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Nastasia [14]
2 years ago
6

EARTH SCIENCE: PLEASE HELP ME !!! Which tools do astronomers use to determine the effects of collisions between galaxies? Select

the two correct answers.(1 point)
computer simulations

print encyclopedia entries

space telescopes

solar system models
Chemistry
1 answer:
topjm [15]2 years ago
6 0

In studying effects of collisions between galaxies, astronomers usually make use of space telescopes.

Astronomy deals with the study of outer space. The study focuses on planets and other celestial bodies. Generally telescopes are used to study outer space. The particular type of telescope that is used depends on the need.

In studying effects of collisions between galaxies, astronomers usually make use of space telescopes.

Learn more: brainly.com/question/556195

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Identify the conjugate acid/base pairs in each of the following equations:
Valentin [98]

Answer:

(a) Pair 1: H₂S and HS⁻

    Pair 2: NH₃ and NH₄⁺

(b) Pair 1: HSO₄⁻ and SO₄⁻

    Pair 2: NH₃ and NH₄⁺

(c) Pair 1: HBr and Br⁻

    Pair 2: CH₃O⁻ and CH₃OH

(d)  Pair 1: HNO₃ and NO₃⁻

     Pair 2: H₃O⁺

Explanation:

When an acid loses its proton (H⁺), a conjugate base is produced.

When a base accepts a proton (H⁺), it forms a conjugate acid.

(a) H₂S is an acid. When it loses a proton, it forms the conjugate base HS⁻.

    NH₃ is a base. When NH₃ gains a proton, it forms the conjugate acid NH₄⁺

(b) The acid HSO₄⁻ loses a H⁺ ion and forms the conjugate base SO₄²⁻.

     The base NH₃ accepts a H⁺ ion to form the conjugate acid NH₄⁺.

(c) HBr is an acid. When loses the H⁺ ion, it forms the conjugate base Br⁻.

   CH₃O⁻ accepts a H⁺ ion to form the conjugate acid CH₃OH.

(d) HNO₃ loses a proton to form the conjugate base NO₃⁻.

    H₂O gains a proton to form the conjugate acid H₃O⁺.

6 0
3 years ago
Is HNO3 an acid or a base<br><br><br><br>I NEED HELP ASAP​
Artyom0805 [142]

Answer:

HNO3 is a potent acid, a base, a nitrating agent and a heavy oxidising agent at times. In the presence of a stronger acid, it serves as a base.

Explanation:

3 0
2 years ago
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_______A representative element in the fourth period that is an alkaline earth
MariettaO [177]

Answer:calcium

Explanation: calcium is an alkaline earth metal,group 2 period 4

4 0
3 years ago
What was Bohr's model of the atom?
saveliy_v [14]

Answer:

In atomic physics, the Bohr model or Rutherford–Bohr model, presented by Niels Bohr and Ernest Rutherford in 1913, is a system consisting of a small, dense nucleus surrounded by orbiting electrons—similar to the structure of the Solar System, but with attraction provided by electrostatic forces in place of gravity.

Explanation:

4 0
3 years ago
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Isooctane, C8H18, is the component of gasoline from which the term octane rating derives.
lina2011 [118]

Answer:

a) C₈H₁₈ + (23/2)O₂ -----> 8CO₂ + 9H₂O

b) Mass of CO₂ produced annually from this combustion of isooctane gasoline = 1.12 × 10⁵ Kg

c) CO₂ produced from the combustion of the gasoline in a year will occupy 5.632 × 10⁷ L

d) There needs to be a minimum of 1.752 × 10⁷ moles of air and 3.92 × 10⁸ L of air for the oxygen to be in excess all through the year of gasoline combustion.

Explanation:

a) C₈H₁₈ + (23/2)O₂ -----> 8CO₂ + 9H₂O

b) C₈H₁₈ has a density of 0.792 mg/L.

Since density = mass/volume;

mass = density × volume

Mass of C₈H₁₈ with 4.6 x 10^10 L volume = 0.792 × 4.6 x 10^10 = 3.643 × 10^10 mg = 3.643 × 10⁷ g.

To obtain the mass of CO₂ produced, we need the number of moles of C₈H₁₈ that burned.

Number of moles = mass/molar mass

Molar mass of C₈H₁₈ = (8×12) + 18 = 114g/mol

Number of moles of C₈H₁₈ = (3.643 × 10⁷)/114 = (3.2 × 10⁵) moles.

From the chemical reaction,

1 mole of C₈H₁₈ burns to give 8 moles of CO₂

(3.2 × 10⁵) moles will give 8 × 3.2 × 10⁵ = (2.56 × 10⁶) moles of CO₂

Mass of CO₂ produced = number of moles × Molar mass

Molar mass of CO₂ = 44 g/mol

Mass of CO₂ produced = 2.56 × 10⁶ × 44 = 1.12 × 10⁸ g = 1.12 × 10⁵ kg

c) 1 mole of any gas at stp occupies 22.4L

2.56 × 10⁶ moles of CO₂ will occupy 2.56 × 10⁶ × 22.4 = 5.632 × 10⁷ L

d) 1 mole of C₈H₁₈ requires 23/2 moles of O₂ for complete combustion yearly.

3.2 × 10⁵ moles would require 3.2 × 10⁵ × 23/2 = 3.68 × 10⁶ moles of O₂

O₂ makes up 21% of the air

That is,

0.21 moles of O₂ would be contained in 1 mole of air

3.68 × 10⁶ moles of O₂ would be contained in (3.68 × 10⁶ × 1)/0.21 = 1.752 × 10⁷ moles of air.

1 mole of any gas at stp occupies 22.4L

1.752 × 10⁷ of air will occupy

1.752 × 10⁷ × 22.4/1 = 3.92 × 10⁸ L of air!

3 0
3 years ago
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