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iren2701 [21]
3 years ago
8

A block is pulled 0.90 m to the right in 4.5 s. What is the block's average speed to the nearest tenth of m/s?

Chemistry
2 answers:
pshichka [43]3 years ago
8 0

Answer:

-

Explanation:

jeka57 [31]3 years ago
5 0

Answer:

The average speed of the blocks are 0.36 m/s.

Explanation:

Average speed is defined as the ratio of distance covered per unit time. So if it is said that blocks are pulled to 0.9 m in the right side. This means the blocks cover a distance of 0.9 m from the origin and that distance is covered in 2.5 s. Thus, the average speed can be calculated from the change in speed with respect to time. As at time t = 0 , the speed is also zero, and at time t = 2.5 s , the speed will be.

Since, in this case, the speed is equal to the average speed of blocks. So the average speed of the blocks will be 0.36 m/s.

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Value of Δ H ∘ rxn for the equation NH 3 ( g ) + 2 O 2 ( g ) ⟶ HNO3 ( g ) + H 2 O ( g )
koban [17]

Answer: - 894.6 kJ/mol.

Explanation:

Hess law is states that the changes in enthalpies in a chemical reactions is independent of the pathway between the initial and final states.

∆H is the change in the sum of the internal energy of a system.

We are to find the Value of ΔH°(rxn) for the equation:

NH3 (g) + 2 O2 (g) ⟶ HNO3 (g) + H2O (g). ----------------------------------(**).

From the series of equations given;

==> 4NH3 (g) + 5O2 (g) -------->4 NO(g) + 6H2O (l). ∆H = -1166.0 kJ/mol.--------------------------------------(1).

===> 2NO(g) + O2 (g) ------> 2NO2 (g). ∆H = -116.2 kJ/mol.---------------(2).

===> 3NO2 (g) + H2O (l) ---------> 2HNO3 (aq) + NO (g). ∆H = -137.3 kJ/mol.-------------------------------------(3).

The first thing to do is to multiply equation (2) by 3. Also, multiply equation (3) by 2. This will give us equation (4) and (5) respectively.

6NO + 3 O2 ----------------> 6NO2. ∆H= 3 × (-116.2 kJ/mol) = -348.6 kJ/mol. ------------------------------------(4).

6NO2 + 2 H2O ----------------> 4HNO3 + 2 NO. ∆H= 2 × (-137.3kj/mol) = -274.6 kJ/mol ---------------------------(5).

Next, add equations (4) and (5) to give;

4NO +3O2 +2H2O -------------> 4HNO3. ∆H = -623.2 kJ/mol. -----(6).

Add this equation to the equation (1) from above, we have;

4NH3 + 8O2 --------------> 4HNO3 + 4H2O. ∆H= -1789.2 kJ/mol. --------(7).

Then, divide the equation (7) above by 2 to give us back the equation (**).

NH3 (g) + 2 O2 (g) ⟶ HNO3 (g) + H2O (g). ∆H= -894.6 kJ/mol.

Δ H^∘ (rxn)= - 894.6 kJ/mol.

5 0
2 years ago
Use the equation to determine what mass of FeS must react to form 326 g of FeCl2.
iogann1982 [59]

Answer:

We need 226 grams of FeS

Explanation:

Step 1: Data given

Mass of FeCl2 = 326 grams

Molar mass FeCl2 = 126.75 g/mol

Step 2: The balanced equation

FeS + 2 HCl → H2S + FeCl2

Step 3: Calculate moles FeCl2

Moles FeCl2 = 326 grams / 126.75 grams

Moles FeCl2 = 2.57 moles

Step 4: Calculate moles FeS needed

For 1 mol H2S and 1 mol FeCl2 produced, we need 1 mol FeS and 2 moles HCl

For 2.57 moles FeCl2 we need 2.57 moles FeS

Step 5: Calculate mass FeS

Mass FeS = 2.57 moles * 87.92 g/mol

Mass FeS = 226 grams FeS

We need 226 grams of FeS

4 0
3 years ago
Determine the final temperature of sample with a specific heat of 1.1 J/g°C and a mass of 385 g if it starts out at a temperatur
Assoli18 [71]

Answer:

T2 =21.52°C

Explanation:

Given data:

Specific heat capacity of sample = 1.1 J/g.°C

Mass of sample = 385 g

Initial temperature = 19.5°C

Heat absorbed = 885 J

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - initial temperature

885J = 385 g× 1.1 J/g.°C×(T2 - 19.5°C )

885 J = 423.5 J/°C× (T2 - 19.5°C )

885 J / 423.5 J/°C = (T2 - 19.5°C )

2.02°C = (T2 - 19.5°C )

T2 = 2.02°C + 19.5°C

T2 =21.52°C

8 0
3 years ago
Rolls of foil are 302 mm wide and 0.018 mm thick. (The density of foil is 2.7 g/cm3 .)
Kruka [31]

Answer:

The length of foil will be 8107.81 cm or 81.7081 m.

Explanation:

Given data:

Width of roll of foil = 302 mm

Height or thickness = 0.018 mm

Density of foil = 2.7 g/cm³

Mass of foil = 1.19 Kg

Length of foil = ?

Solution:

d = m/ v

v = length (l) × width (w) × height (h)

First of we will convert the Kg into gram and mm into cm.

one Kg = 1000 g

1.19 × 1000 = 1190 g

one cm = 10 mm

302 / 10 = 30.2 cm

0.018 / 10 =  0.0018 cm

Now we will put the values in formula:

d = m/ l× h× w

l = m / d × h× w

l = 1190 g / 2.7 g/cm³× 30.2 cm × 0.0018 cm

l = 1190 g/ 0.146772 g/cm

l = 8107.81 cm or 81.7081 m

3 0
3 years ago
The mass number of an element is equal to
klasskru [66]

Answer:

The mass number is defined as the total number of protons and neutrons in an atom.

Explanation:

8 0
3 years ago
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