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Alborosie
3 years ago
15

Which occurs whenever valence electrons are shared or transferred between atoms

Chemistry
1 answer:
Andrews [41]3 years ago
7 0
D is my answer. hope this helps.

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What is the destiny of an object that has a mass of 20 grams and a volume of 10 millimeters
blondinia [14]

Answer:

2gcm⁻³

Explanation:

Given parameters:

Mass of the object  = 20g

Volume = 10mL

Unknown:

Density of the object  = ?

Solution:

Density of a body is its mass per unit volume;

   Density  = \frac{mass}{volume}  

We need to convert mL to cm³ for the volume

     1mL  = 1cm³;

        10mL is therefore, 10cm³  

So;

       Density  = \frac{20}{10}   = 2gcm⁻³

7 0
3 years ago
Yasir wants to determine if there is a relationship between room color and sleep. Based on his research, Yasir makes an educated
Helen [10]

Answer: B. An experiment that directly tests the hypothesis

Explanation:

Yasir didn’t really test his experiment he only ask people for the opinion on which color they like so he forget to actually do the experiment therefore the answer is b.

8 0
3 years ago
Phosphorous pentoxide, P2O5(s), is produced from the reaction between pure oxygen and pure phosphorous (P, solid). What is the v
BartSMP [9]

Answer:

4190.22 L = 4.19 m³.

Explanation:

  • For the balanced reaction:

<em>2P₂ + 5O₂ ⇄ 2P₂O₅. </em>

It is clear that 2 mol of P₂ react with <em>5 mol of O₂ </em>to produce <em>2 mol of P₂O₅.</em>

  • Firstly, we need to calculate the no. of moles of 6.92 kilograms of P₂O₅ produced through the reaction:

no. of moles of P₂O₅ = mass/molar mass = (6920 g)/(283.88 g/mol) = 24.38 mol.

  • Now, we can find the no. of moles of O₂ is needed to produce the proposed amount of P₂O₅:

<u><em>Using cross multiplication:</em></u>

5 mol of O₂ is needed to produce → 2 mol of P₂O₅, from stichiometry.

??? mol of O₂ is needed to produce → 24.38 mol of P₂O₅.

∴ The no. of moles of O₂ needed = (5 mol)(24.38 mol)/(2 mol) = 60.95 mol.

  • Finally, we can get the volume of oxygen using the general law of ideal gas:<em> PV = nRT.</em>

where, P is the pressure of the gas in atm (P = 606.1 mm Hg/760 = 0.8 atm).

V is the volume of the gas in L (V = ??? L).

n is the no. of moles of the gas in mol (n = 60.95 mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (396.90°C + 273 = 669.9 K).

∴ V of oxygen needed = nRT/P = (60.95 mol)(0.0821 L.atm/mol.K)(669.9 K)/(0.8 atm) = 4190.22 L/1000 = 4.19 m³.

3 0
3 years ago
Cu2O + 2 HCl → 2 CuCl + H2O
ivann1987 [24]
What is your question please write properly>-< >_
8 0
2 years ago
Consider the following system at equilibrium:
alexgriva [62]

Answer:

A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)

B - Increase(P), Increase(q), Decrease (R)

C - Triple (P) and reduce (q) to one third

Explanation:

<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>

P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.

In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.

Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.

6 0
3 years ago
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