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Oduvanchick [21]
2 years ago
11

What do lysosomes and Golgi bodies have in common?

Chemistry
2 answers:
ser-zykov [4K]2 years ago
6 0
The answer is B ribosomes
Drupady [299]2 years ago
5 0
The first question would be B) they are examples of cell organelles. The second question would be A) mitochondria. Hope this helps!
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What is the solution to the problem expressed to the correct number of significant figures
quester [9]

I don't know what the problem is, but here are some rues to help you out:

  1. All non-zero figures are significant
  2. When a zero falls between non-zero digits, that zero is significant.
  3. When a zero falls after a decimal point, that zero is significant.
  4. When multiplying and dividing significant figures, the answer is limited to the number of sig figs equal to the least number of sig figs in the problem.
  5. When adding and subtracting, the answer is limited to the number of decimal places in the number with the least number of decimal places.
4 0
3 years ago
If a reaction taking place in a flask shows a drop in temperature, which type of reaction is it? Options: exothermic reactions e
liubo4ka [24]

Answer:

exothermic reaction

Explanation:

If there is a drop in temperature, then energy was lost to the surroundings because temperature is the average measure of kinetic energy. An exothermic reaction would result in this lost of energy. An endothermic reaction would absorb energy and make the temperature rise.

4 0
3 years ago
Read 2 more answers
What is true when a reaction has reached equilibrium?
goldfiish [28.3K]

Answer:

In a chemical equilibrium, the forward and reverse reactions occur at equal rates, and the concentrations of products and reactants remain constant. A catalyst speeds up the rate of a chemical reaction, but has no effect upon the equilibrium position for that reaction.

Explanation:

7 0
3 years ago
Read 2 more answers
5. given the ratio of reactants, what is the possibility of obtaining di- and polychlorinated product? explain.
Vladimir79 [104]

The ratio of reactants is chlorination of <u>2,3</u> dimethyl butane the possibility of obtaining do and the polychlorinated product is not seen.

When a mixture of methane and chlorine is exposed to ultraviolet light a substitution reaction occurs and the organic product is chloromethane. Because there are various hydrogen atoms that can be extracted in the first propagation step.

Abstraction of a hydrogen atom from the middle carbon of propane results in 2-chloropropane. In the presence of sunlight, methane reacts with chlorine to form chloromethane. The chlorination of methane is a free radical substitution reaction. Chlorine cannot turn into free radicals in the dark, so no reaction takes place. Therefore, the presence of sunlight is essential for the reaction to proceed.

Learn more about The reactants here:- brainly.com/question/6421464

#SPJ4

3 0
1 year ago
If 15.6 grams of copper (ii) chloride react with 20.2 grams of sodium nitrate how many grams of sodium chloride can be formed? W
olasank [31]

Answer:

- 13.56 g of sodium chloride are theoretically yielded.

- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.

- 0.50 g of sodium nitrate remain when the reaction stops.

- 92.9 % is the percent yield.

Explanation:

Hello!

In this case, according to the question, it is possible to set up the following chemical reaction:

CuCl_2+2NaNO_3\rightarrow 2NaCl+Cu(NO_3)_2

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

m_{NaCl}^{by\ CuCl_2}=15.6gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2} *\frac{2molNaCl}{1molCuCl_2} *\frac{58.44gNaCl}{1molNaCl} =13.56gNaCl\\\\m_{NaCl}^{by\ NaNO_3}=20.2gNaNO_3*\frac{1molNaNO_3}{84.99gNaNO_3} *\frac{2molNaCl}{2molNaNO_3} *\frac{58.44gNaCl}{1molNaCl} =13.89gNaCl

Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

m_{NaNO_3}^{by\ NaCl}=13.56gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{2molNaNO_3}{2molNaCl} *\frac{84.99gNaNO_3}{1molNaNO_3}=19.72gNaNO_3

Therefore, the leftover of sodium nitrate is:

m_{NaNO_3}^{leftover}=20.2g-19.7g=0.5gNaNO_3

Finally, the percent yield is computed via:

Y=\frac{12.6g}{13.56g} *100\%\\\\Y=92.9\%

Best regards!

6 0
2 years ago
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