In your OWN words...HOW is Hydropower produced?

How do we know which elements<br> are more reactive?

Natasha2012 [34]

For metals, reactivity increases down a group and from right to left across a period. Non metals, reactivitt increases up a grouo and from left to right across a period. Francium is the most reactive metal and fluorine is the most reactive non-metal.

5 0

3 years ago

As the mass of an object increases, what happens to the gravitational force created by that object?

alexandr1967 [171]

Answer:

the gravitational force is proportional to the mass of both interacting objects, bigger objects will attract each other with a greater gravitational force. So the mass of either object increases, the force of gravitational attraction between them also increases.

4 0

3 years ago

Read 2 more answers

_____ NaPO4+_____KOH

Artist 52 [7]

NaPO4 + KOH -> KPO4 + NaOH
already balance

6 0

3 years ago

An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in jou

nevsk [136]

Question:

Part D) An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in joules per kelvin.

Part E) An object at 500 K dissipates 25.0 kJ of heat into the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.

Part F) An object at 400 K absorbs 25.0 kJ of heat from the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.

Part G) Two objects form a closed system. One object, which is at 400 K, absorbs 25.0 kJ of heat from the other object,which is at 500 K. What is the net change in entropy ΔSsys of the system? Assume that the temperatures of the objects do not change appreciably in the process. Express your answer numerically in joules per kelvin.

Answer:

D) 85 J/K

E) - 50 J/K

F) 62.5 J/K

G) 12.5 J/K

Explanation:

Let's make use of the entropy equation: ΔS = $\frac{Q}{T}$

Part D)

Given:

T = 20°C = 20 +273 = 293K

Q = 25.0 kJ

Entropy change will be:

ΔS = $\frac{25*1000}{293}$

= 85 J/K

Part E)

Given:

T = 500K

Q = -25.0 kJ

Entropy change will be:

ΔS = $\frac{-25*1000}{500}$

= - 50 J/K

Part F)

Given:

T = 400K

Q = 25.0 kJ

Entropy change will be:

ΔS = $\frac{25*1000}{400}$

= 62.5 J/K

Part G:

Given:

T1 = 400K

T2 = 500K

Q = 25.0 kJ

The net entropy change will be:

ΔS = $(\frac{25*1000}{400}) + (\frac{-25*1000}{500}$

= 12.5 J/K

7 0

3 years ago

Five ?L of a 10-to-1 dilution of a sample were added to 5mL of Bradford reagent. The absorbance at 595 nm was 0.78 and,according

Tomtit [17]

Answer:

0.03g/mL

Explanation:

Given parameters include:

Five μL of a 10-to-1 dilution of a sample; This implies the Volume of dilute sample is given as 5 μL

Dilution factor = 10-to-1

The absorbance at 595 nm was 0.78

Mass of the diluted sample = 0.015 mg

We need to first determine the concentration of the diluted sample which is required in calculating the protein concentration of the original solution.

So, to determine the concentration of the diluted sample, we have:

concentration of diluted sample = $\frac{mass}{volume}$

= $\frac{0.015 mg}{ 5 \alpha L}$ (where ∝ was use in place of μ in the expressed fraction)

= 0.003 mg/μL

The dilution of the sample is from 10-to-1 indicating that the original concentration is ten times higher; as such the protein concentration of the original solution can be calculated as:

protein concentration of the original solution = 10 × concentration of the diluted sample.

= 10 × 0.003 mg/μL

= 0.03 mg/μL

$= \frac{0.03*10^{-3g}}{10^{-3}mL}$

= 0.03g/mL

Hence, the protein concentration of the original solution is known to be 0.03g/mL

8 0

3 years ago