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kvv77 [185]
3 years ago
6

A chemist prepares a solution of silver(I) nitrate by measuring out of silver(I) nitrate into a volumetric flask and filling the

flask to the mark with water.
Chemistry
1 answer:
leva [86]3 years ago
4 0

Answer:

3.33 M

Explanation:

It seems your question is incomplete, however, that same fragment has been found somewhere else in the web:

" <em>A chemist prepares a solution of silver nitrate (AgNO3) by measuring out 85.g of silver nitrate into a 150.mL volumetric flask and filling the flask to the mark with water.</em>

<em>Calculate the concentration in mol/L of the chemist's silver nitrate solution. Be sure your answer has the correct number of significant digits.</em> "

In this case, first we <u>calculate the moles of AgNO₃</u>, using its molecular weight:

  • 85.0 g AgNO₃ ÷ 169.87 g/mol = 0.500 mol AgNO₃

Then we<u> convert the 150 mL of the volumetric flask into L</u>:

  • 150 / 1000 = 0.150 L

Finally we <u>divide the moles by the volume</u>:

  • 0.500 mol AgNO₃ / 0.150 L = 3.33 M
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How many valence electrons do the halogens possess? group of answer choices
jok3333 [9.3K]

The valence electron does the halogens possess are 7

  • Valence electrons are found in the outermost energy level of an atom
  • They are involved in the formation of chemical bonding with other atoms.
  • The halogens elements are found in group 17 on the periodic table
  • The halogens include fluorine, chlorine, bromine, iodine and astatine.
  • They have seven valence electrons, so they are extremely reactive as they only need one more to fill their outer shell.
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Hence the halogens posses 7 valence electron

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Unit conversions.(a) Convert 10,000 dynes to units of lbm ∙ ft/s^2 and lbf(b) Convert 0.2 atm to units of Kpa and lbf/in^2(c) Co
Maru [420]

Answer:

(a)  0.72 lbm· ft/s²

(b)  20.3 kPa,  2.94 lbf / in²

(c)  98.6 ºF,  310 K

(d)  1.5 x 10⁻² J,  6.1 x 10⁻² cal

Explanation:

Our strategy here will be to find the conversion factors for the quantities we are asked in each part, and perform the calculations.

(a) 10,000 dynes to lbm ·ft/s²

here we are asked to convert  the  force of 10,000 dynes to lbm ·ft/s². Recall that F= ma ( m= mass, a = acceleration), thus

10,000 dynes = 10 g cm/s²

converting the force

10,000 g cm/s² x (1 lbm/454 g) x (1 ft / 30.48 cm ) /s² = 0.72 lbm· ft/s²

(b)

1 atm = 101.33 pa

0.2 atm x ( 101.33 kPa ) = 20.3 kPa

1 atm = 14.7 lbf / in²

0.2 atm x ( 14.7 lbf / in² /atm ) = 2.94 lbf / in²

(c) The formula for the conversion from ºC to ºF is:

ºF = 9/5 ºC +32

ºF = 9/5 ( 37ºC) + 32 = 98.6 ºF

K = ºC + 273

K = (37 + 273) K = 310 K

(d) 50 in²·lbm/s² to joules and calories

Since the unit in² ·lbm/s² is not that common, lets convert it using their definition.

These are energy units, and we know the energy is the force times distance. In turn force is mass times acceleration so that the units of energy are mass time distance per time squared.

Joules is the unit of energy  in the metric system.

50 in² lbm/s² = 50 in²x ( 2.54 cm/in x 1m /100cm)² x (1lbm x 0.454 Kg/lbm)/s²

= 1.5 x 10⁻² Kg m²/² =  1.5 x 10⁻² J

To convert to cal it wilñl be easier to use the value in joules just calculated:

1.5 x 10⁻² J x  (4.184 cal/J) = 6.1 x 10⁻² cal

4 0
4 years ago
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