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3 years ago

A company wants to sell more sports drinks so they decide to add more sugar. For Group A they

Chemistry

1 answer:

ycow [4]3 years ago

8 0

Answer:

Group B

Explanation:

The control group of an experiment is considered to be the "normal" because it does not receive the expeiemental treatment and hence is used to compare with the experimental group. The control and experimental groups are similar in every other aspect with the exception of the "INDEPENDENT VARIABLE". The independent variable is not changed in the control group.

In this experiment, the control group is GROUP B, which was given a plain gatorade without adding sugar. Sugar is the independent variable here, and it was not included in GROUP B, meaning that GROUP B is the CONTROL GROUP. On the other hand, the experimental group is the GROUP A.

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, the elements present in common salt are

katrin2010 [14]

Answer:

Chemically, table salt consists of two elements, sodium (Na) and chloride (Cl). Neither element occurs separately and free in nature, but are found bound together as the compound sodium chloride.

Explanation:

please mark me brainliest.

7 0

3 years ago

Read 2 more answers

Calculate the electronegativity difference of CH3OH

Maru [420]

You can't calculate the electronegativity difference of a single molecule you'll need a frame of reference to explain the electronegativity difference and properties arising due to the difference. Another term arise here is modified electronegativity that is calculated within the molecule between two atoms of different electronegativity and use to explain the nature of molecule wether they are electron withdrawing group or electron donating group and also used to compare the electronegativity with other molecule!

5 0

3 years ago

If 18.7ml of 0.01500M aqueous HCl is required to titrâtes 15.00ml of an aqueous solution of NaOH to the equivalence point, what

sweet [91]

Answer:

0.0187 M

Explanation:

Step 1: Write the balanced neutralization reaction

NaOH + HCl ⇒ NaCl + H₂O

Step 2: Calculate the reacting moles of HCl

18.7 mL of 0.01500 M HCl react.

0.0187 L × 0.01500 mol/L = 2.81 × 10⁻⁴ mol

Step 3: Calculate the reacting moles of NaOH

The molar ratio of HCl to NaOH is 1:1. The reacting moles of NaOH are 1/1 × 2.81 × 10⁻⁴ mol = 2.81 × 10⁻⁴ mol.

Step 4: Calculate the molarity of NaOH

2.81 × 10⁻⁴ moles are in 15.00 mL of NaOH.

[NaOH] = 2.81 × 10⁻⁴ mol/0.01500 L = 0.0187 M

6 0

3 years ago

2NO (g) + O2 (g) →2NO2 (g) At equilibrium [NO] = 2.4 × 10 -3 M, [O2] = 1.4 × 10 -4 M, and [NO2] = 0.95 M.

azamat

Answer:

$K=1.12x10^9$

Explanation:

Hello there!

Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

$K=\frac{[NO_2]^2}{[NO]^2[O_2]}$

Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

$K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9$

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.

Best regards!

4 0

3 years ago

Calcium carbonate decomposes when heated to solid calcium oxide and carbon dioxide gas. The balanced equation is: CaCO3(s) → CaO

wlad13 [49]

Answer:

The mass of CaCO3 reacted is 5.00 grams

Explanation:

Step 1 :Data given

Before the reaction, the container, including the CaCO3, had a mass of 24.20 g

After the reaction the container with product had a mass of only 22.00 g because the CO2 gas produced did not remain in the container.

Molar mass of CO2 = 44.01 g/mol

Molar mass CaCO3 = 100.09 g/mol

Step 2: The balanced equation

CaCO3 → CaO + CO2

Step 3: Calculate mass of CO2

Mass of CO2 = 24.20 grams - 22.00 grams

Mass of CO2 = 2.20 grams

Step 4: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 2.20 grams / 44.01 g/mol

Moles CO2 = 0.0500 moles

Step 5: Calculate moles CaCO3

For 1 mol CaCO3 we'll have 1 mol CaO and 1 mol CO2

For 0.0500 moles CO2 we need 0.0500 moles CaCO3

Step 6: Calculate mass CaCO3

Mass CaCO3 = moles CaCO3 * molar mass CaCO3

Mass CaCO3 = 0.0500 moles * 100.09 g/mol

Mass CaCO3 = 5.00 grams

The mass of CaCO3 reacted is 5.00 grams

8 0

3 years ago