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Ronch [10]
3 years ago
7

Jamal is working with three ionic compounds: sodium chloride, calcium sulfide, and barium oxide. His teacher asks him which are

the positive ions in each compound. What should Jamal’s answer be? sodium, calcium, and barium barium, sulfur, and oxygen calcium, barium, and chlorine chlorine, sulfur, and oxygen
Chemistry
1 answer:
Lubov Fominskaja [6]3 years ago
4 0

Answer:

option A is correct  ( sodium, calcium and barium)

Explanation:

Given compounds:

Sodium chloride , calcium sulfide, barium oxide

We know that metals form positive ions. In order to solve the problem we must identify the metals from given compounds.

Na⁺Cl⁻

Ca²⁺S²⁻

Ba²⁺O²⁻

We can see that sodium, calcium and barium contain positive charges.

Thus option A is correct.

Because sodium have one valance electron. When it combine with chlorine sodium lose its one electron to complete the octet and chlorine accept it to complete its octet. Thus sodium form positive ion and chlorine form negative ion.

Similarly barium and calcium are present in group 2. Both have two valance electron. When they lose them cation are formed.

Other option are incorrect because,

Option B have sulfur and oxygen which are anion.

Option C have chlorine which is also anion

Option D have chlorine, sulfur and oxygen that are anions.

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If 45.0 mL of ethanol (density = 0.789 g/mL) initially at 8.0 ∘C is mixed with 45.0 mL of water (density = 1.0 g/mL) initially a
strojnjashka [21]

Answer : The final temperature of the mixture is, 22.14^oC

Explanation :

First we have to calculate the mass of ethanol and water.

\text{Mass of ethanol}=\text{Density of ethanol}\times \text{Volume of ethanol}=0.789g/mL\times 45.0mL=35.5g

and,

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1.0g/mL\times 45.0mL=45.0g

Now we have to calculate the final temperature of the mixture.

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of ethanol = 2.42J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of ethanol = 35.5 g

m_2 = mass of water = 45.0 g

T_f = final temperature of mixture = ?

T_1 = initial temperature of ethanol = 8.0^oC

T_2 = initial temperature of water = 28.6^oC

Now put all the given values in the above formula, we get:

35.5g\times 2.42J/g^oC\times (T_f-8.0)^oC=-45.0g\times 4.18J/g^oC\times (T_f-28.6)^oC

T_f=22.14^oC

Therefore, the final temperature of the mixture is, 22.14^oC

3 0
4 years ago
The reaction is first order in cyclopropane and has a measured rate constant of k=3.36×10−5 s−1 at 720 k. if the initial cyclopr
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Answer:

0.0277 M.

Explanation:

The integral rate law of a first order reaction:

<em>Kt = ln ([A₀]/[A]),</em>

where, k is the rate constant of the reaction <em>(k = 3.36 × 10⁻⁵ s⁻¹)</em>,

t is the time of the reaction <em>(t = 235.0 min = 14100 s)</em>,

[A₀] is the initial concentration of cyclopropane <em>([A₀] = 0.0445 M)</em>

<em>∵ Kt = ln ([A₀]/[A]),</em>

∴ (3.36 × 10⁻⁵ s⁻¹)(14100 s) = ln (0.0445 M)/[A]

Taking the exponential of both sides:

1.6 = (0.0445 M)/[A]

<em>∴ [A] = (0.0445 M)/1.6 = 0.0277 M.</em>

<em />

6 0
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The answer to your question is,

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I think it is 89.3%?
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