What type of learner am i

What is the density of an object that has a mass of 6.5 g and when placed in water displaces the volume from 4.5mL to 11.8mL? ro

KengaRu [80]

Answer:

$\rho =0.9g/mL$

Explanation:

Hello,

In this case, due to the volume displacement caused the by the object's submersion, it's volume is:

$V=11.8mL-4.5mL=7.3 mL$

In such a way, considering the mathematical definition of density, it turns out:

$\rho =\frac{m}{V}=\frac{6.5g}{7.3mL}\\ \\\rho =0.89g/mL$

Rounding to the nearest tenth we finally obtain:

$\rho =0.9g/mL$

Regards.

3 0

4 years ago

Fe(NO3)2 not sure how to get the oxidation numbers of all elements

Shtirlitz [24]

You must remember that oxidation number of hydrogen in acids is always +1, oxidation number of oxygen in oxides & acids is always -2... metals has always oxidation number on plus!

group NO3 comes from HNO3...and oxidation number of whole acid group is always on minus and equal to the amount of hydrogen atoms in this acid... so oxidation number of NO3 = -1

we have 2 NO3 groups so 2*(-1) = -2 and that is the reason why oxidation number of Fe in this formula must be +2... because sum of all elements always gives 0!

Now we could count of oxidation number for nitrogen... we write HNO3 and start counting from right to left:
3*(-2) from oxygens + 1 from hydrogen = -5
so nitrogen must have +5 oxidation number... because sum all in formula must be 0.

4 0

3 years ago

A 0.879 g sample of a CaCl2 ∙ 2 H2O / K2C2O4 ∙ H2O solid salt mixture is dissolved in 150 mL of deionized water. A precipitate f

Alecsey [184]

Answer:

a. CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)

b. Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted = 0.326 g

e. Moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted = 0.408 g

g. mass of K₂C₂O₄.H₂O remaining unreacted = 0.145 g

h. Percent by mass CaCl₂.2 H₂O = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 62.9%

Explanation:

a. Molecular equation of the reaction is given below :

CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)

b. The net ionic equation is given below

Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. mass CaC₂O₄ produced = 0.284 g, molar mass of CaC₂O₄ = 128 g/mol

moles CaC₂O₄ produced = 0.284 g / 128 g/mol = 0.00222 moles

Mole ratio of CaC₂O₄ and CaCl₂.2 H₂O is 1 : 1, therefore moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted in the mixture = number of moles × molar mass

Molar mass of CaCl₂.2 H₂O = 147 g/mol

Mass of CaCl₂.2 H₂O reacted = 0.00222 moles × 147 g/mol = 0.326 g

e. Mole ratio of K₂C₂O₄.2 H₂O and CaC₂O₄ is 1 : 1, therefore, moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted in the mixture = number of moles × molar mass

Molar mass of K₂C₂O₄.H₂O = 184 g/mol

grams K2C2O4-H2O reacted = 0.00222 moles 184 g/mole = 0.408 g

g. Mass of sample = 0.879 g

mass of CaCl₂.2 H₂O in sample completely used up = 0.326 g

mass of K₂C₂O₄.H₂O in sample = 0.879 g - 0.326 g = 0.553 g

mass of K₂C₂O₄.H₂O remaining unreacted = 0.553 g - 0.408 g = 0.145 g

h. Percent by mass CaCl₂.2 H₂O = 0.326 /0.879 x 100% = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 0.553/0.879 × 100% = 62.9%

6 0

3 years ago

1 Point

AleksandrR [38]

Answer:

Explanation:

A or B

I would say B do

4 0

3 years ago

Choose the molecule(s) that will only show two signals, with an integration ratio of 2:3, in their 1H NMR spectrum.

mrs_skeptik [129]

The question is missing the molecules in which the integration ratio of 2:3 will be observed. The complete question is given in the attachment.

Answer:

Molecule (a), (c), and (f) will show two peaks with the integration ratio of 2:3 in their 1H NMR spectrum

Explanation:

In the 1H NMR spectrum, the peak area is dependent on the number of hydrogen in a specific chemical environment. Hence, the ratio of the integration of these signals provides us with the relative number of hydrogen in two peaks. This rationale is used for the assignment of molecules that will give 2:3 integration ratio in the given problem.

Molecule (a) have two CH₂ and three CH₃ groups. Hence, it will give two peaks and their integration ratio becomes 2:3 (Answer)
Molecule (b) contains three chemical environments for its hydrogen atoms
Molecule (c) have a single CH₂ and CH₃ group giving integration ratio of 2:3 (Answer)
Molecule (d) will give two peaks but their ratio will be 1:3 because of two hydrogens of CH₂ and six hydrogens from two CH₃ groups
Molecule (e) have three CH and three CH₃ groups, so their ratio will become 1:3
Molecule (f) contains four CH and two CH₃ groups, giving two peaks. So, the integration ratio of their peaks is 2:3 (Answer)
Molecules
(g)
and
(h)
both have two CH and two CH₃ groups giving two peaks with the integration ratio of 1:3

3 0

4 years ago