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balandron [24]
3 years ago
13

If you start with 10ml of 0.75 m cu(no3)2 how much cu (s) in grams should be recovered in step #7

Chemistry
1 answer:
LenaWriter [7]3 years ago
4 0
<span>0.48 grams. Not a well worded question since it's assuming I know the reactions. But I'll assume that since there's just 1 atom of copper per molecule of Cu(NO3)2, that the reaction will result in 1 atom of copper per molecule of Cu(NO3)2 used. With that in mind, we will have 0.010 l * 0.75 mol/l = 0.0075 moles of copper produced. To convert the amount in moles, multiply by the atomic weight of copper, which is 63.546 g/mol. So 0.0075 mol * 63.546 g/mol = 0.476595 g. Round the results to 2 significant figures, giving 0.48 grams.</span>
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Is C5H10 ionic or covalent?
brilliants [131]
Covalent because it is 5 and 10 so there even numbers I think
7 0
3 years ago
Read 2 more answers
A student has 2.99 g of Potassium Permaganate. How many mL of a .532 M solution can the student make?​
Anika [276]

Answer:

Explanation:

you must be know the formula of potassium permaganate is KMnO₄

now you have already in question 2 things ,

1. the molarity of solution = 0.532 M

2. the mass of KMnO₄ = 2.99 g

the molarity law is  M = moles of solute  / volume (L)

so use this equation to calculate the volume in liter and convert the result to milliliter.

( if the mass unit is gram (g) = volume must be litter (L) and if the mass unit is milligram = volume must be milliliter (ml) )

** first you must use this equation to calculate how many moles do you have in KMnO₄

number of moles = mass (g) / molecular weight "MW" (g/mole )  

now you have 2.99 g ( Mentioned in Question )

and you can calculate MW if you now the Chemical formula

so we have KMnO₄  

Mw for KMnO₄  = 39 + 55 + 4x16 = 158 g/mole

**( K=39 , Mn=55 , O= 16 )

so number of moles = 2.99 / 158 = 0.018 = 0.02 moles

and if you go back to the molarity law : 

Molarity = moles of solute  / volume (L)

so  0.532 = 0.02  / volume (L)

Volume (L) = 0.532 x 0.02 = 0.01064 L

convert to ml =  0.01064 x 1000 = 10.64 ml

Good Luck

4 0
3 years ago
Fe2o3 + 3C —&gt; 2Fe +3CO
Anika [276]

Answer:

1. iron(III) oxide

2. 8

3. 8

Explanation:

Know the naming rules

Reactants on left side products on right side

3 0
4 years ago
Use the Rydberg equation to calculate the wavelength (in nm) of the photon emitted when a hydrogen atom undergoes a transition f
natka813 [3]

Answer:

The answer is 434nm

Explanation:

The Rydberg equation is an empirical relationship expressed by Balmer and Rydberg which is stated as:

1/λ =      R_{H} (\frac{1}{n_{f} ^{2} }-\frac{1}{n_{i} ^{2} } ).............................. (1)

where λ is the wavelength, R_{H} is the Rydberg constant equal to 1.097 x 10^{7}m^{-1}, n is the transition level number, the subscript f and i are the final and initial levels respectively. Therefore for final transition n = 2 and for initial transition n = 5.

Making substitutions into equation (1), gives

1/λ =  1.097 x 10^{7} (\frac{1}{2^{2} }-\frac{1}{5^{2} })

   = 1.097 x 10^{7} (\frac{1}{4} - \frac{1}{25} )

     = 1.097 x 10^{7} (0.25 - 0.04)

   = 1.097 x 10^{7} x 0.21

   = 2303700

∴ λ = \frac{1}{2303700} = 4.34 x 10^{-7}m

Converting to nm, we have

  λ   = \frac{4.34 x 10^{-7} }{10^{-9} } = 434nm

Therefore, the wavelength of the emitted photon is 434nm

6 0
4 years ago
A 151.6-g sample of a metal at 75.3°C is added to 151.6 g at 15.6°C. The temperature of the water rises to 18.9°C. Calculate the
denis-greek [22]

Answer:

4.60 J/°C g

Explanation:

This a simple calorimetry excersise. If all the heat, which is lost by the metal, is gained by the water, we assume that

Q from water = Q from metal

Q = m . C . ΔT

where C is heat capacity and ΔT, the differences between the temperatures. Let's determine the heat gained by water.

Q = 151.6 g . 4.18 J /°C g . (18.9°C - 15.6°C)

Q (+) = 2091 Joules

As this heat, was gained by the water, this heat was lost by the metal (-)

- 2091 Joules  = 151.6 g  . C  . (Final T° - 75.3°C)

We do not know at what T° was the meta, by the end, but all the heat was gained from the water, as water was increased by 3°C, metal decreases -3°C

- 2091 Joules  = 151.6 g  . C  . - 3°C

-2091 J / -3°C . 151.6g = 4.60 J/°C g

5 0
3 years ago
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