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kakasveta [241]

3 years ago

11

Science fair projects for 7th grade ideas?

2 answers:

allsm [11]3 years ago

8 0

Try the classic volcano project, grab construction paper and special liquid and some paint and etc and u good to go or u can try the twister idea of where u connect two soda cans and pour it out to make a twister tornado.

klio [65]3 years ago

4 0

Volcano you make the volcano with baking soda and vinegera

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The element in group 4 and period 3

sladkih [1.3K]

The three group 4 elements that occur naturally are titanium, zirconium, and hafnium. The first three members of the group share similar properties; all three are hard refractory metals under standard conditions.

8 0

3 years ago

Read 2 more answers

Start with 100.00 mL of 0.10 M acetic acid, CH3COOH. The solution has a pH of 2.87 at 25 oC. a) Calculate the Ka of acetic acid

jasenka [17]

Answer: a) The $K_a$ of acetic acid at $25^0C$ is $1.82\times 10^{-5}$

b) The percent dissociation for the solution is $4.27\times 10^{-3}$

Explanation:

$CH_3COOH\rightarrow CH_3COO^-H^+$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 0.10 M and $\alpha$ = ?

Also $pH=-log[H^+]$

$2.87=-log[H^+]$

$[H^+]=1.35\times 10^{-3}M$

$[CH_3COO^-]=1.35\times 10^{-3}M$

$[CH_3COOH]=(0.10M-1.35\times 10^{-3}=0.09806M$

Putting in the values we get:

$K_a=\frac{(1.35\times 10^{-3})^2}{(0.09806)}$

$K_a=1.82\times 10^{-5}$

b) $\alpha=\sqrt\frac{K_a}{c}$

$\alpha=\sqrt\frac{1.82\times 10^{-5}}{0.10}$

$\alpha=4.27\times 10^{-5}$

$\% \alpha=4.27\times 10^{-5}\times 100=4.27\times 10^{-3}$

5 0

3 years ago

If we were to continue to add more and more naoh from our original stock solution, what is the highest ph that could be reached?

Novay_Z [31]

If its a acid it will be 3 or if its a alkali then 14 or neutral then 7

7 0

3 years ago

In the modern periodic table, the elements are arranged according to

DedPeter [7]

Answer:

Explanation:

d

4 0

3 years ago

An inflated balloon contains X air molecules.After some time the volume of the balloon is found to be the half at the same tempe

astra-53 [7]

<u>Answer:</u>

(a): The balloon will contain $\frac{X}{2}$ number of molecules now.

(b): The gas law associated with this is Avogadro's law.

<u>Explanation:</u>

Avogadro's law states that the volume of the gas is directly proportional to the number of moles of the gas at constant pressure and temperature

Mathematically,

$V\propto n$ (At constant pressure and temperature)

OR

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$ .....(1)

where,

$V_1\text{ and }n_1$ = Initial volume and number of moles

$V_2\text{ and }n_2$ = Final volume and number of moles

We are given:

$V_2=\frac{V_1}{2}$

Putting values in equation 1, we get:

$\frac{V_1}{n_1}=\frac{V_1}{2\times n_2}\\\\n_2=\frac{n_1}{2}$

(a):

If $n_1$ number of moles of gas contains X number of molecules

So, $\frac{n_1}{2}$ number of moles of gas will contain = $\frac{X}{n_1}\times {n_1}{2}=\frac{X}{2}$ number of molecules

Hence, the balloon will contain $\frac{X}{2}$ number of molecules now.

(b): The law used is Avogadro's law.

8 0

3 years ago