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Elodia [21]
3 years ago
8

GUYS PLEASE HELP IM CRYNG

Chemistry
1 answer:
erica [24]3 years ago
5 0

Answer:

A.) 8.796 B.) 234780 C.) 25.8

Explanation:

thats all i can read the rest is blurry

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43. Use Table 3-1 to identify a substance that undergoes a
LuckyWell [14K]

Answer:

Oxygen, and from solid to liquid. This is because oxygen's melting point is at -218°C. Melting point refers to the temperature where heat causes particles to vibrate with sufficient energy to break the solid structure, so for oxygen this means it's being turned into a liquid.

4 0
3 years ago
How to make Sodium thiosulfate
Arada [10]

Explanation:

Method of prepration of sodium thiosulphate - definition

In the laboratory, this salt can be prepared by heating an aqueous solution of sodium sulphite with sulphur or by boiling aqueous NaOH and sulfur according to this equation:

6NaOH + 4 S _{2} Na _{2} S ->Na _{2} S_{2}O_{3}+ 3H _{2}O.

3 0
3 years ago
0.105 g of an unknown diprotic acid is titrated with 0.120 M NaOH. The first equivalence point occurred at a volume of 7.00 mL o
adell [148]

Answer:

2,53x10⁻³ moles of NaOH

Explanation:

The reactions of a diprotic acid with NaOH are:

H₂X + NaOH → HX⁻ + H₂O + Na⁺

HX⁻ + NaOH → X²⁻ + H₂O + Na⁺

Where the complete first reaction gives the first equivalence point and the complete second reaction gives the second equivalence point.

The total volume spent of NaOH to reach the second equivalence point is:

7,00mL + 14,07mL = 21,07 mL = <em>0,02107L</em>

As molar concentration of NaOH is 0,120M, the moles used to reach the second equivalence point are:

0,02107L×(0,120mol/L) = <em>2,53x10⁻³ moles of NaOH</em>

I hope it helps!

6 0
3 years ago
What volume in L of a 0.375 M solution of NaOH is needed to provide 15.3 g NaOH?
anyanavicka [17]

Answer:

V= 10.2L

Explanation:

NB , n= m/M= CV

Molar mass of NaOH= 40gmol-1

15.3/40= 0.0375×V

Simplify

V= 10.2L

3 0
4 years ago
2. What ions are present in what ratio in a solution of aqueous calcium chloride?
Alenkasestr [34]

Answer:

\mathrm{Ca}^{2+} \text { and } \mathrm{Cl} \text { - ions are present in } 1: 2 \text { ratio in a solution of aqueous calcium chloride. }

Explanation:

Here in Calcium Chloride ionic bond is present in between calcium and chlorine atoms. As we know according to Octet rule calcium have two excess atoms and for matching nearest noble gas electronic configuration. It donate two electrons to gain more stability and form \mathrm{Ca}^{2+}, while chlorine is deficient from one electron to meet nearest noble gas electronic configuration therefore two chlorine atoms accept excess electron from calcium individually and form two\mathrm{Cl}^{-} ions.

\text { Equation is as follows: } \mathrm{Ca}^{2+}+2 \mathrm{Cl}^{-} \rightarrow \mathrm{CaCl}_{2}

Hence aqueous solution of calcium chloride breaks the ionic bond pairing in one \mathrm{Ca}^{2+}and two\mathrm{Cl}^{-}ions: \mathrm{CaCl}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O} \quad \mathrm{Ca}^{2+}(\mathrm{ag})+2 \mathrm{Cl}(\mathrm{ag})

5 0
4 years ago
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