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3 years ago

The variable that a scientist changes when conducting an experiment is called the _____variable.

Chemistry

2 answers:

Flauer [41]3 years ago

6 0

He variable that a scientist changes when conducting an experiment is called the manipulated variable

vaieri [72.5K]3 years ago

4 0

Answer: The correct answer is option (a).

Explanation:

Dependent variables are the variables whose value depend on another variable. Its value changes as the independent variable changes.

Independent variables are the variables which do not depend on any other variable beside . Its value does not change with a change in other dependent variables.

While conducting experiment scientist changes dependent variable in order to record observations.

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If a reaction mixture contains 24 g of Mg and 32 g of O2 , what is the limiting reactant?

Alexxx [7]

This problem is providing the mass of both magnesium metal and oxygen gas and involved in a chemical reaction and asks for the limiting reactant. At the end, it turns out to be identified as magnesium.

<h3>Stoichiometry</h3>

In chemistry, stoichiometry is a widely-used tool we use in order to relate the mass and moles of different chemical substances involved in a chemical reaction. Thus, we consider the following chemical equation between magnesium and oxygen to produce magnesium oxide.

$2Mg+O_2\rightarrow 2MgO$

However, when the mass of the both of the reactants is given, one must identify the limiting reactant as the one producing the least of the moles of the product, which means we can use the given grams of the both of the reactants, their molar masses and mole ratios with the product to obtain the aforementioned:

$24gMg*\frac{1molMg}{24.3gMg}*\frac{2molMgO}{2molMg}=0.988molMgO\\ \\ 32gO_2*\frac{1molO_2}{32.0gO_2}*\frac{2molMgO}{1molO_2}=2molMgO$

Thus, we can evidence how 24 g of magnesium produce the least of the moles of magnesium oxide, fact validating the magnesium as the limiting reactant and the oxygen as the excess one.

Learn more about stoichiometry: brainly.com/question/9743981

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2 years ago

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2 years ago

Read 2 more answers

For lunch, a patient consumed 3 oz of skinless chicken, 3 oz of broccoli, 1 medium apple, and 1 cup of nonfat milk (see Table 3.

Volgvan

Lunch of a patient has 3 oz skinless chicken, 3 oz of broccoli, 1 medium apple, and 1 cup of nonfat milk

Energy content of 3 oz skinless chicken is = 110 kcal

Energy content of 3 oz broccoli = 30 kcal

Energy content of 1 medium apple = 60 kcal

Energy content of 1 cup non-fat milk = 90 kcal

So the kilocalories of energy patient obtained from lunch

= 110 kcal+ 30 kcal + 60 kcal + 90 kcal = 290 kcal

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3 years ago

Assuming that the experiments performed in the absence of inhibitors were conducted by adding 5 μl of a 2 mg/ml enzyme stock sol

prohojiy [21]

Hey there!:

From the given data ;

Reaction volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )

Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:

[E]t in molar = g/L * mol/g

[E]t = 0.01 g/L * 1 / 45,000

[E]t = 2.22*10⁻⁷

Vmax = 0.758 umole/min/ per mL

= 758 mmole/L/min

=758000 mole/L/min => 758000 M

Therefore :

Kcat = Vmax/ [E]t

Kcat = 758000 / 2.2*10⁻⁷ M

Kcat = 3.41441 *10¹² / min

Kcat = 3.41441*10¹² / 60 per sec

Kcat = 5.7*10¹⁰ s⁻¹

Hence kcat of xyzase is 5.7*10¹⁰ s⁻¹

Hope that helps!

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3 years ago