All categories

3 years ago

Lysosomes are important for _____.

Chemistry

2 answers:

kirza4 [7]3 years ago

8 0

Digestion would be correct. Hope this helps

Ilia_Sergeevich [38]3 years ago

3 0

Answer:

Digestion

Explanation:

You might be interested in

Which substance is the reducing agent in this reaction? 2KMnO4+3Na2SO3+H2O→2MnO2+3Na2SO4+2KOH

LuckyWell [14K]

2KMnO4+3Na2SO3+H2O→2MnO2+3Na2SO4+2KOH
In a reaction, the reducing agent is the element or compound that donates electron or the one tht loses electrons. The oxidized species. The opposite is called the oxidizing agent. It is the one who accepts the electrons lost. For this reaction KMnO4 is reduced into MnO2.

5 0

3 years ago

Read 2 more answers

A sample of gas contains 0.1700 mol of OF2(g) and 0.1700 mol of H2O(g) and occupies a volume of 19.5 L. The following reaction t

andreev551 [17]

Answer: The volume of the sample after the reaction takes place is 29.25 L.

Explanation:

The given reaction equation is as follows.

$OF_{2}(g) + H_{2}O(g) \rightarrow O_{2}(g) + 2HF(g)$

So, moles of product formed are calculated as follows.

$\frac{3}{2} \times 0.17 mol \\= 0.255 mol$

Hence, the given data is as follows.

$n_{1}$ = 0.17 mol, $n_{2}$ = 0.255 mol

$V_{1}$ = 19.5 L, $V_{2} = ?$

As the temperature and pressure are constant. Hence, formula used to calculate the volume of sample after the reaction is as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}$

Substitute the values into above formula as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{19.5 L}{0.17 mol} = \frac{V_{2}}{0.255 mol}\\V_{2} = \frac{19.5 L \times 0.255 mol}{0.17 mol}\\= \frac{4.9725}{0.17} L\\= 29.25 L$

Thus, we can conclude that the volume of the sample after the reaction takes place is 29.25 L.

8 0

2 years ago

You have a red ballooe that has a volume of 20 liters, a pressure of 1.5 atmospheres and a temperature of 28 C. What is the volu

ki77a [65]

Explanation:

The given data is as follows.

$P_{1}$ = 1.5 atm, $V_{1}$ = 20 L, $T_{1}$ = (28 + 273) K = 301 K

$P_{2}$ = 5 atm, $V_{2}$ = ?, $T_{2}$ = (50 + 273) K = 323 K

Formula to calculate the volume will be as follows.

$\frac{P_{1} \times V_{1}}{T_{1}}$ = $\frac{P_{2} \times V_{2}}{T_{2}}$

Putting the given values into the above formula as follows.

$\frac{P_{1} \times V_{1}}{T_{1}}$ = $\frac{P_{2} \times V_{2}}{T_{2}}$

$\frac{1.5 atm \times 20 L}{301 K}$ = $\frac{5 atm \times V_{2}}{323 K}$

$V_{2}$ = 0.64 L

Thus, we can conclude that the change in volume of the balloon will be 0.64 L.

6 0

3 years ago

Which choice below would affect the rate of reaction in the opposite way from the other four?

bulgar [2K]

Adding a catalyst as this would speed up the reaction and the rest would slow it down

4 0

3 years ago

How many electrons will be found in the “p” orbitals in the ground state of chlorine atom?

emmasim [6.3K]

Chlorine, Cl , is located in period 3, group 17 of the periodic table, and has an atomic number equal to 17 . This tells you that a neutral chlorine atom will have a total of 17 electrons surrounding its nucleus. Now, notice that the first energy level doesn't not contain a p-subshell,

7 0

3 years ago