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mr Goodwill [35]

2 years ago

5

Mr. Bullis wants to know if he gives teachers candy, it will make them smile more. He gives half the teachers candy and the othe

r half get nothing. He then records the number of times the teachers smiled.
Experimental Group:

Control Group:

1 answer:

mote1985 [20]2 years ago

5 0

Answer:

the experimental group would be the teachers who recieved the candy

control group would be the teachers who dont get any candy

<em>Explanation:</em>

experimental group is the group that receives the test variable being tested in this case the variable being the candy

and the control group is the group who doesn't receive the variable

purpose of having a control is to rule out other factors which may influence the results of an experiment.

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Name two compounds with more exothermic lattice energies than scandium oxide and justify your choice.

N76 [4]

Answer:

1 . $Al_2O_3$

2. $TiO_2$

Explanation:

The more stable the ionic compound, the more is it lattice energy.

The more the charge on the cation and the anion, the greater is the lattice energy.
The less the size of the cation and the anion, the greater is the lattice energy.

Scandium oxide ( $Sc_2O_3$ ) is an oxide in which $Sc^{3+}$ behaves as cation and $O^{2-}$ behaves as anion.

The compounds which has higher lattice energy than scandium oxide are:

1 . $Al_2O_3$

This is because the charge are same on the cation and the anion as in the case of the Scandium oxide but the size of the cation $Al^{3+}$ is smaller than $Sc^{3+}$ . Thus, this corresponds to higher lattice energy.

2. $TiO_2$

This is because the charge on the cation $Ti^{4+}$ is greater than that of $Sc^{3+}$ and also the size of the cation $Ti^{4+}$ is smaller than $Sc^{3+}$ . Thus, this corresponds to higher lattice energy.

3 0

3 years ago

A solution of malonic acid, H2C3H2O4 , was standardized by titration with 0.1000 M NaOH solution. If 21.17 mL of the NaOH soluti

MrRa [10]

Answer: The molarity of the malonic acid solution is 0.08335 M

Explanation:

$H_2C_3H_2O_4 +2NaOH\rightarrow Na_2C_3H_2O_4+2H_2O$

To calculate the molarity of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2C_3H_2O_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=?\\V_1=12.70mL\\n_2=1\\M_2=0.1000M\\V_2=21.17mL$

Putting values in above equation, we get:

$2\times M_1\times 12.70=1\times 0.1000\times 21.17\\\\M_1=0.08335M$

Thus the molarity of the malonic acid solution is 0.08335 M

5 0

3 years ago

A metal weighing 7.101 g is placed in a graduated cylinder containing 33.0 mL of water. The water level rose to the 37.4 mL mark

musickatia [10]

Volume of the metal = change in volume reading
Volume = 37.4 - 33
Volume = 4.4 ml

Density = mass / volume
Density = 7.101 / 4.4
Density = 1.61 g/ml

7 0

3 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

7 0

3 years ago

How are biology and Chemistry are related give any two points or reasons

k0ka [10]

Answer:

Both of them are related most of the structures of human body .The working of cells, the cellular respiration processes ,the proteins synthesizing carbohydrates syntheses lipids synthesis and many other reactions involves chemistry.

•Chemistry and biology both are branches of science so they are much more related to each other.

Explanation:

<h2>I hope it's helpful for you</h2>

6 0

2 years ago

Read 2 more answers