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2 years ago

How many outer planets are there?

Chemistry

1 answer:

Ivenika [448]2 years ago

3 0

Answer:

four outer planetsThe Outer Planet the four outer planets and the Sun, with sizes to scale. From left to right, the outer planets are Jupiter, Saturn, Uranus, and Neptune. The gas giants are mostly made of hydrogen and helium. These are the same elements that make up most of the Sun.

Explanation:

I hope this helps you

Ari

mwah

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A 100.0 g sample of sugar contains carbon, hydrogen, and oxygen. The sample contains 40.0 g of carbon and 53.3 g of oxygen. What

Lynna [10]

Answer:

I think this is it.

Explanation:

As all the mass of other molecules are given and we have to find the mass of hydrogen so total subtracted by carbon and oxygen.

6 0

2 years ago

How many grams of ammonium chloride (NH4Cl) would be required to make a saturated solution in 1000 grams of water at 50 degrees

Pavel [41]

Since,
1000 grams of water = 1000 mL of water
So,
At any of the given temperature:
1000 mL = 10 x 100 mL

moles of NH4Cl = 53.5/53.49
= 1.0 m
= 1.0 mol/Kg
Delta T = 2 x 1.86 x 1.0
= 3.72 c
= - 3.72 °C

3 0

2 years ago

Determine the elements present in sodium hydrogen carbonate

Ilia_Sergeevich [38]

Sodium, hydrogen, carbon, and oxygen

6 0

2 years ago

What is photosynthesis?

tamaranim1 [39]

the process by which green plants and some other organisms use sunlight to synthesize foods from carbon dioxide and water. Photosynthesis in plants generally involves the green pigment chlorophyll and generates oxygen as a byproduct.

7 0

3 years ago

quien y en que cantidad sera el reactivo limite, si utilizamos 125 g de ácido, H (CH3COO) y 275 g de hidróxido , Al(OH)3

Deffense [45]

Answer:

The limiting reactant is acetic acid. All 125 g will react.

Explanation:

1. Assemble the information

We will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.

Mᵣ: 60.05 78.00

3CH₃COO-H + Al(OH)₃ ⟶ (CH₃COO)₃Al + 3H₂O

Mass/g: 125 275

2. Calculate the moles of each reactant

$\text{Moles of CH$_{3}$COOH} = \text{125 g CH$_{3}$COOH} \times \dfrac{\text{1 mol CH$_{3}$COOH}}{\text{60.05 g CH$_{3}$COOH}} = \text{2.082 mol CH$_{3}$COOH }\\\\\text{Moles of Al(OH)}_{3} = \text{275 g Al(OH)}_{3} \times \dfrac{\text{1 mol Al(OH)}_{3}}{\text{78.00 g Al(OH)}_{3}} = \text{3.526 mol Al(OH)}_{3}$

3. Calculate the moles of (CH₃COO)₃Al from each reactant

$\textbf{From CH$_{3}$COOH:}\\\text{Moles of (CH$_{3}$COO)$_{3}$Al} = \text{2.082 mol CH$_{3}$COOH} \times \dfrac{\text{1 mol (CH$_{3}$COO)$_{3}$Al}}{\text{3 mol CH$_{3}$COOH}}\\\\= \text{0.6939 mol (CH$_{3}$COO)$_{3}$)Al}\\\textbf{From Al(OH)}_{3}:\\\text{Moles of (CH$_{3}$COO)$_{3}$Al } =\text{3.526 Al(OH)}_{3} \times \dfrac{\text{1 mol (CH$_{3}$COO)$_{3}$Al }}{\text{1 mol Al(OH)}_{3}}\\\\= \text{3.526 mol (CH$_{3}$COO)$_{3}$Al}$

$\text{Acetic acid is the $\textbf{limiting reactant}$ because it gives fewer moles of} \\\text{(CH$_{3}$COO)$_{3}$Al. All $\textbf{125 g}$ will react.}$

5 0

2 years ago