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2 years ago

about the lung with balloons what respiratory organ do the small balloons represents? the big balloons?

Chemistry

1 answer:

butalik [34]2 years ago

7 0

Answer:

The plastic bottle represents the chest cavity, the straws act as the bronchi, the small balloons inside the plastic bottle serves as the lungs and the big balloon at the bottom of the plastic bottle is the diaphragm.

Explanation:

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Consider 5.00 L of a gas at 365 mmHg and 20. ∘C . If the container is compressed to 2.50 L and the temperature is increased to 3

pickupchik [31]

First convert celcius to Kelvin.

20 + 273 = 293K

31 + 273 = 304K

Now we can set up an equation based on the information we have.

V1 = 5

P1 = 365

T1 = 293

V2 = 5

P1 = x

T2 = 304

The equation be: $\frac{(5)(365)}{293} = \frac{5x}{304}$

Now just solve.

1825/293 = 5x/304

Cross multiply.

554800 = 1465x

Divide both sides by 1465

x = 378.7030717 which can then be rounded to 378.7 mmHg

4 0

2 years ago

What tendency is the standard reduction potential of an element based on?

Y_Kistochka [10]

I think the answer is B :)

3 0

2 years ago

Read 2 more answers

Put the substance in a 25 mL beaker.

kkurt [141]

Answer: 1. Liquid

2. oily

3. no (not a solid)

Explanation:

5 0

3 years ago

Read 2 more answers

How much time is required to plate out 20.0 g of Zn from a solution containing Zn2+ ions, using a current of 2.00 amperes?

lyudmila [28]

Answer: 494.872mins

Explanation:

M= MrIt÷n

M=20g

Mr=65g/mol

I=2A

1 farad= 96500

For zinc with 2ions

2×96500= 193000

M×n= MrIt

T= M×n÷MrI

t= 20×193000/2×65

t= 3860000/130

t= 29692.307sec

t=29692.307/60

t= 494.872 mins

7 0

2 years ago

If a 95.27 mL sample of acetic acid (HC2H3O2) is titrated to the equivalence point with 79.06 mL of 0.113 M KOH, what is the pH

KATRIN_1 [288]

Answer:

8.73

Explanation:

The concentration of acetic acid can be determined as follows:

$M_1V_1 = M_2V_2\\(KOH) = (CH_3COOH)$

$M_{KOH}=0.113 M\\V_{KOH}=79.06 mL$

$V_{CH_3COOH}=95.27 \\\\M_{CH_3COOH)=?????$

$M_{CH3COOH} = \frac{M_{KOH}*V_{KOH}}{V_{CH_3COOH}}$

$M_{CH3COOH} = \frac{0.113*79.06}{95.27}$

$M_{CH3COOH} = 0.094 M$

Moles of $CH_3COOH$ = $95.27* 10^{-3}* 0.094$

=0.0090 moles

Moles of $KOH = 79.06*10^{-3}*0.113$

= 0.0090 moles

The equation for the reaction can be expressed as :

$CH_3COOH$ $+$ $KOH$ -----> $CH_3COO^{-}K^+$ $+$ $H_2O$

Concentration of $CH_3COO^{-$ ion = $\frac{0.0090}{Total volume (L)}$

= $\frac{0.0090}{(95.27+79.06)} *1000$

= 0.052 M

Hydrolysis of $CH_3COO^{-$ ion:

$CH_3COO^{-$ $+$ $H_2O$ -----> $CH_3COOH$ $+$ $OH^-$

$K = \frac{K_w}{K_a} = \frac{x*x}{0.052-x}$

⇒ $\frac{10^{-14}}{1.82*10^{-5}}= \frac{x*x}{0.052-x}$

= $0.5494*10^{-9}= \frac{x*x}{0.052-x}$

As K is so less, then x appears to be a very infinitesimal small number

0.052-x ≅ x

$0.5494*10^{-9}= \frac{x^2}{0.052}$

$x^2 = 0.5494*10^{-9}*0.052$

$x^2 = 0.286*10^{-10$

$x = \sqrt{0.286*10^{-10$

$x =0.535*10^{-5}M$

$[OH] = x =0.535*10^{-5}$

$pOH = -log[OH^-]$

$pOH = -log[0.535*10^{-5}]$

$pOH = 5.27$

pH = 14 - pOH

pH = 14 - 5.27

pH = 8.73

Hence, the pH of the titration mixture = 8.73

8 0

2 years ago