Ask question

Ask question

All categories

zloy xaker [14]

3 years ago

7

Help fill in the blanks this is for 8th-grade science btw.

2 answers:

babymother [125]3 years ago

8 0

Answer:

i know that 10 x is .22300

Dvinal [7]3 years ago

4 0

I know that 10x is .22300

You might be interested in

Help with number 3 plz help due tomorrow!!!!!!!!!!

IceJOKER [234]

Answer:

which one?

Explanation:

???????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????

4 0

3 years ago

Read 2 more answers

A flexible container holds 32.4L of gas at 55.0c. What will be the volume if the temperature falls to 17.0c

mash [69]

The answer will be 25.92 L for 17.0 c

5 0

3 years ago

Why it’s possible to walk through air but not a concrete wall

Sunny_sXe [5.5K]

well you can't walk through the air but you can't walk through a concrete wall because all of the solid atoms holding it together

3 0

4 years ago

An unknown amount of mercury (II) oxide was decomposed in the lab. Mercury metal was formed and 5.20 L of oxygen was released at

Yakvenalex [24]

Answer:

<u>68.3g</u>

Explanation:

<u>1. Word equation:</u>

<em>mercury(II) oxide → mercury + oxygen </em>

<u>2. Balanced molecular equation:</u>

2HgO → 2Hg + O₂(g)

<u>3. Mole ratio</u>

Write the ratio of the coefficients of the substances that are object of the problem:

$2molHgO/1molO_2$

<u>4. Calculate the number of moles of O₂(g)</u>

Use the equation for ideal gases:

$pV=nRT\\\\\\n=\dfrac{pV}{RT}\\\\\\n=\dfrac{0.970atm\times5.20L}{0.08206atm.L/K.mol\times 390.0K}\\\\\\n=0.1576mol$

<u>5. Calculate the number of moles of HgO</u>

$\dfrac{2molHgO}{1molO_2}\times 0.1576molO_2=0.315molHgO$

<u>6. Convert to mass</u>

mass = # moles × molar mass

molar mass of HgO: 216.591g/mol

mass = 0.315mol × 216.591g/mol = 68.3g

7 0

3 years ago

What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

$P_H =- log[H^{+} ]$

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

$c=0.0014\times \frac{1}{10}$

$c=1.4\times 10^{-4}$

6 0

3 years ago