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katrin [286]
3 years ago
8

What kind of lens does a nearsighted person need

Chemistry
2 answers:
love history [14]3 years ago
5 0
Near/shortsighted people need concave lenses. I myself am nearsighted
Troyanec [42]3 years ago
5 0

I know I am 5 years late but the answer is concave lens

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There are three subatomic particles found inside atoms. Two of the subatomic particle have a charge of positive or negative. The
NARA [144]

Neutrons don't have a charge. They are neutral

7 0
2 years ago
Read 2 more answers
A 150.0 mL solution of 2.888 M strontium nitrate is mixed with 200.0 mL of a 3.076 M sodium fluoride solution. Calculate the mas
Lelechka [254]

Answer:

Mass SrF2 produced = 38.63 g SrF2 produced

[Na^+]:  = 1.758 M

[NO3^-]:  = 1.238 M

[Sr^2+] = 0.3589 M

[F^-] = 2.36*10^-5 M

Explanation:

Step 1: Data given

Volume of 2.888M strontium nitrate = 150.0 mL = 0.150 L

Volume of 3.076 M sodium fluoride = 200.0 mL = 0.200 L

Step 2 : The balanced equation

Sr(NO3)2(aq) + 2NaF(aq) → SrF2(s) + 2NaNO3(aq) → Sr2+ + 2F- + 2

Step 3: Calculate moles strontium nitrate

Moles Sr(NO3)2 = Molarity * volume  

Moles Sr(NO3)2 = 2.888 M * 0.150 L

Moles Sr(NO3)2 = 0.4332 moles

Step 4: Calculate moles NaF

Moles NaF = 3.076 M * 0.200 L

Moles NaF = 0.6152 moles

It takes 2 moles F^- to precipitate 1 mole Sr^2+, so F^- is limiting.

Step 5: Calculate limiting reactant

For 1 mol of Sr(NO3)2 we need 2 moles of NaF to produce 1 mol of SrF2 and 2 moles of NaNO3

NaF is the limiting reactant. It will completely be consumed (0.6152 moles).

Sr(NO3)2 is in excess. There will react 0.6152/2 = 0.3076 moles

Moles Sr^2+ precipitated by F^- = 0.3076

There will remain 0.4332 - 0.3076 = 0.1256 moles of Sr(NO3)2

Moles Sr^2+ no precipitated (left over) = 0.1256 moles

Step 6: Calculate moles SrF2  

For 1 mol of Sr(NO3)2 we need 2 moles of NaF to produce 1 mol of SrF2 and 2 moles of NaNO3

For 0.6152 moles NaF we have 0.6152/2 = 0.3076 moles of SrF2

Mass SrF2 produced:  0.3076 mol * 125.6 g/mol = 38.63 g SrF2 produced

Step 7: Calculate concentration of [Na+] and [NO3-]

Since both Na^+ and NO3^- are spectator ions, and the final volume is 150 ml + 200 ml = 350 ml (0.350 L), the concentrations of Na^+ and NO3^- can be calculated as follows:

[Na^+]:  (200 ml)(3.076 M) = (350 ml)(x M) and x = 1.758 M

[NO3^-]:  (150 ml)(2.888 M)(2) = (350 ml)(x M) = 1.238 M

Step 8: Calculate [Sr^2+] and [F^-]

[Sr^2+] = 0.1256 moles/0.350 L = 0.3589 M

To find [F^-], one needs the Ksp for SrF2.  There are several values listed in the literature. I am using a value of 2x10^-10.

SrF2(s) <==> Sr^2+(aq) + 2F^-(aq)

Ksp = [Sr^2+][F^-]²

2x10^-10 = (0.3589)(x)²

x² = 5.57*10^-10

x = [F^-] = 2.36*10^-5 M

4 0
3 years ago
How much heat must be removed from 456 g of water at 25.0°c to change it into ice at?
N76 [4]
This question can be simply solved by using heat formula,
    Q = mCΔT

Q = heat energy (J)
m = Mass (kg)
C = Specific heat capacity (J / kg K)
ΔT = Temperature change (K)

when water freezes, it produces ice at 0°C (273 K)
hence the temperature change is 25 K (298 K - 273 K)
C for water is 4186 J / kg K or 4.186 J / g K
By applying the equation,
 Q = 456 g x 4.186 J / g K x 25 K
     = 47720.4 J
     = 47.72 kJ

hence 47.72 kJ of heat energy should be removed.
5 0
3 years ago
lead can react with oxygen gas. If lead (IV) oxide is the product of the reaction, how would the reaction be classified
slamgirl [31]

Answer:

this reaction is an oxidation reaction

7 0
3 years ago
5 meters
Feliz [49]

Answer:

9

Explanation:

7 0
2 years ago
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