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san4es73 [151]
3 years ago
14

Anita was using steel to make rusted sculptures. After building each sculpture, she caused the steel in the sculpture to rust by

placing it into a tub filled
with salt water for eight hours. Anita wondered if steel would rust faster submerged in vinegar instead of salt water. To find out, Anita cut ten squares of

steel sheet metal and split them into two equal groups. She put one group of squares into a tub filled with salt water and the other group of squares

into a tub filled with vinegar. Once an hour for eight hours, Anita counted the number of rusted steel squares in each group.

What is the manipulated independent variable in Anita's experiment?
Chemistry
1 answer:
alexandr1967 [171]3 years ago
5 0

Answer:

The type of liquid in the tub (salt water or vinegar)

Explanation:

<em>The manipulated independent variable in Anita's experiment is </em><em>the type of liquid in the tub. </em>

The independent variable is the controlled or manipulated variable in the course of an experiment. It can also be referred to as the 'cause' variable which has the capacity to produce 'effects' on another variable - the dependent variable.

In this case, the type of liquid the tub is filled (salt water or vinegar) will hypothetically affect the rusting period of the steel. Hence, the dependent variable is the type of liquid the tub is filled while the dependent variable would be the time it takes for the steel to get rusted.

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If a black ball is denser than a white ball of the same size, what does the black ball have??
yKpoI14uk [10]

Answer: The black ball will have more mass than the white ball.

Explanation: Density of a substance is defined as the ratio of mass and volume occupied by the substance.

Mathematically,

Density=\frac{Mass}{Volume}

We are given that the two balls are of same size, which means that the volume of both the balls are same. We are also given that the black ball is more dense than the white ball, which means that the black ball will have more mass.

As, Density\propto Mass

7 0
3 years ago
A student dissolves of glucose in of a solvent with a density of . The student notices that the volume of the solvent does not c
nikitadnepr [17]

Answer:

0.052 M

0.059 m

Explanation:

There is some missing info. I think this is the complete question.

<em>A student dissolves 4.6 g of glucose in 500 mL of a solvent with a density of 0.87 g/mL. The student notices that the volume of the solvent does not change when the glucose dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to 2 significant digits.</em>

Step 1: Calculate the moles of glucose (solute)

The molar mass of glucose is 180.16 g/mol.

4.6 g × 1 mol/180.16 g = 0.026 mol

Step 2: Calculate the molarity of the solution

0.026 moles of glucose are dissolved in 500 mL (0.500 L) of solution. We will use the definition of molarity.

M = moles of solute / liters of solution

M = 0.026 mol / 0.500 L = 0.052 M

Step 3: Calculate the mass corresponding to 500 mL of the solvent

The solvent has a density of 0.87 g/mL.

500 mL × 0.87 g/mL = 435 g = 0.44 kg

Step 4: Calculate the molality of the solution

We will use the definition of molality.

m = moles of solute / kilograms of solvent

m = 0.026 mol / 0.44 kg = 0.059 m

4 0
3 years ago
150 ml of 0.1 m naoh is added to 200 ml of 0.1 m formic acid, and water is added to give a final volume of 1 l. what is the ph o
N76 [4]

Number of moles of NaOH = V(NaOH) * M(NaOH)= 0.150 L * 0.1 moles/L = 0.015 moles

Number of moles of formic acid, HCOOH = V(HCOOH) * M(HCOOH) = 0.200 L * 0.1 moles/L = 0.020 moles

Here, the limiting reagent is NaOH

The reaction is represented as:

HCOOH + NaOH ↔HCOONa + H2O

Moles of HCOONa formed = Moles of the limiting reagent, NaOH = 0.015 moles

Moles of HCOOH remaining = 0.020-0.015 = 0.005 moles

Total final volume is given as 1 L

Therefore: [HCOOH] = 0.005 moles/1 L = 0.005 M

[HCOONa] = 0.015/1 = 0.015 M

pKa of HCOOH = 3.74

As per Henderson-Hasselbalch equation

pH = pka + log[HCOONa]/[HCOOH] = 3.74+log[0.015/0.005] = 4.22

Therefore, pH of the final solution = 4.22


                       


3 0
3 years ago
Convert the following<br> 1)<br> 52.1g NaNO3<br> To moles
ivann1987 [24]

Answer:

0.6129796138981438

5 0
3 years ago
Which radioisotopes have the same decay mode and have half-lives greater than
AveGali [126]

The correct answer is (3)


I-131 and P-32


The explanation:


according to attached table:


- we can see that the half life of p 32 is 14.28d (more than one hour)

- and the half life of I-131 is 8.021 d (more than one hour)


and They both have β- decay mode and with half-lives greater than hour.


4 0
3 years ago
Read 2 more answers
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