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navik [9.2K]
3 years ago
7

List the physical properties for the element Kr

Chemistry
1 answer:
Tanzania [10]3 years ago
3 0
Uranium fission is a property
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Balance the following chemical reaction:
mrs_skeptik [129]

Answer:

what the freak is thisnenejehey

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Consider the following reaction: 3Fe(s) + 4H₂O(g) ➞ Fe₃O₄(s) + 4H₂(g). To answer the following question: "How many moles of hot
UkoKoshka [18]
X moles of hot water vapor(H2O) * 1 mol Fe3O4 / 4 mol H2O * 233.55g Fe3O4 / 1 mol = 275 g

Solve the equation

x = 4.71 mol of hot water vapor (steam)

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2 years ago
The elements in Groups 1A(1) and 7A(17) are all quite reactive. What is a major difference between them?
matrenka [14]

The elements in Groups 1A(1) and 7A(17) are all quite reactive.

<h3>Major difference between Groups 1A(1) and 7A(17) : </h3>

Group 7's halogens, which are non-metal elements, become less reactive as you move down the group. In contrast to the alkali metals in Group 1 of the periodic table, this trend is the opposite. The most reactive element in Group 7 is fluorine.

Alkali metals are soft and reactive metals. They react vigorously with water and become more reactive. And other hand halogens are reactive non metals.

  • Elements of group 1A are known as alkali metals. Elements of this group are lithium, sodium, potassium, rubidium, cesium.
  • Reactivity increase down group 1 but decrease up group 7 this is because group 7 elements react by gaining an electron. As one move down the group, the amount of electron shielding increases, meaning that the electron is less attracted to the nucleus.

To know more about Groups 1A(1) and 7A(17) please click here :

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6 0
1 year ago
What percentage of oxygen is attributed fossil fuel combustion
Oxana [17]

Answer:

C. 4%

Explanation:

The answer is 4%

6 0
2 years ago
2Al + 6HCl --&gt; 2AlCl3 + 3H2 Aluminium reacts with hydrochloric acid. How many grams of aluminum are necessary to produce 11 L
dem82 [27]

Answer:

8.8g of Al are necessaries

Explanation:

Based on the reaction, 2 moles of Al are required to produce 3 moles of hydrogen gas.

To solve this question we must find the moles of H2 in 11L at STP using PV = nRT. With these moles we can find the moles of Al required and its mass as follows:

<em>Moles H2:</em>

PV = nRT; PV/RT = n

<em>Where P is pressure = 1atm at STP; V is volume = 11L; R is gas constant = 0.082atmL/molK and T is absolute temperature = 273.15K at STP</em>

Replacing:

1atm*11L/0.082atmL/molK*273.15K = n

n = 0.491 moles of H2 must be produced

<em />

<em>Moles Al:</em>

0.491 moles of H2 * (2mol Al / 3mol H2) = 0.327moles of Al are required

<em />

<em>Mass Al -Molar mass: 26.98g/mol-:</em>

0.327moles of Al * (26.98g / mol) = 8.8g of Al are necessaries

7 0
3 years ago
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