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Feliz [49]
3 years ago
12

PLEASE HELP!!!! 60 points

Chemistry
2 answers:
Tom [10]3 years ago
8 0

Answer:

d

d.d.d.d.d.d..d.d.d.d.d.d.d.d..d.d.d.d.d.d.d

Explanation:

lina2011 [118]3 years ago
7 0

Answer:

Have a nice Day , I would appreciate it if you could mark my answer brainliest

if 1 teacher cant teach all subjects how come 1 student can learn all subjects ?

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Balance<br> Fe(s) + Ca(NO3)2(aq)-&gt;
Bess [88]

Fe(s)+Ca(NO₃)₂(aq)⇒no reaction

<h3>Further explanation</h3>

In voltaic series  

Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au  

The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent  

The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent  

So that the element located on the left can push the element on the right in the redox reaction  

Reaction

Fe(s)+Ca(NO₃)₂(aq)⇒no reaction

Fe cannot reduce Ca because Ca is more reactive, so the reaction does not occur

On the contrary, this reaction can occur

3Ca(s) + 2Fe(NO₃)₃(aq) = 3Ca(NO₃)₂(aq) + 2Fe(s)

6 0
3 years ago
The pH of a softdrink is determined to be 4.0. What is the [OH-] of the drink?
irakobra [83]

Answer: A) 1.0 x 10-10 M

Explanation: The pH is 4.0, and pOH + pH = 14, so pOH = 10. [OH-] = 10^-pOH, so for this drink, [OH-] = 10^-10 M, or 1.0 x 10^-10 M.

6 0
3 years ago
2H2 + O2 → 2H2O
Bogdan [553]
Mole ratio:

<span>2 H2 + O2 =  2 H2O
</span>
2 mol H2 ---------- 1 mol O2
1.67 mol H2 ------ ??

1.67 x 1 / 2 => 0.835 moles of O2


5 0
3 years ago
Read 2 more answers
In order to prepare 50.0 mL of 0.100 M NaOH you will add _____ mL of 1.00 M NaOH to _____ mL of water
FinnZ [79.3K]

The question requires us to complete the sentence regarding the preparation of a more dilute NaOH solution (0.100 M, 50.0 mL) from a more concentrated NaOH solution (1.00 M).

Analyzing the blank spaces that we need to fill in the sentence, we can see that we must provide the volume of the more concentrated solution and the volume of water necessary to prepare the solution.

We can use the following equation to calculate the volume of more concentrated solution required:

\begin{gathered} C_1\times V_1=C_2\times V_2 \\ V_1=\frac{C_2\times V_2}{C_1} \end{gathered}

where C1 is the concentration of the initial solution (C1 = 1.00 M), V1 is the volume required of the inital solution (that we'll calculate), C2 is the concentration of the final solution (C2 = 0.100 M) and V2 is the volume of the final solution (V2 = 50.0 mL).

Applying the values given by the question to the equation above, we'll have:

\begin{gathered} V_1=\frac{C_2\times V_2}{C_1} \\ V_1=\frac{0.100M_{}\times50.0mL_{}}{1.00M_{}}=5.00mL \end{gathered}

Thus, we would need 5.00 mL of the more concentrated solution.

Since the volume of the final solution is 50.0 mL and it corresponds to the volume of initial solution + volume of water, we can calculate the volume of water necessary as:

\begin{gathered} \text{final volume = volume of initial solution + volume of water} \\ 50.0mL=5.00mL\text{ + volume of water} \\ \text{volume of water = 45.0 mL} \end{gathered}

Thus, we would need 45.0 mL of water to prepare the solution.

Therefore, we can complete the sentence given as:

<em>"In order to prepare 50.0 mL of 0.100 M NaOH you will add </em>5.00 mL<em> of 1.00 M NaOH to </em>45.0 mL<em> of water"</em>

5 0
1 year ago
What is the h oh ph and poh of a 0.005m solution of calcium hydroxide?
fiasKO [112]

Answer: [OH^{-}]= 0.01M or 1.0\times 10^{-2}M

[H^{+}]= 1.0\times 10^{-12}M

pH = 12

pOH = 2

Explanation: Calcium hydroxide (Ca(OH)_{2}) is a strong base that dissociates completely.

Dissociation equation of Calcium hydroxide is :

Ca(OH)_{2} \rightarrow Ca^{+2} + 2OH^{-}

1. Concentration of [OH-]

1 mol Ca(OH)_{2} produces 2 mol OH- ions.

The given solution is 0.005M Ca(OH)_{2} , then  concentration of OH- would be twice the concentration of Ca(OH)_{2}

[OH^{-}] = 0.005\times 2 = 0.01M or 1.0\times 10^{-2}M

2.Concentration of [H+]

Concentration of [H+] can be calculated by the formula: [H^{+}] = \frac{Kw}{[OH^{-}]}

kw = ionic product of water and its values is (1\times 10^{-14})

[OH-] = 0.01 M or 1.0\times 10^{-2}M

[H^{+}] = \frac{(1\times 10^{-14})}{[OH^{-}]}

[H^{+}] = \frac{(1\times 10^{-14})}{[0.01]}

[H^{+}] = 1.0\times 10^{-12}M

3. pH value

pH is calculated by the formula : pH = -log[H^{+}]

pH = -log[1.0\times 10^{-12}]

pH = 12

4. pOH value

pOH is calculated by the formula : pOH = 14 - pH

pOH = 14 - 12

pOH = 2

pOH can also be calculated by using a different formula which is :

pOH = -log(OH^{-})

pOH = -log(0.01)

pOH = 2.


3 0
3 years ago
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