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dmitriy555 [2]
3 years ago
9

What is the kinetic energy of a 2,000-kilogram ball that is on the ground

Chemistry
1 answer:
Finger [1]3 years ago
6 0
Kinetic energy is associated with the motion of an object.
Since the object is at rest on the ground, its velocity is zero.
Since kinetic energy is directly proportional to the square of velocity, kinetic energy of the ball on the ground is zero.
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In the presence of a strong acid, litmus paper exhibits a(n) _________________ color.
tino4ka555 [31]
The answer is blue!!!!!!
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2 years ago
How many liters are in 5.42 mols of O2 gas?<br><br> (2 decimal places)
Debora [2.8K]

Explanation:

1 mol = 22.4 l

5.42 mol = 22.4 × 5.42 = 121.408

in two decimal place it is 121.41

7 0
3 years ago
Group 17 is called the halogen family, and the group to its right is called the noble gases. How are the elements in these group
Firlakuza [10]

Answer: they all have 7  electrons they have different atom sizes

Explanation: hope it helps

7 0
3 years ago
In a coffee-cup calorimeter, 1.55 g KOH is added to 125 ml of 0.80 M HCl. The following reaction occurs. KOH(s) + HCl(aq) → H2O(
DENIUS [597]

Answe1.55 gr:

Explanation:

yes

4 0
3 years ago
7.00 of Compound x with molecular formula C3H4 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25c.
beks73 [17]

Answer:

\Delta H_{f,C_3H_4}=276.8kJ/mol

Explanation:

Hello!

In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:

\Delta H_{rxn} =- m_wC_w\Delta T

We plug in the mass of water, temperature change and specific heat to obtain:

\Delta H_{rxn} =- (35000g)(4.184\frac{J}{g\°C} )(2.316\°C)\\\\\Delta H_{rxn}=-339.16kJ

Now, this enthalpy of reaction corresponds to the combustion of propyne:

C_3H_4+4O_2\rightarrow 3CO_2+2H_2O

Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:

\Delta H_{rxn}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{f,C_3H_4}

However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:

\Delta H_{rxn} =-339.16kJ*\frac{1}{7.00g}*\frac{40.06g}{1mol}=-1940.9kJ/mol

Now, we solve for the enthalpy of formation of C3H4 as shown below:

\Delta H_{f,C_3H_4}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{rxn}

So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):

\Delta H_{f,C_3H_4}=3(-393.5kJ/mol)+2(-241.8kJ/mol)-(-1940.9kJ/mol)\\\\\Delta H_{f,C_3H_4}=276.8kJ/mol

Best regards!

7 0
3 years ago
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