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Ivahew [28]
2 years ago
12

In a laboratory, Carlos places one plate with grape jelly and another plate with cooked brown rice inside an ant farm. The next

day, he finds that both plates are filled with ants. Carlos wonders if there are certain household foods that are more effective than others at attracting ants. How could Carlos use the scientific inquiry process to determine which foods are most effective at attracting ants? Check all that apply.
Chemistry
2 answers:
Orlov [11]2 years ago
6 0

Answer:

The Answers are: A. C. D.

Tems11 [23]2 years ago
6 0

Answer:

by making quantitative observations about the number of ants found on each plate of food

by comparing a plate of cooked brown rice to another household food such as peanut butter

by designing an experiment to test different foods and the number of ants each food attracts

Explanation:

It was right on edge

You might be interested in
Identify the precipitate (if any) that forms when KOH and Cu(NO3)2 are mixed
iris [78.8K]
When KOH and Cu(NO3)2 are mixed it yields copper(II) hydroxide and potassium nitrate. The balanced reaction is:

Cu(No3)2(aq) + 2 KOH(aq)<span> = Cu(OH)2(s) + 2 KNo3(aq)
</span>

As we can see from the equation a solid is formed therefore a precipitate is formed which is the copper (II) hydroxide which has a blue to purple appearance. This can be observed since copper (II) hydroxide has a low solubility in aqueous solution.

3 0
2 years ago
It takes 330 j of energy to raise the temperature of 14.6 g of benzene from 21.0 °c to 28.7 °c at constant pressure. What is the
Akimi4 [234]

The mathematical expression for heat capacity at constant pressure is given as:

Q=n\times C_{p}\times \Delta T   (1)

where, Q = heat capacity

C_{p} =  molar heat capacity at constant pressure

\Delta T = change in temperature

n = number of moles

Therefore, \Delta T = 28.7^{o}C-21^{o}C

= 7.7 ^{o}C

Number of moles  =\frac{given mass in g}{molar mass}

= \frac{14.6 g }{78.11 g/mol}

= 0.186 mole

Put the values in formula (1)

330 J=0.186 mole\times C_{p}\times (7.7 ^{o}C+ 273) (conversion of degree Celsius into kelvin)

C_{p} = \frac{330 J}{0.186 mole\times 280.7 K}

= \frac{330 J}{52.2102 mole K}

= 6.32 J /mol K

Hence, molar heat capacity of benzene at constant pressure  = 6.32 Jmol^{-1} K^{-1}

6 0
2 years ago
A gem has a mass of 4.50 g. When the gem is placed in a graduated cylinder 12.00 mL of water, the water level rises to 13.45 mL.
Mandarinka [93]
<span>Displaced volume :

</span>Final volume - <span>Initial volume

</span>13.45 mL - 12.00 mL => 1.45 mL

Mass =  4.50 g

Therefore:

density = mass / volume

D = 4.50 / 1.45

<span>D = 3.103 g/mL </span>
6 0
2 years ago
Lithium has an atomic mass of 6.941 amu. Lithium has two common isotopes. The one isotope has a mass of 6.015 amu and a relative
koban [17]

Answer:

The atomic mass of second isotope is 7.016

Explanation:

Given data:

Average Atomic mass of lithium = 6.941 amu

Atomic mass of first isotope = 6.015 amu

Relative abundance of first isotope = 7.49%

Abundance of second isotope = ?

Atomic mass of other isotope = ?

Solution:

Total abundance = 100%

100 - 7.49 = 92.51%

percentage abundance of second isotope = 92.51%

Now we will calculate the mass if second isotope.

Average atomic mass of lithium = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

6.941 = (6.015×7.49)+(x×92.51) /100

6.941 =  45.05235 + (x92.51) / 100

6.941×100 = 45.05235 + (x92.51)

694.1 - 45.05235   = (x92.51)

649.04765 = x 92.51

x = 485.583 /92.51

x = 7.016

The atomic mass of second isotope is 7.016

3 0
2 years ago
Write the electron configurations for the following ions:
Ket [755]

Answer:

Co²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷

Sn²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰

Zr⁴⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶

Ag⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4d¹⁰

S²⁻ : 1s² 2s² 2p⁶ 3s² 3p⁶

Explanation:

Cobalt (Co): atomic number 27

<u>The electronic configuration of Co in ground state: </u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁷

<u>The electronic configuration of Co in +2 oxidation state (Co²⁺) :</u>

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷

Tin (Sn): atomic number 50

<u>The electronic configuration of Sn in ground state: </u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p²

<u>The electronic configuration of Sn in +2 oxidation state (Sn²⁺) </u>:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰

Zirconium (Zr): atomic number 40

<u>The electronic configuration of Zr in ground state:</u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d²

<u>The electronic configuration of Zr in +4 oxidation state (Zr⁴⁺) :</u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶

Silver (Ag): atomic number 47

<u>The electronic configuration of Ag in ground state:</u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹ 4d¹⁰

<u>The electronic configuration of Ag in +1 oxidation state (Ag⁺) :</u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4d¹⁰

Sulphur (S): atomic number 16

<u>The electronic configuration of S in ground state:</u>

1s² 2s² 2p⁶ 3s² 3p⁴

<u>The electronic configuration of S in -2 oxidation state (S²⁻) :</u>

1s² 2s² 2p⁶ 3s² 3p⁶

8 0
2 years ago
Read 2 more answers
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