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3 years ago

After concluding his research, which statements would Virchow agree with? Check all that apply.

Chemistry

2 answers:

djyliett [7]3 years ago

5 0

Answer:

Cells come from pre-existing cells.

Explanation:

The cell theory asserts that cells are the morphological and functional units of living things and that they originate only from another pre-existing cell.

Nowadays the idea that living beings are made up of cells, except for viruses, is widespread. This idea, however, arose only after several observations and later formulation of the so-called Cell Theory. After accepting that living things were made up of cells, it began to study how these structures emerged. According to Rudolph Virchow, one cell could only emerge from another pre-existing cell. A very famous phrase from this researcher is: "Omnis cellula ex cellula", which means "Every cell originates from another cell."

Rudolf Virchow was a German pathologist, archaeologist and anthropologist, founder of cell pathology. He was the first to demonstrate that cell theory applies to both diseased and healthy tissues. That is, that diseased cells derive from healthy cells of normal tissues.

wlad13 [49]3 years ago

3 0

Cells come from pre existing cells

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jekas [21]

The third question requires you to solve for the weight of sodium (Na) and weight of Chloride (Cl) from the calculated moles of each element Na, and Cl.

So, you need to multiply the calculated moles of Na with its molar mass (23 g/ mol) to get the answer for Na. And multiply the calculated moles of Cl with its molar mass (35.45 g/mol) to get the answer for Cl.

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3 years ago

Does a precipitate form when a solution of calcium chloride and a solution of mercury(I) nitrate are mixed together? Write the n

olganol [36]

Answer : Yes, a precipitate form when a solution of calcium chloride and a solution of mercury(I) nitrate are mixed together.

The net ionic equation will be,

$2Hg^{+}(aq)+2Cl^{-}(aq)\rightarrow Hg_2Cl_2(s)$

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The given balanced ionic equation will be,

$CaCl_2(aq)+2HgNO_3(aq)\rightarrow Ca(NO_3)_2(aq)+Hg_2Cl_2(s)$

The ionic equation in separated aqueous solution will be,

$Ca^{2+}(aq)+2Cl^{-}(aq)+2Hg^{+}(aq)+2NO_3^{-}(aq)\rightarrow Hg_2Cl_2(s)+Ca^{2+}(aq)+2NO_3^{-}(aq)$

In this equation, $Ca^{2+}\text{ and }NO_3^-$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$2Hg^{+}(aq)+2Cl^{-}(aq)\rightarrow Hg_2Cl_2(s)$

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3 years ago

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Sav [38]

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4 years ago

What must happen before a body cell can begin mitotic cell division

Musya8 [376]

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It occupies the second stage of cell cycle, the first being interphase ( which has three stages that is G1, S, and G2 phase). Replication of DNA takes place in the interphase ( particulary in the S phase) so that cell is ready to divide in the mitotic phase.

Thus, replication of DNA ( deoxyribonucleic acid) must happen before a body cell can begin mitotic cell division.

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3 years ago

When 125 mL of 0.150 M Pb(NO3)2 is mixed with 145 mL of 0.200 M KBr, 4.92 g of PbBr2 is collected. Calculate the percent yield.

Semenov [28]

Answer:

Y = 92.5 %

Explanation:

Hello there!

In this case, since the reaction between lead (II) nitrate and potassium bromide is:

$Pb(NO_3)_2+2KBr\rightarrow PbBr_2+2KNO_3$

Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:

$0.125L*0.150\frac{molPb(NO_3)_2}{L} *\frac{1molPbBr_2}{1molPb(NO_3)_2} =0.01875molPbBr_2\\\\0.145L*0.200\frac{molKBr}{L} *\frac{1molPbBr_2}{2molKBr} =0.0145molPbBr_2$

Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:

$0.0145molPbBr_2*\frac{367.01gPbBr_2}{1molPbBr_2} =5.32gPbBr_2$

And the resulting percent yield:

$Y=\frac{4.92g}{5.32g} *100\%\\\\Y=92.5\%$

Regards!

4 0

3 years ago