A scientist thinks that ducks swim faster when they are in cold water. He tests two groups of ducks, one in cold water and one in room temperature water. He records how fast they swim.Experimental Group:

Control Group:

Answers

Answer 1

Answer: Experimental group is the ducks in cold water
And the control group is the ducks in room temperature water

Answer 2

Answer:

Final answer:

In the experiment, the Experimental Group represents ducks in cold water, while the Control Group represents ducks in room temperature water. The scientist is testing whether water temperature affects duck swimming speed.

Explanation:

In the experiment described, the scientist is exploring whether the temperature of the water affects the speed of swimming of ducks. The Experimental Group in this scenario represents the ducks that are swimming in cold water. The scientist expects that these ducks will swim faster, based on the original hypothesis. On the other hand, the Control Group represents the ducks that are swimming in room temperature water. This group serves a basis for comparison to help the scientist determine what changes (if any) occur when the variable (water temperature) is altered.

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At STP, oxygen exists in two forms, O2(g) and O3(g). These two forms of oxygen have 1. different molecular structures and different properties 2. different molecular structures and the same properties 3. the same molecular structure and different properties 4. the same molecular structure and the same properties

Answers

Answer: option 1. different molecular structures and different properties.

Explanation:

1) O₂ and O₃ are different forms of the same element, in the same physical state (gaseous).

2) That is what is called allotropes.

3) The definition of allotropes is different forms of an element with different structures and different properties in the same physical state.

Graphite and diamond is other example of common allotropes.

4) O₃ and O₂ have the same kind of atoms (oxygen), but they are bonded differently (different structure, one molecule has 3 atoms, other 2 atoms) which conferes them different properties (e.g. different odor, different chemical reactivity).

At STP, oxygen exists in two forms, O2(g) and O3(g). These two forms of oxygen have different molecular structures and different properties. The answer is number 1. The rest of the numbers do not answer the question above.

What is the speed of a man that travels 2 meters in 6.5 seconds?

Answers

Answer:

≈ 0.31 m/s

Explanation:

v = d/t

= 2m/6.5s

= 0.3079 m/s

≈ 0.31 m/s

Answer:

0.305556 m / s

Explanation:

Speed = distance / time

2 metres = 0.002 km

6.5 sec = 0.00180556 hrs

Therefore,

Speed = 0.002 / 0.00180556

= 1.1 km / hr

= 0.305556 m / s

The reaction between NaOH(aq) and H2SO4(aq) was studied in a constant-pressure calorimeter: 100.0 mL portions of 1.00 M aqueous NaOH and H2SO4, each at 24.0C, were mixed. The volume of the mixture was 200.0 mL. The maximum temperature achieved was 30.6C. Neglect the heat capacity of the calorimeter and the thermometer, and assume that the solution of products has a density of 1.00 g/mL and a specific heat capacity of 4.18 J/(g·K). Calculate H, the heat (enthalpy) of reaction, in kJ per mole of acid. (Hint: You may not have the equivalent amounts of NaOH and H2SO4 in this experiment.)

Answers

Answer:

110,4 kJ / mole of acid

Explanation:

The reaction of NaOH with H₂SO₄ is:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Moles of NaOH and H₂SO₄ are:

0,1000L×1,00M = 0,100 moles of NaOH and 0,100 moles of H₂SO₄. As 2 moles of NaOH reacts with 1 mole of H₂SO₄, moles of acid that react are 0,100/2 = 0,0500moles of acid.

The produced heat is:

Q = C×m×ΔT

Where C is specific heat capacity (4,18J/gK)

m is mass: 200,0mL×(1,00g/mL) = 200g

ΔT is change in temperature: 30,6°C - 24,0°C = 6,6°C = 6,6K

Thus, Q is:

Q = 4,18J/gK×200g×6,6K

Q = 5518J = 5,52kJ

ΔH in kJ per mole of acid:

5,52kJ / 0,0500moles of acid = 110,4 kJ / mole of acid

I hope it helps!

Match each element with the number of atoms found in one molecule of this compound: C5H7N (nicotine)

Answers

Carbon (5)
Hydrogen(7)
Nitrogen(1)

A hydrogen ion, H+, in aqueous solution may also be written as(1) H2O (3) H3O+
(2) H2O2 (4) OH–

Answers

Answer: Option (3) is the correct answer.

Explanation:

In an aqueous solution, a compound exists in the form of ions that is in the form of protons and electrons.

Protons are positively charged and electrons are negatively charged. A proton (usually hydrogen ion, $H^(+)$ ) in water (solvent) also exists as $H_(3)O^(+)$ .

For example, $H_(2)O + HCl \rightleftharpoons H_(3)O^(+) + Cl^(-)$

This shows that the conjugate acid of $H_(2)O$ is $H_(3)O^(+)$ .

Thus, we can conclude that a hydrogen ion, H+, in aqueous solution may also be written as $H_(3)O^(+)$ .

The answer is (3) H3O+. In solution, H+ exists interchangeably bound to H2O (as the hydronium ion) and as a free ion. Some books represent strong acids in net ionic equations as simply H3O+

Metallurgy was well known by the Guptas, as can be attested to by the copper iron steel bronze pillar of Delhi.

Answers

The correct answer for the question that is being presented above is this one: "TRUE." Metallurgy was well known by the Guptas, as can be attested to by the copper iron steel bronze pillar of Delhi. It is true that metallurgy was well know in the areas of the Guptas.

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