How many grams of solute are present in 1250 mL of a 1.34 M NaNO3 solution?

Advocard [28]

Answer: Oxidation half: $2Ag\rightarrow 2Ag^++2e^-$

Reduction half: $2H^++2e^-\rightarrow H_2$

Explanation:

Oxidation-reduction reaction or redox reaction is defined as the reaction in which oxidation and reduction reactions occur simultaneously.

$2Ag(s)+H_2S(s)\rightarrow Ag_2S(s)+H_2(g)$

Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.

Oxidation half: $2Ag\rightarrow 2Ag^++2e^-$

Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.

Reduction half: $2H^++2e^-\rightarrow H_2$

8 0

2 years ago

How much does 200 gallons of water weigh?

Jlenok [28]

To be able to obtain the mass of the given volume of a substance, we need data on its density since it is the mass per volume of a substance. We multiply this value to the given volume. For water, it is approximately 10.02 lb/gal.

Mass of water = 200 gal ( 10.02 lb/gal ) = 200.4 lb

Hope this helped.

3 0

3 years ago

What is microorganisms

Wewaii [24]

Answer:

Technically a microorganism or microbe is an organism that is microscopic. The study of microorganisms is called microbiology. Microorganisms can be bacteria, fungi, archaea or protists. The term microorganisms does not include viruses and prions, which are generally classified as non-living.

4 0

2 years ago

PLEASE HELP! List the 5 basic steps of the scientific method in order

Kruka [31]

Step 1: Ask a question or identify a problem. ...Step 2: Background research. ...Step 3: Form a hypothesis. ...Step 4: Experiment and observe. ...Step 5: Draw a conclusion.

3 0

3 years ago

A 60.0 g block of iron that has an initial temperature of 250. °C and 60.0 g bloc of gold that has an initial temperature of 45.

Maslowich

Answer:

The final temperature at the equilibrium is 204.6 °C

Explanation:

Step 1: Data given

Mass of iron = 60.0 grams

Initial temperature = 250 °C

Mass of gold = 60.0 grams

Initial temperature of gold = 45.0 °C

The specific heat capacity of iron = 0.449 J/g•°C

The specific heat capacity of gold = 0.128 J/g•°C.

Step 2: Calculate the final temperature at the equilibrium

Heat lost = Heat gained

Qlost = -Qgained

Qiron = -Qgold

Q=m*c*ΔT

m(iron) * c(iron) *ΔT(iron) = -m(gold) * c(gold) *ΔT(gold)

⇒with m(iron) = the mass of iron = 60.0 grams

⇒with c(iron) = the specific heat of iron = 0.449 J/g°C

⇒with ΔT(iron)= the change of temperature of iron = T2 - T1 = T2 - 250.0°C

⇒with m(gold) = the mass of gold= 60.0 grams

⇒with c(gold) = the specific heat of gold = 0.128 J/g°C

⇒with ΔT(gold) = the change of temperature of gold = T2 - 45.0 °C

60.0 *0.449 * (T2 - 250.0) = -60.0 * 0.128 * (T2 - 45.0 )

26.94 * (T2 - 250.0) = -7.68 * (T2 - 45.0)

26.94T2 - 6735 = -7.68T2 + 345.6

34.62T2 = 7080.6

T2 = 204.5 °C

The final temperature at the equilibrium is 204.6 °C

5 0

3 years ago