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Sergeeva-Olga [200]

3 years ago

9

Did I balance this equation correctly

1 answer:

11111nata11111 [884]3 years ago

7 0

Yep! I typed your equation into this website"http://www.endmemo.com/chem/balancer.php" and it said it was balanced

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The men's world record for swimming 1500 m in a long course pool is 14 min 34.56 s. At this rate how many seconds would it take

vlabodo [156]

Distance = 1500m, time = 14min 34.56s = 14*60 + 34.56 = 874.56s

Speed = Distance / time = 1500/874.56 = 1.715 m/s

In order to swim 0.7 miles, 1 mile = 1609.344 m

Time = Distance / speed = 1609.344 m / 1.715 m/s = 938.393 seconds

938/60 = 15 minutes, 38 seconds.

= 15 minutes 38.393 seconds.

6 0

2 years ago

The shells further away from the nucleus are larger/smaller and can hold more/less electrons.

Sergeu [11.5K]

The shells further away from the nucleus are LARGER and can hold MORE electrons

8 0

2 years ago

I messed up Èeeeeeeeeeeeeeeeeeeee

Triss [41]

Answer:

o

Explanation:

6 0

1 year ago

What information is needed to determine the amount of moles in 1.6g of HCL?

alukav5142 [94]

Answer:

Molar mass and Mass

Explanation:

The relationship between mass and number of moles is given as;

Number of moles = Mass / Molar mass

Mass = 1.6 g

Molar mass of HCl = ( 1 + 35.5 ) = 36.5 g/mol

Number of moles = 1.6 g / 36.5 g/mol

Number of moles = 0.0438 mol

8 0

2 years ago

A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipita

mestny [16]

Answer:

The answer is:

(a) $NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)$

(b) NaCl

(c) 0.211 g

Explanation:

Given:

The mass of NaCl,

= 0.0860 g

The molar mass of NaCl,

= 58.44 g/mol

The volume of $AgNO_3$ ,

= 30.0 ml

or,

= 0.030 L

Molarity of $AgNO_3$ ,

= 0.050 M

Moles of NaCl will be:

= $\frac{Given \ mass}{Molar \ mass}$

= $\frac{0.0860}{58.44}$

= $0.00147 \ mol$

now,

Moles of $AgNO_3$ will be:

$= Molarity\times Volume$

$= 0.050\times 0.030$

$=0.0015 \ mol$

(a)

The reaction is:

⇒ $NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)$

(b)

1 mole of NaCl react with,

= 1 mol of $AgNO_3$

0.0015 mol $AgNO_3$ needs,

= $0.00150 \ mol \ NaCl$

Available mol of NaCl < needed amount of NaCl

So,

The limiting reagent is "NaCl".

(c)

The precipitate formed,

= $0.00147\times \frac{1}{1}\times \frac{143.32}{1}$

= $0.211 \ g \ AgCl$

4 0

2 years ago