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3 years ago

Please help me i will mark brainliest

Chemistry

1 answer:

gulaghasi [49]3 years ago

4 0

Answer:

7) Example: Gravitational force

Gravitational force, also known as gravity, is a type of non- contact force.

10) Unbalanced forces can cause an object to accelerate, make a stationary object move, or cause a moving object to stop.

12) Gravitational force, Frictional force, Magnetic force

13) Friction is beneficial when we are walking as it prevents us from slipping.

For the table, find the net force first and that would help us to find the direction and the type of force (balanced/ unbalanced).

As written in Q11, add all the forces together to determine the net force acting on an object. More specifically, add the forces that act in one direction together. Then subtract the forces acting in the other direction.

Let's look at the teddy bear (Q14 -16):

Take note that we have to look at the direction of the arrow not which side the arrow is on (left/ right of the bear).

Total force acting towards the left

= 16 +7

= 23N

Total force acting on the right= 22N

Net force

= 23 -22

= 1N towards the left

Thus, 14) Unbalanced (since the net force is not 0N)

15) 1N

15) 1N16) left

Vase question:

Total downward force

= 11 +33

= 44N

Total upward force

= 25 +19

= 44N

Net force= 0N

17) Balanced

17) Balanced18) 0N

17) Balanced18) 0N19) No direction

Bicycle question:

Total force acting towards the right

= 65 +115

= 180N

Total force to the left= 170N

Net force

= 180 -170

= 10N to the right

20) Unbalanced

20) Unbalanced21) 10N

20) Unbalanced21) 10N22) To the right

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A 1.28-kg sample of water at 10.0 °C is in a calorimeter. You drop a piece of steel with a mass of 0.385 kg at 215 °C into it. A

Kryger [21]

Answer:

$T_{2}=16,97^{\circ}C$

Explanation:

The specific heats of water and steel are

$Cp_{w}=4.186 \frac{KJ}{Kg^{\circ}C}$

$Cp_{s}=0.49 \frac{KJ}{Kg^{\circ}C}$

Assuming that the water and steel are into an adiabatic calorimeter (there's no heat transferred to the enviroment), the temperature of both is identical when the system gets to the equilibrium $T_{2}_{w}= T_{2}_{s}$

An energy balance can be written as

$m_{w}\times Cp_{w}\times (T_{2}- T_{1})_{w}= -m_{s}\times Cp_{s}\times (T_{2}- T_{1})_{s}$

Replacing

$1.28Kg\times 4.186\frac{KJ}{Kg^{\circ}C}\times (T_{2}-10^{\circ}C)= -0.385Kg\times 0.49 \frac{KJ}{Kg^{\circ}C} \times (T_{2}-215^{\circ}C)$

Then, the temperature $T_{2}=16,97^{\circ}C$

8 0

3 years ago

100 Points and Brainliest

Vedmedyk [2.9K]

Answer:

100 g

Explanation:

We know the concentration is 200 g/L.

Convert 500mL to L: 500mL/1000mL = 0.5 L

Now, multiply this by 200 to cancel out the L units and get grams:

200 g/L * 0.5 L = 100 g

Thus, the answer is 100 grams.

Hope this helps!

5 0

3 years ago

When a 6.00 g sample of KBr is dissolved in water in a calorimeter that has a total heat capacity of 2.71 kJ ⋅ K − 1 , the tempe

Umnica [9.8K]

Answer:

Molar heat enthalpy of KBr = -19.89 kJ/mol

Explanation:

Change in temperature (Δt) = 0.370 K

Heat capacity = 2.71 kJ ⋅ K^-1

Heat absorbed by calorimeter = heat capacity × change in temperature

= 2.71× 0.370

= 1.0027 kJ

Molar mass of KBr = 119 g/mol

No. of moles of KBr = 6.00/119

= 0.0504 mol

Heat absorbed by the calorimeter is given by KBr.

Now calculate the heat released by per mol of KBr as follows:

Heat released by per mol of KBr = 1.0027 kJ / 0.0504 mol

=19.89 kJ/mol

Heat is released therefore, sign will be negative.

Molar heat enthalpy of KBr = -19.89 kJ/mol

4 0

3 years ago

This is due at 4:00 P.M. and I do not want to make a 76%, so please help MEEEEEEEEEEEEE!!!

Ann [662]

Answer:

Hi, There!...

<h2>Answer</h2>

My Answer would be Option B.

Explanation:

nuclear energy protects air quality by producing massive amounts of carbon-free electricity.

Thermal energy from nuclear reactors may also be used to decarbonize other energy-intensive sectors such as transportation – the largest contributor to carbon pollution.

8 0

3 years ago

Read 2 more answers

The equation represents the decomposition of a generic diatomic element in its standard state. 12X2(g)⟶X(g) Assume that the stan

aliina [53]

Answer:

$K^{2000K}=0.774\\\\K^{3000K}=12.56$

Explanation:

Hello,

In this case, considering the reaction, we can compute the Gibbs free energy of reaction at each temperature, taking into account that the Gibbs free energy for the diatomic element is 0 kJ/mol:

$\Delta _rG=\Delta _fG_{X}-\frac{1}{2} \Delta _fG_{X_2}=\Delta _fG_{X}$

Thus, at 2000 K:

$\Delta _rG=\Delta _fG_{X}^{2000K}=4.25kJ/mol$

And at 3000 K:

$\Delta _rG=\Delta _fG_{X}^{3000K}=-63.12kJ/mol$

Next, since the relationship between the equilibrium constant and the Gibbs free energy of reaction is:

$K=exp(-\frac{\Delta _rG}{RT} )$

Thus, at each temperature we obtain:

$K^{2000K}=exp(-\frac{4250J/mol}{8.314\frac{J}{mol\times K}*2000K} )=0.774\\\\K^{3000K}=exp(-\frac{-63120J/mol}{8.314\frac{J}{mol\times K}*3000K} )=12.56$

In such a way, we can also conclude that at 2000 K reaction is unfavorable (K<1) and at 3000 K reaction is favorable (K>1).

Best regards.

4 0

3 years ago