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3 years ago

Large objects that form dense gravity wells in space A. Galaxies B. Star C. Nebulae D. Black holes.

Chemistry

2 answers:

USPshnik [31]3 years ago

6 0

D is the right answer , ur welcome sir

BigorU [14]3 years ago

5 0

Answer:

D. Black Holes

Explanation:

Black holes are large objects that form dense gravity wells in space. Their gravitational pull is so strong that even light cannot escape it.

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4. Assume that the water stream is replaced by a stream of CCI.. Predict what would happen in each case:

RoseWind [281]

Answer:

The answer is c) charged balloon

6 0

3 years ago

What is the difference between nitrite and nitrate

Dahasolnce [82]

Answer:

Nitrite:

Formula : NO₂⁻
1 nitrogen molecule and two oxygen molecules with an excess electron.
Angular shape

Nitrate:

Formula: NO₃⁻

1 nitrogen molecule and three oxygen molecules with an excess electron.
Trigonal planar structure.

$\rule[225]{225}{2}$

Hope this helped!

4 0

2 years ago

A) Calculate the standard free-energy change at 25 ∘C for the following reaction:

Genrish500 [490]

Answer:

A) ΔG° = -3,80x10⁵ kJ

B) E° = 2,85V

Explanation:

A) It is possible to answer this problem using the standard ΔG's of formation. For the reaction:

Mg(s) + Fe²⁺(aq) → Mg²⁺(aq) + Fe(s)

The ΔG° of reaction is:

ΔG° = ΔGFe(s) + ΔGMg²⁺(aq) - (ΔGFe²⁺(aq) + ΔGMg(s) (1)

Where:

ΔGFe(s): 0kJ

ΔGMg²⁺(aq): -458,8 kJ

ΔGFe²⁺(aq): -78,9 kJ

ΔGMg(s): 0kJ

Replacing in (1):

ΔG° = 0kJ -458,8kJ - (-78,9kJ + okJ)

ΔG° = -3,80x10² kJ ≡ -3,80x10⁵ kJ

B) For the reaction:

X(s) + 2Y⁺(aq) → X²⁺(aq) + 2Y(s)

ΔG° = ΔH° - (T×ΔS°)

ΔG° = -629000J - (298,15K×-263J/K)

ΔG° = -550587J

As ΔG° = - n×F×E⁰

Where n are electrons involved in the reaction (2mol), F is faraday constant (96485 J/Vmol) And E° is the standard cell potential

Replacing:

-550587J = - 2mol×96485J/Vmol×E⁰

E° = 2,85V

I hope it helps!

3 0

3 years ago

s e vapor pressure of the waterretnylene glycol solution at 32 "C? 5. What is the Boiling Point of the solution resulted from th

lyudmila [28]

Answer:

What is the Boiling Point of the solution resulted from the dissolving of 32.5 g of NaCl in 250.0 g of water? 102,31°C

Explanation:

This question involves the Elevation of boiling point

ΔT = Kb . m . i

(T°solution - T°solvent pure) = Ebulloscopic constant . molality . Van 't Hoff factor

Van 't Hoff factor is 2 for NaCl because you have two ions on disociation.

Kb for water is, 0,52 °C . kg/mol - It is a known value.

You know that water pure boils at 100°C so let's build the formula

(T°solution - 100°C) = 0,52 °C.kg/mol . molality . 2

We have to find out the molality (moles of solute/1kg solvent)

In 250 g H2O we have 32,5 g NaCl so the rule of three will be

250 g H2O ______32,5 g NaCl

1000g __________ (1000g . 32,5g) / 250g = 130 g NaCl

Molar mass NaCl : 58,45 g/m

Moles NaCl : mass/ molar mass --> 130g /58,45 g/m = 2,22 m

(T°solution - 100°C) = 0,52 °C.kg/mol . 2,22 mol/gk . 2

T°solution - 100°C = 2,31 °C

T° at boiling point in the solution = 2,31°C + 100°C = 102,31°C

3 0

3 years ago

An unknown concentration of sodium thiosulfate, Na2S2O3, is used to titrate a standardized solution of KIO3 with excess KI prese

Maru [420]

Answer: The molarity of $Na_2S_2O_3$ is 0.108 M

Explanation:

$KIO_3+5KI+3H_2SO_4\rightarrow 3K_2SO_4+3H_2O+3I_2$

$2Na_2S_2O_3+I_2\rightarrow Na_2S_4O_6+2NaI$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

$\text{Moles of }KIO_3=\frac{0.0131mol/L\times 21.55}{1000}=2.8\times 10^{-4}mol$

1 mole of $KIO_3$ produces = 3 moles of $I_2$

$2.8\times 10^{-4}$ moles of $KIO_3$ produces = $\frac{3}{1}\times 2.8\times 10^{-4}=8.4\times 10^{-4}$ moles of $I_2$

Now 1 mole of $I_2$ uses = 2 moles of $Na_2S_2O_3$

$8.4\times 10^{-4}$ moles of $I_2$ uses = $\frac{2}{1}\times 8.4\times 10^{-4}=1.69\times 10^{-3}$ moles of $Na_2S_2O_3$

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution in L}}=\frac{1.69\times 10^{-3}\times 1000}{15.65}=0.108M$

The molarity of $Na_2S_2O_3$ is 0.108 M

6 0

3 years ago

Large objects that form dense gravity wells in space A. Galaxies B. Star C. Nebulae D. Black holes.​

Large objects that form dense gravity wells in space A. Galaxies B. Star C. Nebulae D. Black holes.