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alisha [4.7K]
2 years ago
12

Help meeeeeeee??????????​

Chemistry
1 answer:
kolbaska11 [484]2 years ago
5 0

Answer:

njjii

Explanation:

uujhhhjhhhthnjkk

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What volume of 1.00 m hcl in liters is needed to react completely (with nothing left over) with 0.750 l of 0.100 m na2co3?
kotykmax [81]
The balanced equation for the reaction is as follows
Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O
stoichiometry of Na₂CO₃ to HCl is 1:2
number of Na₂CO₃ moles reacted = molarity x volume
number of Na₂CO₃ moles = 0.100 mol/L x 0.750 L = 0.0750 mol 
according to molar ratio of 1:2
1 mol of Na₂CO₃ reacts with 2 mol of HCl
then 0.0750 mol of Na₂CO₃ mol reacts with - 2 x 0.0750 = 0.150 mol 
molarity of given HCl solution is 1.00 mol/L
molarity is defined as the number of moles of solute in 1 L of solution 
there are 1.00 mol in 1 L of solution 
therefore there are 0.150 mol in - 0.150 mol / 1.00 mol/L = 0.150 L 
volume of HCl required is 0.150 L 
3 0
3 years ago
Which of the following is NOT an allotrope of carbon? A. graphite B. carbon-13 C. a diamond D. a fullerene
egoroff_w [7]
B. carbon-13 is not an allotrope of Carbon.
Allotropes<span> are elements on the periodic table that have more than one crystalline form. </span>Isotopes<span> are atoms of the same element with the same atomic number but have a different mass number.
C-13 is an isotope of carbon, not an allotrope.</span>
5 0
3 years ago
How many of the zeros in the measurement 0.050060 are significant
Feliz [49]
I think 3 of them are its been 1 half years since ive done this i dont take chemistry anymore
6 0
3 years ago
Read 2 more answers
If the concentration of the stock (provided) Cu(NH3)42 was 0.041 M, what concentration will the Cu2 be in beaker?
kodGreya [7K]

Answer:

[Cu^{2+}]=0.041 M

Explanation:

Hello!

In this case, since the molarity of a solution is defined in terms of the moles of the solute and the volume of solution, given that the concentration of Cu(NH₃)₄²⁺ is 0.041 M, and there is only one copper atom per Cu(NH₃)₄²⁺ ion, we can compute the concentration of Cu²⁺ as shown below:

[Cu^{2+}]=0.041\frac{molCu(NH_3)_4^{2+}}{L}*\frac{1molCu^{2+}}{1molCu(NH_3)_4^{2+}} =0.041 \frac{molCu(NH_3)_4^{2+}}{L}

[Cu^{2+}]=0.041 M

Best regards!

6 0
2 years ago
Phuong trinh phan ung ohc cho +02
scZoUnD [109]

Answer:

yes

hope this helps!!!!

5 0
2 years ago
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