4

21.1 moles of P2O5 will yield 21.1 ×2/ 1 = 42.2 moles of H3PO4

7 0
3 years ago
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