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3 years ago

What is another term for anode?

Chemistry

1 answer:

Rashid [163]3 years ago

7 0

Answer:positive electrode

Explanation:

Anode can also be referred to as positive electrode in a cell

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Find the density of an object that has a mass of 12.69 grams and a volume of 3.5cm3

lutik1710 [3]

Answer:

3.62 g/cm³

Explanation:

density = mass ÷ volume

Therefore, do 12.69 divided by 3.5

6 0

2 years ago

A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling

aniked [119]

Answer: The mass of glucose in final solution is 0.420 grams

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL$

Putting values in above equation, we get:

$0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M$

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g$

Hence, the mass of glucose in final solution is 0.420 grams

3 0

3 years ago

How much silver chloride is precipitated when 25.0 g of calcium chloride and 25.0 g of silver nitrate are mixed together in a be

Burka [1]

Answer:

21.09 g of AgCl

Explanation:

Hopefully the picture is clear and the method is understandable.

For more information go to

https://socratic.org/questions/5631d10b11ef6b4609a78ee2

7 0

2 years ago

51.7ml at 27 Celsius and 90kpa to stp

never [62]

STP is the abbreviation of standard condition for temperature and pressure which is 273.15K temperature and 1.013× 10^5 Pa pressure. Since the pressure and temperature changes, I assume the question would ask about the result of the volume. The temperature used in ideal gas should be Kelvin, so 27 Celcius would be 300.15K.
The calculation would be
PV=T
V=T/P

V2/V1= T2*P1/T1*P2
V2/V1=273.15K* 90^10^3Pa/ 300.15K * 1.013× 10^5 Pa
V2= 0.81904 * 51.7ml
V2= 42.34ml

6 0

3 years ago

Which characteristic of minerals is represented in this picture?

Marianna [84]

B is the correct answer to your question

7 0

3 years ago

Read 2 more answers