All categories

3 years ago

Dependent variable? ( what did we measore)?

Chemistry

1 answer:

lana66690 [7]3 years ago

6 0

The thing u change.

You might be interested in

Why is frying an egg an example of a chemical change

skelet666 [1.2K]

A Chemical change is any form of chemical reaction changing the state of the matter involved. When you fry an egg, you are making a reaction between the heat and the chemicals that make up the egg. Therefore once you have fried that egg, you have completed a chemical reaction. This is why it's a chemical change when you fry an egg.

8 0

3 years ago

Read 2 more answers

Fluorine + sodium chloride → sodium fluoride + __________. What is the missing chemical name?

Deffense [45]

Answer:

Chlorine (Cl)

Explanation:

The equation would be:

$F+NaCl$ → $NaF + Cl$

This is because no atom is lost in a reaction. Chlorine would be the missing chemical as it is on the other side of the equation and needs to be accounted for on the products side.

7 0

3 years ago

Select odd one out.<br> I) O<br> II) Na<br> III) Cu<br> IV) Ag

USPshnik [31]

Explanation:

0 is the odd one...............

and rest are metals and they lose electron and 0 gain electron and is non metal

4 0

3 years ago

Read 2 more answers

Identify the f-block element.

GaryK [48]

Answer: E. none of these

Explanation: because the f-block element is n-2

3 0

3 years ago

After balancing the following reaction under acidic conditions, how many mole equivalents of water are required and on which sid

nordsb [41]

Answer:

d. 8 moles of H2O on the product side

Explanation:

Hello,

In this case, we need to balance the given redox reaction in acidic media as shown below:

$MnO_4^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O\\\\2Cl^{1-}\rightarrow Cl_2^0+2e^-\\\\2*[(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O]\\\\5*[2Cl^{1-}\rightarrow Cl_2^0+2e^-]\\\\\\\\2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10e^- \rightarrow 2Mn^{2+}+8H_2O\\\\10Cl^{1-}\rightarrow 5Cl_2^0+10e^-\\$

Then, we add the half reactions:

$2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10Cl^{1-} \rightarrow 2Mn^{2+}+8H_2O+5Cl_2^0$

Thereby, we can see d. 8 moles of H2O on the product side.

Best regards.

4 0

3 years ago