3. compounds in which elements are held together by covalent bonds...keeps their own properties

In what ways can the motion of an object change?

lozanna [386]

Answer:

A force can speed up or slow down an object. A force can change the direction in which an object is moving. A bigger force on an object will produce a bigger change in the motion. A heavier object requires a larger force than a lighter object in order to undergo the same change in motion.

Explanation:

Mark me brainlist, Please.

8 0

2 years ago

150 ml of 0.1 m naoh is added to 200 ml of 0.1 m formic acid, and water is added to give a final volume of 1 l. what is the ph o

N76 [4]

Number of moles of NaOH = V(NaOH) * M(NaOH)= 0.150 L * 0.1 moles/L = 0.015 moles

Number of moles of formic acid, HCOOH = V(HCOOH) * M(HCOOH) = 0.200 L * 0.1 moles/L = 0.020 moles

Here, the limiting reagent is NaOH

The reaction is represented as:

HCOOH + NaOH ↔HCOONa + H2O

Moles of HCOONa formed = Moles of the limiting reagent, NaOH = 0.015 moles

Moles of HCOOH remaining = 0.020-0.015 = 0.005 moles

Total final volume is given as 1 L

Therefore: [HCOOH] = 0.005 moles/1 L = 0.005 M

[HCOONa] = 0.015/1 = 0.015 M

pKa of HCOOH = 3.74

As per Henderson-Hasselbalch equation

pH = pka + log[HCOONa]/[HCOOH] = 3.74+log[0.015/0.005] = 4.22

Therefore, pH of the final solution = 4.22

3 0

3 years ago

Which type of reaction produces the most dangerous radioactive waste?

k0ka [10]

The correct answer to the question above is fussion of uranium. The fussion of uranium is the only type of reaction in which produces the most dangerous radioactive waste. The fusion of uranium means the binding of their atoms and produce a radioactive waste.

7 0

3 years ago

A 155 g sample of an unknown substance was heated from 25°c to 40°c. in the process, the substance absorbed 569 calories of ener

Tasya [4]

Energy= 2381 joules
heat= Mass(kg) *change in temperature(K) * Cp
2381=0.155*(15)*Cp
Cp=1024 J/kg K

5 0

3 years ago

Read 2 more answers

A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35 oC to 195 oC. Calculate the specific heat.

Alja [10]

Answer:

Question 1: 0.2J/(gºC)

Question 2: 6,000J

Question 3: 300J

Question 4: 80g

Question 5: 74ºC

Question 6: 50g

Explanation:

Question 1. A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35º oC to 195º C. Calculate the specific heat.

The thermal energy equation is:

Q = m × C × ΔT

Substitute and solve for C:

566J = 20g × C × (195ºC - 35ºC)

C = 566J / (20g × 160ºC)

C = 0.177 J/(gºC) ≈ 0.2J/(gºC)

You must round to one significant figure because one factor has one significant figure).

Qustion 2. 40g of water is heat at 40ºC and the temperature rise to 75ºC. What is the amount of heat needed for the temperature to rise? (specific heat of water is 4.184 J/gºC)

Use the thermal energy equation again:

Q = m × C × ΔT

Substitute and compute:

Q = 40g × 4.184 J/gºC × (75ºC - 40ºC)

Q = 5,857.6J

Round to one significant figure: 6,000J

Question 3. Graphite has a mass of 50g and a specific heat of 0.420 J/gºC. If graphite is cooled from 50ºC to 35ºC, how much energy was lost?

Q = m × C × ΔT

Q = 50g × 0.420J/gºC × (35ºC - 50ºC)

Q = 315J

Round to one significant figure (because 50g has one significant figure)

Q = 300J

Question 4. Iron has a specific heat of 0.712 J/gºC. A piece of iron absorbs 3000J of energy and undergoes a temperature change totaling 50ºC, What is the mass of iron?

Q = m × C × ΔT

Solve for m:

m = Q / (C × ΔT)

Substitute and compute:

m = 3,000J / (0.712J/gºC × 50ºC)

m = 84.26 g ≈ 80 g (rounded to one significant figure, because the factor 3,000J has one significant figure).

Question 5. If 400g of an unknown solution at 70ºC loses 7500 J of heat, what is the final temperature of the unknown solution. The unknown solution has a specific heat of 4.184 J/gºC.

Q = m × C × ΔT

Q is negative, since it is released.

Substitute and solve for T:

- 7,500J = 400g × 4.184J/gºC × (T - 70ºC)

T = - 7500J / 400g × 4.184J/gºC) + 70ºC

T = 74ºC

If you round to one significant figure you cannot tell the temperature difference, thus leave two significant figures.

Question 6. How many grams of water would require 9500J of heat to raise the temperature from 50ºC to 100ºC

Q = m × C × ΔT

Subsitute:

9,500J = m × 4.184J/gºC × (100ºC - 50ºC)

Solve for m and compute:

m = 9,500J / (4.184J/gºC × 50ºC)

m = 45g

Since the temperatures indicate one singificant figure, the mass should be rounded to one significant figure:

m = 50g.

8 0

3 years ago