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sweet [91]
3 years ago
10

Is it always possible to identify something as an element compound pure substance or mixture

Chemistry
1 answer:
weqwewe [10]3 years ago
7 0
Is it always possible to identify something as an element<span>, </span>compound<span>, </span>pure substance or mixture<span> just by looking at it? </span>
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Spindle fibers break down during what phase?
Rasek [7]
Answer: Prophase.
In prophase, chromosomes condense and become visible. Spindle fibers emerge from the centrosomes.
7 0
3 years ago
Read 2 more answers
A bathtub is filled with 100 L of water. A cat jumps in and the volume rises to 125 L. If the cat has a mass of
kirza4 [7]

Answer:

<h2>The answer is 3.2 g/L</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass of cat = 400 g

volume = final volume of water - initial volume of water

volume = 125 - 100 = 25 L

The density of the cat is

density =  \frac{400}{125}  =  \frac{16}{5}  \\

We have the final answer as

<h3>3.2 g/L</h3>

Hope this helps you

3 0
3 years ago
What is mg is it length, area, volume, mass, density, time or tenperature
Akimi4 [234]
Mg stands for Milligram
8 0
2 years ago
Does sodium satisfy the ostet rule?
Charra [1.4K]

Answer:

The rule is especially applicable to carbon, nitrogen, oxygen, and the halogens, but also to metals such as sodium or magnesium. ... All four of these electrons are counted in both the carbon octet and the oxygen octet, so that both atoms are considered to obey the octet rule.

8 0
3 years ago
In a lab experiment 80.0 g of ammonia [NH3] and 120 g of oxygen are placed in a reaction vessel. At the end of the reaction 72.2
valentinak56 [21]

The percent yield of the reaction : 89.14%

<h3>Further explanation</h3>

Reaction of Ammonia and Oxygen in a lab :

<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>

mass NH₃ = 80 g

mol NH₃ (MW=17 g/mol):

\dfrac{80}{17}=4.706

mass O₂ = 120 g

mol O₂(MW=32 g/mol) :

\tt \dfrac{120}{32}=3.75

Mol ratio of reactants(to find limiting reatants) :

\tt \dfrac{4.706}{4}\div \dfrac{3.75}{5}=1.1765\div 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)

mol of H₂O based on O₂ as limiting reactants :

mol H₂O :

\tt \dfrac{6}{5}\times 3.75=4.5

mass H₂O :

4.5 x 18 g/mol = 81 g

The percent yield :

\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{72.2}{81}\times 100\%=89.14\%

6 0
3 years ago
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