Calculate the energy required to raise the temperature of 230. g of water from 20.0 ° C to 35.0 ° C.

What type of atom is hydrogen with an atomic mass of 1 and 2 electrons

Charra [1.4K]

Answer:

This would be the hydride ion where H gains an electron from a metal such as Na

Explanation:

7 0

3 years ago

3. How many more troops did the Union states have than the Confederate states?

lions [1.4K]

At the beginning of the war the Northern states had a combined population of 22 million people. The Southern states had a combined population of about 9 million. This disparity was reflected in the size of the armies in the field. The Union forces outnumbered the Confederates roughly two to one.

Also this should be listed for history

3 0

3 years ago

1.42 g H2 is allowed to react with 10.4 g N2 , producing 2.14 g NH3 . Part A What is the theoretical yield in grams for this rea

Bad White [126]

Taking into account the reaction stoichiometry, the theorical yield for the reaction is 8.0467 grams of NH₃.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 H₂ + N₂ → 2 NH₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

H₂: 3 moles
N₂: 1 mole
NH₃: 2 moles

The molar mass of the compounds is:

H₂: 2 g/mole
N₂: 28 g/mole
NH₃: 17 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

H₂: 3 moles ×2 g/mole= 6 grams
N₂: 1 mole ×28 g/mole= 28 grams
NH₃: 2 moles ×17 g/mole= 34 grams

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 28 grams of N₂ reacts with 6 grams of H₂, 10.4 grams of N₂ reacts with how much mass of H₂?

$mass of H_{2} =\frac{10.4 grams of N_{2}x 6 grams of H_{2} }{28 grams of N_{2}}$

<u><em>mass of H₂= 2.2286 grams</em></u>

But 2.2286 grams of H₂ are not available, 1.42 grams are available. Since you have less mass than you need to react with 10.4 grams of N₂, H₂ will be the limiting reagent.

<h3>Definition of theorical yield</h3>

The theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

<h3>Theoretical yield in this case</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 6 grams of H₂ form 34 grams of NH₃, 1.42 grams of H₂ form how much mass of NH₃?

$mass of NH_{3} =\frac{1.42 grams of H_{2} x 34 grams of NH_{3}}{6grams of H_{2} }$

<u><em>mass of NH₃= 8.0467 grams</em></u>

Then, the theorical yield for the reaction is 8.0467 grams of NH₃.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

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7 0

1 year ago

Calculate the number of O atoms in 0.364 g of CaSO4 · 2H2O

Nikolay [14]

Answer:

<em>= 7.66 x 10²¹ oxygen atoms in 0.364 grams of CaSO₄·2H₂O</em>

Explanation:

For problems like this posting, one needs an understanding of the following topics:

The definition of the mole

<u>1 mole of substance</u> = mass in grams of substance containing 1 Avogadro's Number ( = 6.023 x 10²³ ) of particles of the specified substance. This is generally one formula weight of the substance of interest. From this, the following equivalent relationships should be memorized:

<em> 1 mole = 1 formula weight = 1 mole weight (g)= 6.023 x 10²³ particles</em>

Converting grams to moles:

<em>Given grams => moles = grams/gram formula wt </em>

Converting moles to grams:

<em>Given moles => grams = moles x gram formula wt</em>

_________________________________________________________

<em>Calculate the number of O atoms in 0.364 g of CaSO₄ · 2H₂O.</em>

Given mass CaSO₄ · 2H₂O = 0.364 grams

Formula Wt CaSO₄ · 2H₂O = 172 g/mole

moles CaSO₄ · 2H₂O = mass <em>CaSO4 · 2H2O / formula Wt. CaSO₄ · 2H₂O</em>

<em>∴ number of Oxy (O) atoms in 0.364 grams CaSO₄·2H₂O </em>

<em>= (2.12 x 10⁻³ mole CaSO₄ · 2H₂O)(6.023 x 10²³ molecules CaSO₄· 2H₂O/ mole)</em>

<em>= 1.276876 x 10²¹molecules CaSO₄· 2H₂O CaSO₄2H₂O </em>

<em>= 1.276876 x 10²¹ molecules CaSO₄· 2H₂O x 6 oxygen atoms / molecule</em>

<em>= 7.661256 x 10²¹ oxygen atoms in 0.364 grams of CaSO₄·2H₂O</em>

<em>= 7.66 x 10²¹ oxygen atoms in 0.364 grams of CaSO₄·2H₂O</em>

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8 0

2 years ago

In the process of eukaryotic pre-mRNA splicing, how is the lariat intermediate formed?

Aleks [24]

Answer:

C

Explanation:

The Eukaryotic pre-mRNA receives a 5' cap and 3'poly(A) tail before Introns are removed and the mRNA is considered ready for translation.

4 0

3 years ago