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Nimfa-mama [501]
3 years ago
7

Classify each of these reactions. A single reaction may fit more than one classification. Ba ( ClO 3 ) 2 ⟶ BaCl 2 + 3 O 2 Ba(ClO

3)2⟶BaCl2+3O2 double‑displacement (a.k.a. exchange or metathesis) combination (synthesis) displacement decomposition redox NaNO 2 + HCl ⟶ NaCl + HNO 2 NaNO2+HCl⟶NaCl+HNO2 decomposition displacement combination (synthesis) redox double‑displacement (a.k.a. exchange or metathesis) CaO + CO 2 ⟶ CaCO 3 CaO+CO2⟶CaCO3 double‑displacement (a.k.a. exchange or metathesis) displacement decomposition redox combination (synthesis) ZnSO 4 + Mg ⟶ Zn + MgSO 4 ZnSO4+Mg⟶Zn+MgSO4 double‑displacement (a.k.a. exchange or metathesis) displacement decomposition combination (synthesis) redox
Chemistry
1 answer:
Vilka [71]3 years ago
4 0

Answer: a) Ba(ClO_3)_2\rightarrow BaCl_2+3O_2:  Decomposition

b) NaNO_2+HCl\rightarrow NaCl+HNO_2: double displacement

c) CaO+CO_2\rightarrow CaCO_3: Synthesis (Combination)

d) ZnSO_4+Mg\rightarrow MgSO_4+Zn: redox

Explanation:

Decomposition is a type of chemical reaction in which one reactant gives two or more than two products.

Ba(ClO_3)_2\rightarrow BaCl_2+3O_2

A double displacement reaction is one in which exchange of ions take place.

NaNO_2+HCl\rightarrow NaCl+HNO_2

Synthesis reaction is a chemical reaction in which two reactants are combining to form one product.

CaO+CO_2\rightarrow CaCO_3

Redox reaction is a type of chemical reaction in which oxidation and reduction takes place in one single reaction. The oxidation number of one element increases and the oxidation number of other element decreases.

ZnSO_4+Mg\rightarrow MgSO_4+Zn

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1. Which of the following statements is a consequence of the equation E=MC2
finlep [7]

Answer:

D. All of the above​

Explanation:

E = MC² is a common equation in physics.

E is energy

M is mass

C is the speed of light

The law was stated by Albert Einstein.

  • From this law, it was shown that energy is released when matter is destroyed.
  • Mass and energy are equivalent as seen in nuclear reactions where mass is converted to energy.
  • Mass and energy is usually conserved in any process and this is a subtle modification of the law of conservation of matter and energy.
  • Most of these postulates apply to nuclear reactions which generally do not follow some precepts of chemical laws.
7 0
3 years ago
Which stage of the scientific process enables a scientist to check the work of other
blsea [12.9K]

Answer:

The answer to this would be communicating.

Explanation:

A scientist would be communicating to his or her fellow colleagues and sharing to them his or her idea.

Hope you find this answer helpful! :)

3 0
3 years ago
Which material is the least likely to be recognized as a mixture by looking at it under a microscope
PSYCHO15rus [73]
A homogenous mixture is uniform and thus hard to recognize as a mixture. An example is water!
5 0
3 years ago
Read 2 more answers
When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H
Debora [2.8K]

Answer:

Y=58.15\%

Explanation:

Hello,

For the given chemical reaction:

Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol  Al_2S_3

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3

Finally, we compute the percent yield with the obtained 2.10 g:

Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

Best regards.

7 0
3 years ago
Which of the following are correct for zero-order reactions?
GREYUIT [131]

Answer:

The answer is "Choice A and Choice B"

Explanation:

The Zero-Order reactions are usually found if a substrate, like a surface or even a catalyst, is penetrated also by reactants. Its success rate doesn't depend mostly on the amounts of the various reaction in this reaction.

Let the Rate = k

As \frac{dx}{dt} \ rate\ \  K_0 doesn't depend on reaction rate, a higher reaction rate does not intensify the reaction.

By the rate k_0 =\frac{dx}{dt}, the created based and the reaction rate is about the same.

5 0
3 years ago
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