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3 years ago

In a neutralization reaction ____

Chemistry

1 answer:

Olenka [21]3 years ago

8 0

Answer:

neutralization reaction

Explanation: Because Neutralization Reactions. ... from a neutralization reaction

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An excess of mg(s) is added to 100.ml of 0.400 m hcl. at 0c and 1 atm pressure, what volume of h 2 (g) can be obtained?

ra1l [238]

The balanced equation for the reaction between Mg and HCl is as follows
Mg + 2HCl --> MgCl₂ + H₂
stoichiometry of HCl to H₂ is 2:1

number of HCl moles reacted - 0.400 mol/L x 0.100 L = 0.04 mol of HCl
since Mg is in excess HCl is the limiting reactant
number of H₂ moles formed - 0.04/2 = 0.02 mol of H₂

we can use ideal gas law equation to find the volume of H₂
PV = nRT
where
P - pressure - 1 atm x 101 325 Pa/atm = 101 325 Pa
V - volume
n - number of moles - 0.02 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 0 °C + 273 = 273 K
substituting these values in the equation

101 325 Pa x V = 0.02 mol x 8.314 Jmol⁻¹K⁻¹ x 273 K
V = 448 x 10⁻⁶ m³
V = 448 mL
therefore answer is
c. 448 mL

5 0

3 years ago

How many liters of salt solution will be needed to provide 15 grams of salt if the concentration of the solution is 50 g/l?

asambeis [7]

Answer:

.30 l

Explanation:

15g/50g/l= .3l

6 0

3 years ago

A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of

Stella [2.4K]

Answer:

See explanation.

Explanation:

Hello,

In this case, we can show how the empirical formula is found by following the shown below procedure:

1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:

$n_C=0.01962molCO_2*\frac{1molC}{1molCO_2} =0.01962molC$

2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:

$n_H=0.01961molH_2O*\frac{2molH}{1molH_2O}=0.03922molH$

3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:

$m_O=0.4647g-0.01962molC*\frac{12gC}{1molC}-0.03922molH*\frac{1gH}{1molH} =0.1900gO$

4. Compute the moles of oxygen by using its molar mass:

$n_O=0.1900gO*\frac{1molO}{16gO}=0.01188molO$

5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:

$C=\frac{0.01962}{0.01188}=1.65\\ \\H=\frac{0.03922}{0.01188} =3.3\\\\O=\frac{0.01188}{0.01188} =1$

6. Search for the closest whole number (in this case multiply by 2):

$C_3H_6O_2$

Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:

$M=12*3+1*6+16*2=74g/mol$

Which is about three times in the molecular formula, for that reason, the actual formula is:

$C_9H_{18}O_6$

It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.

Best regards.

6 0

3 years ago

Why should an acid be handled with caution

tiny-mole [99]

Explanation:

acid should be handled with caution because it is dangerous as it may burn the skin

7 0

2 years ago

What are two things required for something to be matter?

almond37 [142]

Atoms and particles