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Yuliya22 [10]
3 years ago
14

What is the matter of science​

Chemistry
1 answer:
weqwewe [10]3 years ago
3 0
The finagling in the hole
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If 10.0g of powdered iron is heated with 10.0g of sulfur in an open crucible, what is the mass of iron (II) sulfide that is form
Ugo [173]

Answer:

See Explanation

Explanation:

                     8Fe        +        S₈                =>        8FeS

Given:            10g                  10g

moles      10g/56g/mol     10g/256g/mol

                = 0.179mol Fe   = 0.039mol S₈

Reduce => divide mole values by respective coefficients; smaller value is Limiting Reactant.

                 0.179/8 = 0.022      0.039/1 = 0.039  

       => Fe is limiting reactant

                       8Fe        +               S₈                   =>         8FeS    

Given:     10g/56g/mol          10g/256g/mol

               = 0.179mol              = 0.039 mol                  0.179mol FeS produced

                                          1/8(0.179)mol S₈ used        (coefficients are equal,

                                          = 0.022 mol S₈ used       => moles Fe = moles FeS)

                                         = (0.039 - 0.022)mol S₈     = 0.179mol FeS

                                         remains in excess              =(0.179mol)(88g/mol)

                                         = 0.0166 mol S₈ (excess)  = 15.8 g FeS  

                                         = (0.0166mol)(256g/mol)           (Theoretical Yield)

                                         = 4.26g S₈ in excess

7 0
3 years ago
A 1.0 L buffer solution is 0.300 M HC2H3O2 and 0.045 M LiC2H3O2. Which of the following actions will destroy the buffer?
zheka24 [161]

Answer:

b) Adding 0.075 moles of HCl

Explanation:

A buffer is defined as the aqueous mixture of a weak acid and its conjugate base or vice versa (Weak base with its conjugate acid).

The buffer of the problem is the acetic acid / lithium acetate.

The addition of any moles of the acid and the conjugate base will not destroy the buffer, just would change the pH of the buffer. Thus, a and c will not destroy the buffer.

The addition of an acid (HCl) or a base (NaOH), produce the following reactions:

HCl + LiC₂H₃O₂ → HC₂H₃O₂ + LiCl

<em>The acid reacts with the conjugate base to produce the weak acid.</em>

<em />

And:

NaOH + HC₂H₃O₂  →NaC₂H₃O₂ + H₂O

<em>The base reacts with the weak acid to produce conjugate base.</em>

<em />

As the buffer is 1.0L, the moles of the species of the buffer are:

HC₂H₃O₂ = 0.300 moles

LiC₂H₃O₂ = 0.045 moles

The reaction of HCl with LiC₂H₃O₂ consume all LiC₂H₃O₂ -<em>because there are an excess of moles of HCl that react with all </em>LiC₂H₃O₂-

As you will have just HC₂H₃O₂ after the reaction, the addition of b destroy the buffer.

In the other way, 0.0500 moles of NaOH react with the HC₂H₃O₂ but not consuming all HC₂H₃O₂, thus d doesn't destroy the buffer.

5 0
3 years ago
Which is a property of a mineral<br><br> A. Size<br> B. Density <br> C. Odor<br> D. Age
ollegr [7]

B. density is the right answer

3 0
3 years ago
Which action is an example of an organism maintaining homeostatis
VLD [36.1K]

Answer:

\color{Blue}\huge\boxed{Answer}

<em>The body regulates those levels in an example of homeostasis. When levels decrease, the parathyroid releases hormones. If calcium levels become too high, the thyroid helps out by fixing calcium in the bones and lowering blood calcium levels. The nervous system helps keep homeostasis in breathing patterns.</em>

7 0
3 years ago
The half-life of gold-198 is 2.7 days. After
Pie

Answer: 8.1 days

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant = x

a - x = amount left after decay process= \frac{x}{4} 

a) to find rate constant

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{0.693}{k}

k=\frac{0.693}{2.7days}=0.257days^{-1}

b) for completion of one fourth of reaction

t=\frac{2.303}{k}\log\frac{x}{\frac{x}{4}}

t=\frac{2.303}{0.257}\log{4}

t=8.1days

Thus after 8.1 days , one fourth of original amount will remain.

8 0
4 years ago
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