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motikmotik
2 years ago
6

Help, please I don't understand what I have to do

Chemistry
1 answer:
8_murik_8 [283]2 years ago
4 0

Answer:

Is there a picture To go with this?

Explanation:

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Im hvaing a hard time getting the right answer
Tanya [424]

Answer:

V=23.9mL

Explanation:

Hello!

In this case for the solution you are given, we first use the mass to compute the moles of CuNO3:

n=2.49g*\frac{1mol}{125.55 g}=0.0198mol

Next, knowing that the molarity has units of moles over liters, we can solve for volume as follows:

M=\frac{n}{V}\\\\V=\frac{n}{M}

By plugging in the moles and molarity, we obtain:

V=\frac{0.0198mol}{0.830mol/L}=0.0239L

Which in mL is:

V=0.0239L*\frac{1000mL}{1L}\\\\V=23.9mL

Best regards!

6 0
3 years ago
I need chemistry help! How would I set up these problems?
Bond [772]

Answer: -

1) 8.33 minutes

2) 118.39 in/ s

180.43 m/min

10.83 km/ hr

Explanation: -

Speed of light = 3 x 10⁸ m/s

Distance of the earth from the sun= 93 million miles

We know 1 million = 1,000,000

Also 1 mile = 1609 m

Distance of the earth from the sun= 93 million miles

= 93,000,000 miles.

= 1.5 x 10^{11} m

Time taken = \frac{Distance}{Speed}

= \frac{1.5 x [tex] 10^{11} m}{3 x 10⁸ m/s} [/tex]

= 500 s

= 500/ 60

= 8.33 minutes

2) Distance = 1 mile = 63360 inches

Time taken = 8.92 min

= 8.92 x 60

= 535.2 s

Speed = \frac{distance}{time}

= \frac{63360 inches}{535.2 s}

= 118.39 in/ s

Distance = 1 mile = 63360 inches = 63360 x 2.54 cm = 63360 x 2.54 x 10^{-2} m

Time taken = 8.92 min

Speed = \frac{distance}{time}

= \frac{63360 x 2.54 x [tex] 10^{-2} m}{8.92 min} [/tex]

= 180.43 m/ min

1 m = 10⁻³ Km

1 min = 1/60 hour

1 m /min = 10⁻³ km/ \frac{1}{60 hour}

= 60/1000

=0.06 km/hr

180.43 m / min = 180 x 0.06 km / hr

= 10.93 km / hr

4 0
3 years ago
Currents at the surface of the ocean are
lakkis [162]

Answer:

D. The mixing of warm and cold water

Explanation:

5 0
3 years ago
142Pm &gt; 142 Nd +<br> what is the answer to this equation
erastova [34]

Explanation:

nnbbnmkmknn bnnnbbtbbbbn' nn' t

4 0
2 years ago
Calculate the solubility of mn(oh)2 in grams per liter when buffered at ph=7.0. assume that buffer capacity is not exhausted
Vanyuwa [196]
When PH + POH = 14 
∴ POH = 14 -7 = 7

when POH = -㏒[OH-]

          7    = -㏒ [OH-]
∴[OH-] = 10^-7

by using ICE table:

           Mn(OH)2(s) ⇄  Mn2+ (aq)  + 2OH-(aq)
initial                              0                     10^-7
change                           +X                      +2X
Equ                                 X                  (10^-7 + 2X)

when Ksp = [Mn2+][OH-]^2

when Ksp of Mn(OH)2 = 4.6 x 10^-14

by substitution:

4.6 x 10^-14 = X*(10^-7+2X)^2  by solving this equation for X

∴ X =2.3 x 10-5 M

∴ The solubility of Mn(OH)2 in grams per liter (when the molar mass of Mn(OH)2 = 88.953 g/mol
= 2.3 x10^-5 moles/L * 88.953 g/mol

= 0.002 g/ L
3 0
3 years ago
Read 2 more answers
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