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4 years ago

14

In an experiment, the control group is used to test the effect of the independent

1 answer:

olchik [2.2K]4 years ago

3 0

True, the control group is the thing to compare the end result to, to see if the project worked
hoped this helped :)

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Suppose you have a 250 ml container completely filled with ice. what volume of liquid water will form when the ice melts?

defon

The volume. is 2500 ml

4 0

3 years ago

The energy of flowing electrons

Komok [63]

Answer:

C. ElectricalElectrical

Explanation:

( it might be wrong pls dont report me just let me kno y its wrong )

7 0

3 years ago

In which reaction does the oxidation number of hydrogen change? In which reaction does the oxidation number of hydrogen change?

dedylja [7]

<u>Answer:</u> The correct answer is $2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)$

<u>Explanation:</u>

Oxidation number is defined as the number which is given to an atom when it looses or gains electron. When an atom looses electron, it attains a positive oxidation state. When an atom gains electron, it attains a negative oxidation state.

Oxidation state of the atoms in their elemental state is considered as 0. Hydrogen is present as gaseous state.

For the given chemical reactions:

<u>Reaction 1:</u> $2HClO_4(aq.)+CaCO_3(s)\rightarrow Ca(ClO_4)_2(aq.)+H_2O(l)+CO_2 (g)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

<u>Reaction 2:</u> $CaO(s)+H_2O(l)\rightarrow Ca(OH)_2(s)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

<u>Reaction 3:</u> $HCl(aq.)+NaOH(aq.)\rightarrow NaCl(aq.)+H_2O(l)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

<u>Reaction 4:</u> $2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: 0

Thus, the oxidation state of hydrogen is changing.

<u>Reaction 5:</u> $SO_2(g)+H_2O(l)\rightarrow H_2SO_3(aq.)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

Hence, the correct answer is $2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)$

6 0

3 years ago

An ion has a mass number of 65, 36 neutrons, and a charge of +1 . Identify the element symbol, and determine the number of proto

docker41 [41]

Answer:

Element symbol Cu

Number of protons 29

Number of electrons 28

Explanation:

To get the element symbol, we need the name of the element. To correctly identify the name of the element, we need the proton number.

Now the mass number is 65 and the number of neutrons is 36. The number of protons is this the mass number minus the number of neutrons. This equals 65 - 36 = 29 protons.

The element with 29 protons is copper Cu.

Now to get the number of electrons, for an electrically neutral atom , the number of electrons equals the number of protons. But here, the atom is not electrically neutral anymore as it has now formed a univalent positive ion of +1.

To form a positive ion, you have to do so by losing electrons. The atom in question here has just lost one electron. Thus, our of the 29, it is left with only 28 electrons.

8 0

3 years ago

Read 2 more answers

Predict the sign of the entropy change of the system for each of the following reactions. (a) N2(g)+3H2(g)→2NH3(g) (b) CaCO3(s)→

Andrew [12]

Answer:

a. Negative

b. Positive

c. Negative

d. zero

Explanation:

Entropy is measure of disorder. Positive entropy implies that a system is becoming more disordered. The opposite is true.

(a) N2(g)+3H2(g) → 2NH3(g) Negative because the system is becoming less disordered since the number of gaseous moles is decreasing

(b) CaCO3(s)→CaO(s)+CO2(g) Positive because a solid produces a gas which is more disorder therefore there is an increase in entropy

(c) 3C2H2(g)→C6H6(g) Negative because the number of moles of a gas decrease meaninng there is less disorder

(d) Al2O3(s)+3H2(g) → 2Al(s)+3H2O(g) zero because the gaseous moles do not change

5 0

4 years ago