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Paladinen [302]
3 years ago
7

Write the net ionic equation for the reaction that takes places between aqueous copper and nitrate.

Chemistry
1 answer:
PIT_PIT [208]3 years ago
7 0

Answer:

Cu+HNO3

Explanation:

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Determine the empirical formula for succinic acid that is composed of 40.60% carbon, 5.18% hydrogen, and 54.22% oxygen.
aleksklad [387]

Answer:

C₂H₂O₃

Explanation:

The empirical formula of a compound is derived bu finding the whole ratios of the constituent elements.

In succinic acid, the ratios of carbon to hydrogen to oxygen is calculated as follows:

<u>% mass</u>

Carbon- 40.60

Hydrogen - 5.18

Oxygen - 54.22

<u>RAM</u>

Carbon -12

Oxygen - 15.994

Hydrogen -1.008

<u>No of moles elements in the compound</u>

Carbon = 40.60/12=3.3833

Oxygen = 54.22/15.994= 3.39

Hydrogen= 5.18/1.008 = 5.1389

Mole ratios of the individual elements we divide by the smallest value of the number of moles.

Carbon: Hydrogen : Oxygen

3.3833/3.3833:3.39/3.3833:5.1389/3.3833

=1:1:1.5

We can multiply the value by 2 to get the whole number ratio.

=2:2:3

The empirical formula will be:

C₂H₂O₃

5 0
3 years ago
Does anyone know how to do this?
rjkz [21]

Answer:

I know

Explanation:

Sike i dont know. Sorry man, but i dont know what that means - are you on crack? do you need therapy because I am a great therapist. Hmu if you'd like.

3 0
3 years ago
Determine the [OH-], pH, and pOH of a solution with a [H+] of 0.090 M at 25 °C. [OH-] = pH = pOH = 0 POH =
ValentinkaMS [17]

Answer : The concentration of OH^- ion, pH and pOH of solution is, 1.12\times 10^{-13}M, 1.05 and 12.95 respectively.

Explanation : Given,

Concentration of H^+ ion = 0.090 M

pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.

The expression used for pH is:

pH=-\log [H^+]

First we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (0.090)

pH=1.05

The pH of the solution is, 1.05

Now we have to calculate the pOH.

pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-1.05=12.95

The pOH of the solution is, 12.95

Now we have to calculate the OH^- concentration.

pOH=-\log [OH^-]

12.95=-\log [OH^-]

[OH^-]=1.12\times 10^{-13}M

The OH^- concentration is, 1.12\times 10^{-13}M

5 0
3 years ago
Read 2 more answers
How many formula units make up 24.2 g of magnesium chloride (MgCl2)?<br><br> Help!!
NNADVOKAT [17]

Answer:

Approximately 1.53 \times 10^{23} formula units (0.254\; \rm mol).

Explanation:

Refer to a modern periodic table for the relative atomic mass of magnesium (\rm Mg) and chlorine (\rm Cl):

  • \rm Mg: 24.305.
  • \rm Cl: 35.45.

In other words, the mass of 1\; \rm mol of \rm Mg atoms would be (approximately) 24.305\; \rm g.

Likewise, the mass of 1\; \rm mol of \rm Cl atoms would be approximately 35.45\; \rm g.

One formula unit of the ionic compound \rm MgCl_{2} includes exactly as many atoms as there are in the given formula. The formula mass of a compound is the mass of 1\; \rm mol of the formula units of this compound.

The formula \rm MgCl_{2} includes one \rm Mg atom and two \rm Cl atoms.

Hence, every formula unit of \rm MgCl_{2} \! would include the same number of atoms: one \rm Mg\! atom and two \rm Cl\! atoms. There would be 1\; \rm mol of \rm Mg atoms and 2\; \rm mol of \rm Cl atoms in 1\; \rm mol\! of \rm MgCl_{2} formula units.

Thus, the mass of 1\; \rm mol\! of \rm MgCl_{2} formula units would be equal to the mass of 1\; \rm mol of \rm Mg atoms plus the mass of 2\; \rm mol of \rm Cl atoms. (The mass of 1\; \rm mol\!\! of each atom could be found from the relative atomic mass of each element.)

\begin{aligned}& M({\rm MgCl_{2}}) \\ =\; & 24.305\; {\rm g \cdot mol^{-1}} + 2\times {\rm 35.45 \; \rm g \cdot mol^{-1}} \\ =\; & 95.205\; \rm g \cdot mol^{-1}\end{aligned}.

In other words, the formula mass of \rm MgCl_{2} is 95.205\; \rm g \cdot mol^{-1}.

Therefore, the number of formula units in m = 24.2\; \rm g of \rm MgCl_{2} would be:

\begin{aligned}n &= \frac{m({\rm MgCl_{2}})}{M({\rm MgCl_{2}})} \\ &= \frac{24.2\; \rm g}{95.205\; \rm g\cdot mol^{-1}} \\ & \approx 0.254\; \rm mol\end{aligned}.

Multiple n by Avogadro's Number N_{A} \approx 6.022 \times 10^{23}\; \rm mol^{-1} to estimate the number of formula units in 0.254\; \rm mol:

\begin{aligned}N &= n \cdot N_{A} \\ &\approx 0.254\; \rm mol \times 6.022 \times 10^{23}\; \rm mol^{-1} \\ &\approx 1.53\times 10^{23}\end{aligned}.

6 0
3 years ago
Write three primary alcohol with the formula C4H8O​
pishuonlain [190]

Answer:

CH2 will be CH-CH2-CH2OH

CH3-CH will beCH-CH2OH

CH2 will be C(CH3)-CH2OH

3 0
3 years ago
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