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3 years ago

Question 3 of 11

Chemistry

1 answer:

daser333 [38]3 years ago

6 0

Answer:

Explanation:

b and d are out, the variables are changed. a would be a repetition, not a replication. c uses the same method and variables with a different control group

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Plz answer quick :) it would make me happy

dangina [55]

Answer:

Single replacement reaction

Explanation:

By right, it should be called displacement reaction or redox reaction. Metal ion A in solution is reduced to form solid metal A while metal solid B is oxidised to form metal ion B.

Only 1 type of replacement occurred so it should be single replacement reaction

5 0

3 years ago

Read 2 more answers

Rasheed calculates that his chemical reaction should produce 4 moles of product, but when he does the experiment, he gets only 3

Brut [27]

Answer:

The answer to your question is 75%

Explanation:

Data

Theoretical production = 4 moles

Experimental production = 3 moles

Percent yield = ?

Formula

$Percent yield = \frac{Experimental production}{Theoretical production} x 100$

Substitution

$Percent yield = \frac{3}{4} x 100$

Result

Percent yield = 75 %

8 0

3 years ago

Consider the intermediate chemical reactions. 2 equations. First: upper C a (s) plus upper C upper O subscript 2 (g) plus one ha

DochEvi [55]

<u>Answer:</u> When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The overall chemical reaction follows:

$CaO(s)+CO_2\rightarrow CaCO_3(s)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $Ca(s)+CO_2(g)+\frac{1}{2}O_2(g)\rightarrow CaCO_3(s)$ $\Delta H_1=-812.8kJ$

(2) $2Ca(s)+O_2(g)\rightarrow 2CaO(s)$ $\Delta H_2=-1269kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[\frac{1}{2}\times (-\Delta H_2)]$

Hence, when the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed.

3 0

2 years ago

A sample of the chiral molecule limonene is 79% enantiopure. what percentage of each enantiomer is present? what is the percent

Degger [83]

Answer : The % of (+) limonene isomer = 79%

The % of (-) limonene isomer = 0%

The % of enantiomeric excess = 58%

Explanation : Enantiomeric excess (ee) is the measurement of purity used for chiral substances.

Given,

% of pure limonene enantiomer = The % of (+) limonene isomer = 79%

Therefore, The % of (-) limonene isomer = 0%

Formula used :

$\%(+)\text{ isomer}=\frac{ee}{2}+50\%$

Where, ee → enantiomeric excess

Now, put all the values in above formula, we get the value of enantiomeric excess (ee).

${ee}=\frac{\%(+)-50\%}{2}$

$=\frac{79\%-50\%}{2}$

= 58%

7 0

3 years ago

Read 2 more answers

Please help!!!! ⚠️⚠️⚠️⚠️⚠️⚠️⚠️⚠️

Bumek [7]

Polyatomic ions: $CH_3COO^-$ , $NO_3^-$ , $NH^+$ , $CN^-$ , $Li^+$ , and $OH^-$

Monatomic ions: $Ca^{2+$ , $O^{2-$ , and $Fe^{3+$

<h3>Monoatomic vs Polyatomic Ions</h3>

In chemistry, monoatomic ions are ions that consist of only a single type of atom. They are usually positive or negatively charged and are otherwise known as simple ions. Examples include $Ca^{2+$ , $O^{2-$ , and $Fe^{3+$

Polyatomic ions, on the other hand, are ions that consist of more than one atom, unlike monoatomic ions. The two or more atoms are covalently bonded and the entire structure behaves like a single chemical entity in reactions. Polyatomic ions are otherwise known as molecular ions.

Examples of polyatomic ions are $CH_3COO^-$ , $NO_3^-$ , $NH^+$ , $CN^-$ , $Li^+$ , and $OH^-$

Thus, from the diagram:

Polyatomic ions: $CH_3COO^-$ , $NO_3^-$ , $NH^+$ , $CN^-$ , $Li^+$ , and $OH^-$

Monatomic ions: $Ca^{2+$ , $O^{2-$ , and $Fe^{3+$

More on ions can be found here: brainly.com/question/14982375

#SPJ1

6 0

9 months ago