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2 years ago

ELEMENTS ARE CLASSIFIED AS METALS, NONMETALS, AND METALLOIDS. USE THE PERIODIC TABLE AND IDENTIFY TWO OF EACH CLASSIFICATION.

Chemistry

1 answer:

alekssr [168]2 years ago

3 0

Answer:

Elements are classified into- metals – Nonmetals- Metalloids – Noble gases. State which of A, B, C, D is a:

1) Metallic element

2) Non-metallic element

3) Metalloid

4) Noble gas.

A) Is non-malleable, non-ductile and a poor conductor of electricity

B) Has lustre, is malleable and ductile and a good conductor of electricity

C) Is unreactive and inert and present in traces in air

D) Shows properties of both metals and nonmetals

Explanation:

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Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m

Bond [772]

Answer:

$m_{H_2O}=4.86gH_2O$

Explanation:

Hello,

In this case, the described chemical reaction is:

$C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O$

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

$n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6$

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

$n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6$

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

$m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O$

Best regards.

3 0

3 years ago

Write balanced equations for the following reactions

Arte-miy333 [17]

Answer:

See explanation

Explanation:

The shorthand nuclear reaction equations have been given; the first particle in the parentheses is a reactant particle while the second particle is a product particle. These can now be rewritten as the longhand equations as follows;

238/92U + 4/2 He -------> 241/94Pu + 1/0 n

238/92U + 4/2 He ------> 241/94Pu + 1/0 n

14/7N + 4/2 He------> 17/8O + 1/1 p

56/26Fe + 2 4/2 He----> 60/29Cu + 4/2 He

4 0

3 years ago

What part of cellular respiration is affected by arsenic

LenaWriter [7]

Answer:

It effects all of the cellular respiration process

Explanation:

It inhibits the Glycolysis. It replaces the phosphate groups that is needed for making Pyruvate and ATP.

5 0

3 years ago

Word elements can be used to show chemical reactions. When hydrogen burns in oxygen. Water is produced. The word equation is hyd

frez [133]

Iron + sulphur - iron sulphide

8 0

3 years ago

Read 2 more answers

The road map in the text shows that to convert from one unit to another, you must use the ____ as an intermediate step. ___ conv

jonny [76]

Answer:

Conversion factor;

Molar mass;

Avogadro's constant and molar mass

Explanation:

Firstly, an intermediate step is to define the conversion factor that will be then used in a conversion technique called dimensional analysis in order to convert from one unit to another. An example of a conversion factor would be, for example, 1 L = 1000 mL, which can be manipulated as a fraction, either $\frac{1 L}{1000 mL}$ or $\frac{1000 mL}{1 L}$ ;
Secondly, in order to convert mass to moles, we need to know the molar mass of a compound which has a units of g/mol (that is, it shows how many grams we have per 1 mole of substance.
Thirdly, Avogadro's constant, $N_A = 6.022\cdot 10^{23} mol^{-1}$ tells us that there is $N_A$ number of molecules or atoms in 1 mole of substance. We need two conversion factors to convert the number of molecules to a mass: firstly, we need to convert the number of molecules into the number of moles using Avogadro's constant and then we need to use the molar mass to convert the moles obtained into mass.

5 0

3 years ago