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2 years ago

3. What are the planets called that are mostly made of gas?

Chemistry

2 answers:

Ksju [112]2 years ago

6 0

Answer:

<h3>A gas giant is a large planet composed mostly of gases, such as hydrogen and helium, with a relatively small rocky core. The gas giants of our solar system are Jupiter, Saturn, Uranus and Neptune. These four large planets, also called jovian planets after Jupiter, reside in the outer part of the solar system past the orbits of Mars and the asteroid belt. Jupiter and Saturn are substantially larger than Uranus and Neptune, and each pair of planets has a somewhat different composition.</h3><h3 /><h3>Although there are only four large planets in our solar system, astronomers have discovered thousands outside of it, particularly using NASA's Kepler Space Telescope. These exoplanets (as they are called) are being examined to learnmore about how our solar system came to be.</h3>

Explanation:

<h2>hopeit helps...</h2>

miskamm [114]2 years ago

5 0

Explanation:

The gas planets are Jupiter, Saturn, Uranus and Neptune.

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100.mL of a .795 M solution of KBr is diluted to 500.mL. what is the new concentration of the solution?

kap26 [50]

Answer:

0.159 M

Explanation:

convert from mL to L then use the equation:

M1V1 = M2V2

rearrange to find M2

$\frac{M1V1}{V2} = M2$

$\frac{(0.795 M)(0.100 L)}{0.500 L} = 0.159 M$

4 0

2 years ago

A gas with a volume of 4.0 L at a pressure of 2.02 atm is allowed to expand to a volume of 12.0

Maurinko [17]

<h3>Answer:</h3>

P₂ = 0.67 atm

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality<u> </u>

<u>Chemistry</u>

Boyle's Law: P₁V₁ = P₂V₂

P₁ is pressure 1
V₁ is volume 1
P₂ is pressure 2
V₂ is volume 2

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] P₁ = 2.02 atm

[Given] V₁ = 4.0 L

[Given] V₂ = 12.0 L

[Solve] P₂

<u>Step 2: Solve</u>

Substitute in variables [Boyle's Law]: (2.02 atm)(4.0 L) = P₂(12.0 L)
[Pressure] Multiply: 8.08 atm · L = P₂(12.0 L)
[Pressure] [Division Property of Equality] Isolate unknown: 0.673333 atm = P₂
[Pressure] Rewrite: P₂ = 0.673333 atm

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs as our smallest.</em>

0.673333 atm ≈ 0.67 atm

4 0

2 years ago

40 Points!!! Match the correct definition with the correct keyword.

My name is Ann [436]

Answer:

Okay the Answers are top to bottom.

5, 4, 1, 3, 2

Explanation:

8 0

2 years ago

The powder mixture (Cu, Al and Fe) = 10g was oxidized from sufficient chloride acid. 1) What are the possible reactions to the p

Mandarinka [93]

Answer:

1) 2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g

Explanation:

1) Possible reactions

2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Mass of each metal

a) Mass of Cu

The waste was the unreacted copper.

Mass of Cu = 2.5 g

b) Masses of Al and Fe

We have two relations :

Mass of Al + mass of Fe = 10 g - 2.5 g = 7.5 g

H₂ from Al + H₂ from Fe = 6.38 L at NTP

i) Calculate the moles of H₂

NTP is 20 °C and 1 atm.

$\begin{array}{rcl}pV & = & n RT\\\text{1 atm} \times \text{6.38 L} & = & n \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{293.15 K}\\6.38 & = & 24.06n \text{ mol}^{-1} \\n & = & \dfrac{6.38}{24.06 \text{ mol}^{-1} }\\\\ & = & \text{0.2652 mol}\\\end{array}$

(ii) Solve the relationship

Let x = mass of Al. Then

7.5 - x = mass of Fe

Moles of Al = x/27

Moles of Fe = (7.5 - x)/56

Moles of H₂ from Al = (3/2) × Moles of Al = (3/2) × (x/27) = x /18

Moles of H₂ from Fe = (1/1) × Moles of Fe = (7.5 - x)/56

∴ x/18 + (7.5 - x)/56 = 0.2652

56x + 18(7.5 - x) = 267.3

56x + 135 - 18x = 267.3

38x = 132.3

x = 3.5 g

Mass of Al = 3.5 g

Mass of Fe = 7.5 g - 3.5 g = 4.0 g

The masses of the metals are Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g

4 0

2 years ago

What is the molarity of ZnCl, that forms when 20.0 g of zinc completely reacts with CuCl2, according to the following reaction?

laila [671]

1.61m :) use the photo below

8 0

2 years ago