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mamaluj [8]
2 years ago
9

Is the aqueous solution of each of these salts acidic, basic, or neutral? (a) cr(no3)3 acidic basic neutral (b) nahs acidic basi

c neutral (c) zn(ch3coo)2 acidic basic neutral?
Chemistry
2 answers:
melisa1 [442]2 years ago
5 0
1) chromium(III) nitrate is acidic,  because it is the salt of weak base (chromium(III) hydroxide Cr(OH)₃) and strong acid (nitric acid HNO₃).
2) sodium hydrosulfide is basic, because it is the salt of strong base (sodium hydroxide MaOH) and weak acid (hydrogen sulfide H₂S).
3) zinc acetate is little basic, because zinc hydroxide (Zn(OH)₂) is stronger base than acetic acid (CH₃COOH).

lesya [120]2 years ago
5 0

Answer:

(a) Neutral.

(b) Basic.

(c) Basic.

Explanation:

Hello,

In this case, we must take a look of both the cation's and anion's nature, it means, if they come from a strong or weak acid or base, thus:

(a) Chromium (III) hydroxide is a strong base and nitric acid is a strong acid, consequently, chromium (III) nitrate is a neutral base since every ion is neutralized.

(b) Sodium hydroxide is a strong base and the hydrosulfuric acid is a weak acid, so there will be unreacted OH- ions, consequently, sodium hydrosulfide is a basic salt.

(c) Zinc hydroxide is a strong base and acetic acid is a weak acid, therefore, as in the previous salt, this is a basic salt.

Best regards.

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A 52.8-g sample of glucose (a nondissociated, nonvolatile solute with the formula C6H12O6) is dissolved in 158.0 g of water. Wha
LUCKY_DIMON [66]
Calculate  the  mole  of  glucose  and  water
 The  moles  of water =158g/18g/mol=8.778 moles
moles of  glucose =52.8g/180g/mol=0.293 moles
determine  the mole  fraction  of  the  solvent
that  is   x solvent = 8.778/ (8.778+0.293)=0.9677
use  the  Raults  law  to  determine  the  vapor  pressure
100  degree  of  water  has   a  vapor  pressure  of  760 mmhg
p solution=(x  solvent) (p^o solvent)
p solution=0.9677 x760 =735.45 mmHg

7 0
2 years ago
ANSWER Soon!!
vichka [17]

Answer:

  • 1. 0.1683 mol
  • 2. 1.191 g
  • 3. 0.02695 mol
  • 4. Na₂Cl₃
  • 5. The empirical formula obtained is not correct. This is likely due to experimental errors, since much precision was required (the masses are determined in thousandths of grams).

Explanation:

<em><u></u></em>

<em><u>1. How many moles of elemental sodium were used in the reaction?</u></em>

Since all of the solid sodium is used up by the reaction, you can cancluate the number of moles of elemental sodium used dividing the mass by the molar mass:

  • number of moles = mass in grams / atomic mass

  • mass in grams = 0.3870 g (given)

  • atomic mass = 22.990 g/mol

  • number of moles = 0.3870 g / 22.990 g/mol = 0.1683 mol

<u><em>2. What is the mass of chlorine gas used in the reaction?</em></u>

a) Mass of chlorine gas introduced in the flask = mass of the stoppered flask after filling it with chlorine gas - mass of the empty flask with the sopper

  • Mass of chlorine gas introduced = 158.1743g - 156.1870g = 1.9873 g

b) Mass of chlorine gas unreacted = 0.7962 g (given)

c) Mass of chlorine gas used = mass of chorine gas introduced in the flask - mass of chlorine gass un reacted

  • Mass of chlorine gas used = 1.9873g - 0.7962g = 1.1911g

<u><em>3. How many moles of chlorine were used in the reaction?</em></u>

  • molar mass of chlorine gas, Cl₂ = 2 × 35.453 g/mol = 70.906 g/mol

  • number of moles = mass in grams / molar mass = 1.911g / 70.906g/mol = 0.02695 mol

<u><em>4. What is the empirical formula of sodium chloride based on the experimental data?</em></u>

Divide the number of moles of each element by the smalles number of moles:

  • Na: 0.01683 / 0.01683 = 1
  • Cl = 0.02695 / 0.01683 = 1.6

Multiply by 2 to obtain whole numbers:

  • Na = 2
  • Cl = 3.2 = 3

  • Empirical formula Na₂Cl₃

<u><em></em></u>

<u><em>5. Was the empirical formula you obtained correct using the chemists data correct? Why? </em></u>

<u><em></em></u>

No, the empirical formula you obtained using the chemists data is not correct, because the correct empirical formula of sodium chloride is NaCl.

That is, there is 1 atom of sodium per every atom of chlorine in one chemical formula of NaCl, but that is not reflected by the empirical formula Na₂Cl₃.

That is a demostration of big experimental errors. You can speculate that the errors are likely due to problems of procedure collecting the gas or errors in measuring the masses.

As you see, the masses are measured to thousandths of grams, which requires much precision; thus smalls absolute errors could produce huge relative errors.

7 0
2 years ago
A large flask is evacuated and weighed, filled with argon gas, and then reweighed. When reweighed, the flask is found to have ga
Elden [556K]

Answer:

100.52

Explanation:

from the ideal gas equation PV=nRT

for a given container filled with any ideal gas P and V remains constant.So T is also constant.R is as such a constant.

So n i.e no of moles will also be constant.

no of moles of Ar=3.224/40=0.0806

no of moles of unknown gas=0.0806

molecular wt of unknown gas=8.102/0.0806=100.52

8 0
2 years ago
A kettle of water is at 14.5°C. Its temperature is then raised to 50.0°C by supplying it with 5,680 joules of heat. The specific
Tamiku [17]

Answer:- 38.2 g.

Solution:- The equation used for solving this type of calorimetry problems is:

q=mc\Delta T

where, q is the heat energy, m is mass, c is specific heat and delta T is the change in temperature.

Water temperature is increasing from 14.5 degree C to 50.0 degree C.

\Delta T=50.0-14.5  = 35.5 degree C

q is given as 5680 J and specific heat value is 4.186\frac{J}{g.^0C} .

The equation could be rearranged for m as:

m=\frac{q}{c*\Delta T}

Let's plug in the values in it:

m=\frac{5680}{4.186*35.5}

m = 38.2 g

So, the mass of water in the kettle is 38.2 g.


4 0
2 years ago
Read 2 more answers
N a football game, two players tackle each other so hard that they both fly in opposite directions after they hit each other. �W
evablogger [386]

Answer:

Newton's Third Law

Explanation:

Newton's Third Law stipulates that for every action there is an equal and opposite reaction.

So when the two players are tackling they exert a force on each other.

If player 1 tackles (exerts a force) player 2,  player 2 will exert an equal and opposite reaction on player 1 as stated in Newton's Third Law.

Therefore when they tackle each other so hard they both experience reaction forces so powerful that they fly in opposite directions.

Thus this is an example of the Newton's Third Law.

4 0
2 years ago
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