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3 years ago

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mage result for alz... Warm-Up Exercises for Chapter 2 - Acid-Base Reactions roblem 2.3 H2O is the acid and NH2 is the base. Exp

ress your answers as a chemical expression. = AXO O ? H2O + NH4= Submit Previous Answers Request Answer X Incorrect; Try Again; 19 attempts remaining

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laila [671]3 years ago

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Type this question into apex learning! Gives you answer and step by step explanation

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What is the name of this unstable isotope from number 9?

raketka [301]

Is there a picture of the isotope or?- becaue I can’t help if I don’t have a visual.

6 0

3 years ago

Read 2 more answers

Sulfuric acid dissolves aluminum metal according to the following reaction:

Fiesta28 [93]

Answer:

$m_{H_2SO_4}=81.7gH_2SO_4$

$m_{H_2}=1.67gH_2$

Explanation:

Hello,

Based on the given undergoing chemical reaction is is rewritten below:

$2Al (s) + 3H_2SO_4 (aq)\rightarrow Al _2(SO4)_3 (aq) + 3H_2 (g)$

By stoichiometry we find the minimum mass of H2SO4 (in g) as shown below:

$m_{H_2SO_4}=15.0gAl*\frac{1molAl}{27gAl}*\frac{3molH_2SO_4}{2molAl}*\frac{98gH_2SO_4}{1molH_2SO_4} \\m_{H_2SO_4}=81.7gH_2SO_4$

Moreover, mass of H2 gas (in g) would be produced by the complete reaction of the aluminum block turns out:

$m_{H_2}=15.0gAl*\frac{1molAl}{27gAl}*\frac{3molH_2}{2molAl}*\frac{2gH_2}{1molH_2} \\m_{H_2}=1.67gH_2$

Best regards.

3 0

3 years ago

Hunter drew a representation of the atoms of a gas. The illustration shows atoms moving freely at high speeds.

insens350 [35]

hunter did good wih the atoms of a gas

8 0

3 years ago

A mixture contains NaHCO3 together with unreactive components. A 1.75 g sample of the mixture reacts with HA to produce 0.561 g

Lynna [10]

Answer:

$\%NaHCO_3=61.2\%$

Explanation:

Hello.

In this case, since the undergoing chemical reaction is only between the sodium bicarbonate and the acid HA:

$NaHCO_3+HA\rightarrow NaA+H_2O+CO_2$

For 0.561 g of yielded carbon dioxide (molar mass 44 g/mol), the following mass of sodium bicarbonate (molar mass 84 g/mol) that reacted was:

$m_{NaHCO_3}=0.561gCO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molNaHCO_3}{1molCO_2} *\frac{84gNaHCO_3}{1molNaHCO_3} \\\\m_{NaHCO_3}=1.071g$

Considering the 1:1 mole ratio between CO2 and NaHCO3. Finally, the percent by mass of NaHCO3 is computed by dividing the mass of reacted NaHCO3 and t the mixture:

$\%NaHCO_3=\frac{1.071g}{1.75g}*100\%\\ \\\%NaHCO_3=61.2\%$

Best regards.

5 0

3 years ago

you wish to make a 0.161 m hydroiodic acid solution from a stock solution of 3.00 m hydroiodic acid. how much concentrated acid

mestny [16]

5.367 ml of the concentrated acid must be added to obtain a total volume of 100 ml of the dilute solution.

Dilution is defined as the process in which the concentration of a sample is decreased by adding more solvent. The dilution formula is given below.

C₁V₁ = C₂V₂

where C₁ = initial concentration of sample = 3.00 m

V₁ = initial volume of sample

C₂ = final concentration after dilution = 0.161 m

V₂ = total final volume after dilution = 100 ml

Plug in the values to the formula and solve for the volume of the concentrated acid that must be added.

C₁V₁ = C₂V₂

3.00 m (V₁) = 0.161 m (100 ml)

V₁ = 5.367 ml

Learn more about dilution here: brainly.com/question/1615979

#SPJ4

3 0

1 year ago