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zaharov [31]
3 years ago
11

what would you be most likely to measure by immsering and object in water and seeing how much the water level rises

Chemistry
1 answer:
Katena32 [7]3 years ago
4 0
Mass, if you know what element you are working with.
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A or B or C or D, fassst plzzz
olga nikolaevna [1]

Answer:

a.option is the correct answer

4 0
3 years ago
Read 2 more answers
You decide to clean the bathroom. You notice that the shower is covered in a strange green slime . you try to get rid of this sl
Free_Kalibri [48]
  1. The independent variable (IV) is the lemon juice mixture
  2. The dependent variable (DV) is the appearance of the green slime on the shower
  3. The control variable (CV) are time taken to spray, the amount of spray
  4. The experimental group (EG) is the side of the shower sprayed with lemon juice mixture
  5. The control group (CG) is the side of the shower sprayed with water.

INDEPENDENT VARIABLE

  • Independent variable is the variable of an experiment that is changed by the experimenter in order to bring about a change. It is the variable being tested in the experiment. In this case, the IV is the lemon juice mixture tested on the green slime on the shower.

DEPENDENT VARIABLE:

  • Dependent variable is the variable that is observed or measured in an experiment. It is also called responding variable. The DV in this case is the appearance of the green slime on the shower.

CONTROL VARIABLE:

  • Control variable is the variable that is kept constant throughout the experiment for all groups. The CV is the same for all the groups and they include: time taken to spray, the same amount of spray

CONTROL GROUP

  • Control group is the group that does not receive the independent variable or test in an experiment. In this case, the CG is the side of the shower sprayed with water.

EXPERIMENTAL GROUP:

  • Experimental group is the group of ab experiment that receives the experimental treatment or independent variable. In this case, the EG is the side of the shower sprayed with lemon juice mixture.

Therefore, the IV, DV, CV, EG and CG of this experiment are as follows:

  1. The independent variable (IV) is the lemon juice mixture
  2. The dependent variable (DV) is the appearance of the green slime on the shower
  3. The control variable (CV) are time taken to spray, the amount of spray
  4. The experimental group (EG) is the side of the shower sprayed with lemon juice mixture
  5. The control group (CG) is the side of the shower sprayed with water.

Learn more: brainly.com/question/17498238?referrer=searchResults

7 0
3 years ago
Which of the following elements would you expect to be extremely explosive in the presence of water?
Natali5045456 [20]
The answer is a. Sodium
6 0
3 years ago
Read 2 more answers
Calculate the change in enthalpy for the following reaction, using the bond enthalpy data provided below. (Note: you may round t
aliina [53]

Answer:

-514 kJ/mol

Explanation:

The bond enthalpy which is also known as bond energy can be defined as the amount of energy needed to split one mole of the stated bond. The change in enthalpy of a given reaction can be estimated by subtracting the sum of the bond energies of the reactants from the sum of the bond energies of the products.

For the given chemical reaction, the change in enthalpy of the reaction is:

ΔH_{rxn} = [2(409) + 4(388) + 3(496) - 4(630) - 4(463)] kJ/mol = 818 + 1552 + 1488 - 2520 - 1852 = -514 kJ/mol

3 0
4 years ago
the solubility product Ag3PO4 is: Ksp = 2.8 x 10^-18. What is the solubility of Ag3PO4 in water, in moles per liter?
guapka [62]

Answer : The solubility of Ag_3PO_4 in water is, 1.8\times 10^{-5}mol/L

Explanation :

The solubility equilibrium reaction will be:

Ag_3PO_4\rightleftharpoons 3Ag^++PO_4^{3-}

Let the molar solubility be 's'.

The expression for solubility constant for this reaction will be,

K_{sp}=[Ag^{+}]^3[PO_4^{3-}]

K_{sp}=(3s)^3\times (s)

K_{sp}=27s^4

Given:

K_{sp} = 2.8\times 10^{-18}

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

K_{sp}=27s^4

2.8\times 10^{-18}=27s^4

s=1.8\times 10^{-5}M=1.8\times 10^{-5}mol/L

Therefore, the solubility of Ag_3PO_4 in water is, 1.8\times 10^{-5}mol/L

3 0
3 years ago
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