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bulgar [2K]
3 years ago
9

Directions: Show all of your work for each of the following problems.

Chemistry
1 answer:
AysviL [449]3 years ago
8 0

Answer:

<h3>3.Ba=2</h3><h3>4.Ni=2...PO4=3</h3><h3>5.Al=3...CrO4=2</h3><h3>6.NH4=1...SO4=2</h3>
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3. Define isobar 75 points
steposvetlana [31]

Answer:

1.

meteorology A line drawn on a map or chart connecting places of equal or constant pressure.

2.

nuclear physics Either of two nuclides of different elements having the same mass number.

3.

thermodynamics A set of points or conditions at constant pressure.

5 0
3 years ago
The freezing point of benzene C6H6 is 5.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in benzene is DDT . H
Vlad1618 [11]

Answer: 0.028 grams

Explanation:

Depression in freezing point :

Formula used for lowering in freezing point is,

\Delta T_f=k_f\times m

or,

\Delta T_f=k_f\times \frac{\text{ Mass of solute in g}\times 1000}{\text {Molar mass of solute}\times \text{ Mass of solvent in g}}

where,

\Delta T_f = change in freezing point

k_f = freezing point constant  (for benzene} =5.12^0Ckg/mol

m = molality

Putting in the values we get:

0.400^0C=5.12\times \frac{\text{ Mass of solute in g}\times 1000}{354.5\times 209.0}

{\text{ Mass of solute in g}}=0.028g

0.028 grams of DDT (solute) must be dissolved in 209.0 grams of benzene to reduce the freezing point by 0.400°C.

4 0
3 years ago
What happens to the ionizing rate if the steps of ionizing polyprotic acid increases?
Tanzania [10]

Answer:

<h3>An acid that contains more than one ionizable proton is a polyprotic acid. The protons of these acids ionize in steps. The differences in the acid ionization constants for the successive ionizations of the protons in a polyprotic acid usually vary by roughly five orders of magnitude.</h3>
5 0
2 years ago
Read 2 more answers
The following information is to be used for the next 2 questions. In order to analyze for Mg and Ca, a 24-hour urine sample was
Ainat [17]

Answer:

Explanation:

From the given information:

The concentration of metal ions are:

[Ca^{2+}]= \dfrac{0.003474 \ M \times 20.49 \ mL}{10.0 \ mL}

[Ca^{2+}]=0.007118 \ M

[Mg^2+] = \dfrac{0.003474 \ M\times (26.23  - 20.49 )mL}{10.0 \ mL}

=0.001994 \ M

Mass of Ca²⁺ in 2.00 L urine sample is:

= 2.00 L \times 0.001994 \dfrac{mol}{L} \times \dfrac{40.08 \ g}{1 \ mol}

= 0.1598 g

Mass of Ca²⁺ = 159.0 mg

Mass of Mg²⁺ in 2.00 L urine sample is:

= 2.00 L \times 0.007118 \dfrac{mol}{L} \times \dfrac{24.31 \ g}{1 \ mol}

= 0.3461 g

Mass of Mg²⁺ = 346.1 mg

5 0
3 years ago
Can someone please help me!!!!
Ostrovityanka [42]

Answer:

b

Explanation:

science boy

4 0
3 years ago
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