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Lilit [14]
3 years ago
11

A 0.784 g sample of magnesium is added to a 250 ml flask and dissolved in 150 ml of water. magnesium hydroxide obtained from the

reaction required 215.0 ml of 0.300 m hydrochloric acid to completely react. how many moles of hcl were used?
Chemistry
1 answer:
V125BC [204]3 years ago
4 0

A lot of values are given but we must focus only on HCl.

<span>There are 215 mL (0.215 L) of 0.300 M HCl that was completely   used. We must remember the moles is simply the product of volume and molarity, that is:</span>

 

number of moles = 0.215 L * 0.300 M

<span>number of moles= 0.0645 moles HCl</span>

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The image shows the representation of an unknown element in the periodic table. A square is shown. Inside the square twelve is w
Ludmilka [50]

<u>Answer:</u> The correct answer is the mass number of the most common isotope of the element is 24.

<u>Explanation:</u>

We are given:

An element having atomic number 12 is magnesium and atomic mass of the element is 24.305

The image corresponding will be _{24.305}\textrm{Mg}^{12}\\\text{Magnesium}

The number '24.305' is the average atomic mass of magnesium element.

Average atomic mass is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i

Average atomic mass of magnesium = 24.305 amu

As, the average atomic mass of magnesium lies closer to the mass of Mg-24 isotope. This means that the relative abundance of this isotope is the highest of all the other isotopes.

The 'Mg-24' isotope is the most common isotope of the given element.

Hence, the correct answer is the mass number of the most common isotope of the element is 24.

4 0
3 years ago
Balance the equation and state the limiting reagent in the following reaction:
romanna [79]
Hgffvvcbchcgivigcigxgsgixigdfhcgcc fcfuc
8 0
3 years ago
Plz answer ASAP
strojnjashka [21]

Answer:

this should help *not a virus

Explanation:

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6 0
2 years ago
Consider the chemical equation. CuCl2 + 2NaNO3 mc023-1.jpg Cu(NO3)2 + 2NaCl What is the percent yield of NaCl if 31.0 g of CuCl2
lana [24]
CuCl2 + 2NaNO3 ---->  Cu(NO3)2 + 2NaCl

using molar masses:-
Theoretical yields:-
63.54 + 2(35.45) g  of CuCl2  produces  2(22.98 + 35.45) g of NaCl
    134.44  g .................................................... 116.86 g
       31.0 g ....................................................31.0 * 116.86 /134.44=26.95g  
 
 So percentage yield is 21.2* 100 / 26.95    =  78.7%  to nearest tenth                                                                                                              


5 0
3 years ago
Read 2 more answers
1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.
Andrej [43]

Answer:

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

Mg+2HCl\rightarrow MgCl_2+H_2

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

n_{H_2}^{by\  HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl}  =0.0284molH_2\\\\n_{H_2}^{by\  Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg}  =0.690gMg

Thus, the mass of excess magnesium turns out:

m_{Mg}^{excess}=2.38g-0.690g=1.69gMg

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%

Best regards!

8 0
3 years ago
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