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olga nikolaevna [1]
3 years ago
13

Animal cells do not have

Chemistry
2 answers:
Dennis_Churaev [7]3 years ago
7 0

Answer:

chloroplasts and central vacuoles

Explanation:

slamgirl [31]3 years ago
7 0

Answer:

Animal Cells don't have cell walls.

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Enter ionic and net equations: feso4(aq)+ na3po4(aq) arrow fe3(po4)2(s)+na2so4(aq)
stepan [7]

Answer:

<em> ionic equation : </em>3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)

<em> net ionic equation: </em>3Fe(2+)(aq)  + 2PO4 (3-)(aq) → Fe3(PO4)2(s)

Explanation:

The balanced equation is

3FeSO4(aq)+ 2Na3PO4(aq) → Fe3(PO4)2(s)+ 3Na2SO4(aq)

<em>Ionic equations: </em>Start with a balanced molecular equation.  Break all soluble strong electrolytes (compounds with (aq) beside them) into their ions . Indicate the correct formula and charge of each ion. Indicate the correct number of each ion . Write (aq) after each ion .Bring down all compounds with (s), (l), or (g) unchanged. The coefficents are given by the number of moles in the original equation

3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)

<em>Net ionic equations: </em>Write the balanced molecular equation.  Write the balanced complete ionic equation.  Cross out the spectator ions, it means the repeated ions that are present.  Write the "leftovers" as the net ionic equation.

3Fe(2+)(aq)  + 2PO4 (3-)(aq) → Fe3(PO4)2(s)

6 0
3 years ago
Plsss helpppp<br> im confused<br> will mark BRAINLIEST
mr_godi [17]
I am thinking Li & CI :)
6 0
3 years ago
Read 2 more answers
In which or areas on the image dose salinity of the ocean increases
Ronch [10]

Answer and Explanation:

D I believe..

8 0
2 years ago
First-order reaction that results in the destruction of a pollutant has a rate constant of 0.l/day. (a) how many days will it ta
Klio2033 [76]

Rate equation for first order reaction is as follows:

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}

Here, k is rate constant of the reaction, t is time of the reaction, A_{0} is initial concentration and A_{t} is concentration at time t.

The rate constant of the reaction is 0.1 day^{-1}.

(a) Let the initial concentration be 100, If 90% of the chemical is destroyed, the chemical present at time t will be 100-90=10, on putting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{10}=23.03 days

Thus, time required to destroy 90% of the chemical is 23.03 days.

(b) Let the initial concentration be 100, If 99% of the chemical is destroyed, the chemical present at time t will be 100-99=1, on putting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{1}=46.06 days

Thus, time required to destroy 99% of the chemical is 46.06 days.

(c)  Let the initial concentration be 100, If 99.9% of the chemical is destroyed, the chemical present at time t will be 100-99.9=0.1, on putting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{0.1}=69.09 days

Thus, time required to destroy 99.9% of the chemical is 69.09 days.

5 0
2 years ago
What happens inside a can of spoiled food as gas is produced?
ddd [48]

Answer:D the pressure increases

Explanation:

6 0
3 years ago
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