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3 years ago

Can someone please help me with this?

Chemistry

1 answer:

BabaBlast [244]3 years ago

4 0

Answer:

1. Main sequence stars have different masses. The common characteristic they have is their source of energy. They burn fuel in their core through the process of fusing hydrogen atoms into helium.

2. Supergiants are among the most massive and most luminous stars. Supergiant stars occupy the top region of the Hertzsprung–Russell diagram with absolute visual magnitudes between about −3 and −8. The temperature range of supergiant stars spans from about 3,400 K to over 20,000 K.

3. Supergiants develop when massive main-sequence stars run out of hydrogen in their cores.

4. a supernova occur When the pressure drops low enough in a massive star, gravity suddenly takes over and the star collapses in just seconds. This collapse produces the explosion.

5. when a star has reached the end of its life and explodes in a brilliant burst of light

Explanation:

You might be interested in

A crime-scene is roped off with yellow tape. The rectangular area is 5.6 meters long 10.75 meters wide. What is the area of the

maksim [4K]

The area is 60.2, to find area it's length times width.

7 0

3 years ago

Part III.The two reactions involved in quantitatively determining the amount of iodate in solution are: IO3-(aq) 5 I-(aq) 6 H (a

slega [8]

Answer:

$\large \boxed{\math{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}$

Explanation:

The I₂ is the common substance in the two equations.

(1) IO₃⁻ + 5I⁻ + 6H⁺ ⟶ 3I₂ + 3H₂O

{2) I₂ + 2S₂O₃²⁻ ⟶ 2I⁻ + S₄O₆²⁻

From Equation (1), the molar ratio of iodate to iodine is

$\dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{3}{1}$

From Equation (2), the molar ratio of iodine to thiosulfate is

$\dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} = \dfrac{2}{1}$

Combining the two ratios, we get

$\text{Stoichiometric factor} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{IO}_{3}^{-}} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} \times \dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{2}{1} \times \dfrac{3}{1} = \mathbf{\dfrac{6}{1}}\\\\\text{The stoichiometric factor is $\large \boxed{\mathbf{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}$}$

7 0

3 years ago

9474 millimeters to centimeters using scientific notation showing work

puteri [66]

Answer:

9.474 x 10^2

Explanation:

ok. first you have to get the value in the required unit so 9474mm/(10mm/cm) = 947.4 so scientific notation states that the number must be raised to any power of an integer and the value of the number being raised must be less than than 10 and more than or equal to 1

so it must have one digit in front so.. 947.4 becomes 9.474 and because you move 2 places to the left, ur power is positive 2

and proof 10^2 is 100 so multiply 9.474 by 100 and u will get 947.4 cm which is also 9474 mm

8 0

3 years ago

EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?

AlekseyPX

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

$\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq)$ .

The equilibrium constant for this reaction is called the self-ionization constant of water:

$K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}]$ .

Note that water isn't part of this constant.

The value of $K_w$ at 25 °C is $10^{-14}$ . How to memorize this value?

The pH of pure water at 25 °C is 7.
$[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$
However, $[\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$ for pure water.
As a result, $K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14}$ at 25 °C.