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3 years ago

Velocity requires which 2 things?

Chemistry

2 answers:

Nezavi [6.7K]3 years ago

6 0

Its b ()(-(/(:)/(/;:

scoundrel [369]3 years ago

5 0

Direction and speed. Explanation: Velocity is a vector quantity, meaning that direction matters as opposed to speed, which is a scalar quantity.

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Why does the less reactive ion participate in electrolysis, instead of the higher reactive metal??

GuDViN [60]

Answer:

In a reactivity series, the most reactive element is placed at the top and the least reactive element at the bottom. More reactive metals have a greater tendency to lose electrons and form positive ions .

Explanation:

Yes ur well come

3 0

4 years ago

Neon's outer energy level contains ______________ electrons. A) 2 B) 8 C) 10 D) 20

yuradex [85]

B) 8

Neon is a noble gas which means it is in the 18th group which has 8 valence electrons. Thus Neon has 8 valence electrons.

7 0

3 years ago

Read 2 more answers

How many moles of Calcium Oxide are needed to produce 4 moles of Calcium Hydroxide?

Tpy6a [65]

Taking into account the reaction stoichiometry, 2 moles of CaO are required to react with 2 moles of Ca(OH)₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CaO + H₂O → Ca(OH)₂

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:

CaO: 1 mole
H₂O: 1 mole
Ca(OH)₂: 1 mole

<h3>Moles of CaO required</h3>

The following rule of three can be applied: If by stoichiometric reaction 1 mole of Ca(OH)₂ is produced by 1 mole of CaO, 2 moles of Ca(OH)₂ are produced by how many moles of CaO?

$moles of CaO=\frac{2 moles of Ca(OH)_{2}x1 mol of CaO }{1 mole of Ca(OH)_{2}}$

moles of CaO= 2 moles

Finally, 2 moles of CaO are required to react with 2 moles of Ca(OH)₂.

Learn more about the reaction stoichiometry:

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5 0

2 years ago

When silver nitrate is added to an aqueous solution of magnesium chloride, a precipitation reaction occurs that produces silver

Aloiza [94]

Answer:

103.62 g of AgCl.

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AgNO3 + MgCl2 —> 2AgCl + Mg(NO3)2

Step 2:

Determination of the mass of MgCl2 that reacted and the mass of AgCl produced from the balanced equation.

This is illustrated below:

Molar mass of MgCl2 = 24 + (2x35.5) = 95 g/mol

Mass of MgCl2 from the balanced equation = 1 x 95 = 95 g

Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol

Mass of AgCl from the balanced equation = 2 x 143.5 = 287 g

Thus, from the balanced equation above,

95 g of MgCl2 reacted to produce 287 g of AgCl.

Step 3:

Determination of the mass of AgCl produced from the reaction of 34.3 g of MgCl2.

The mass of AgCl produced from the reaction can be obtained as follow:

Form the balanced equation above,

95 g of MgCl2 reacted to produce 287 g of AgCl.

Therefore, 34.3 g of MgCl2 will react to produce = (34.3 x 287)/95 = 103.62 g of AgCl.

Therefore, 103.62 g of AgCl were produced from the reaction.

8 0

4 years ago

A sample of P2Cl5 contains 179 g of phosphorous, how many grams of chlorine are present? Please show step by step thx!

natima [27]

Answer:

About 512 g.

Explanation:

We are given a sample of P₂Cl₅ that contains 179 grams of phosphorus, and we want to determine the grams of chlroine that is present.

Thus, we can convert from grams of phosphorus to moles of phosphorus, moles of phosphorus to moles of chlorine, and moles of chlorine to grams of chlorine.

From the formula, there are two moles of P for every five moles of Cl. The molecular weights of P and Cl are 30.97 g/mol and 35.45 g/mol, respectively. Hence:

$\displaystyle 179\text{ g P} \cdot \frac{1\text{ mol P}}{30.97\text{ g P}} \cdot \frac{5\text{ mol Cl}}{2\text{ mol P}} \cdot \frac{35.45\text{ g Cl}}{1\text{ mol Cl}} = 512\text{ g Cl}$

In conclusion, there is about 512 grams of chlorine present in the sample.

Alternatively, we can mass percentages. The mass percent of phosphorus in P₂Cl₅ is:

$\displaystyle \% \text{P} = \frac{2(30.97)}{5(35.45) + 2(30.97)} = 25.90\%$

Because there are 179 grams of phosphorus, the total amount of sample present is:

$\displaystyle \begin{aligned} 25.90\% \cdot m_T & = 179\text{ g P} \\ \\ m_T & = 691.1 \text{ g}\end{aligned}$

Therefore, the amount of chlorine present is 691.1 g - 179 g, or about 512 g, in agreement with our above answer.

8 0

3 years ago