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3 years ago

In an estuary, _____.

Chemistry

1 answer:

lapo4ka [179]3 years ago

6 0

B; Seawater mixes with freshwater so the water has intermediate salinity

Explanation:

In an estuary, seawater mixes with freshwater so the water has intermediate salinity. Estuaries are usually located in transitional environments.

Estuary is the wide part of a river where it nears the sea.
This is called a transitional zone.
Water from continental rivers usually fresh are brought in close contact with ocean water that is salty.
The water here is said to be brackish as it is intermediate between salt and seawater.
Organisms living in such terrain must be be well adapted to changing salinity.

Learn more:

salinity and density brainly.com/question/10491444

#learnwithBrainly

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What the answer question

Serga [27]

Answer:

it's a trigonal bipyramidial

Explanation:

because NH3 have 3 hydrogen atoms

4 0

2 years ago

10. When a piece of steel wool is burned it gains mass. Which explanation is consistent with the law of conservation of mass?

frez [133]

Answer:

Oh come on. Look at all - then look at A.

Explanation:

4 0

3 years ago

A sample of gaseous arsine (AsH3) in a 460 mL flask at 332 Torr and 223 K, is heated to 437 K, at which temperature arsine decom

Taya2010 [7]

Answer:

28/95 = 29.,5 % of Arsine decomposed

Explanation:A sample of gaseous arsine (AsH3) in a 460 mL flask at 332 Torr and 223 K, is heated to 437 K, at which temperature arsine decom- poses to solid arsenic and hydrogen gas. The flask is then cooled to 273 K, at which tem- perature the pressure in the flask is 488 Torr. What percentage of arsine molecules have de- composed?

Answer in units of %.

initial pressure 332 Torr initial volume 0.46 L initial temperature 223K

final pressure 488 Torr final volume 0.46 L final 273 K

Torr is 1/760 atm 332 torr = 0.437 atm 488 Torr =0.642 atm

PV = nRT so n=RT/PV

INITIAL n= 0.082 X 223/(0.437)(0.46) = 91 moles

final n= 0.082 X 273 / (.437)(488) = 105 moles

2AsH3----------> 2As + 3H2

x moles of Arsine decomposed to make 1.5 moles of H2

the final number of moles was

(91 -X)+ 1.5 X = 105 moles

91 + 0.5 X = 105

0.5 X = 14

X =28

CHECK

if 28 moles of Arsine , then the container would have

91 --28 + 1.5(28) = 91 +14 =105 check

so 28/95 = 29.,5 % of Arsine decomposed

Your answer

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polyalchemVirtuoso

Answer:

Explanation:

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2 years ago

A sample of calcium oxide (CaO) has a mass of 2.80 g. The molar mass of CaO is 56.08 g/mol. How many moles of CaO does this samp

Molodets [167]

Answer is "0.05 mol".

<em>Explanation;</em>

We can do calculation by using a simple formula as

n = m/M

Where, n is the number of moles of the substance (mol), m is the mass of the substance (g) and M is the molar mass of the substance (g/mol).

Here,

n = ?

m = 2.80 g

M = 56.08 g/mol

By substitution,

n = 2.80 g /56.08 g/mol

n = 0.0499 mol ≈ 0.05 mol

4 0

3 years ago

Read 2 more answers

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

3 years ago