Can you please answer this for me please?

A softball is thrown with 145 Joules of kinetic energy. If the ball is moving at 20.0 m/s, what is the mass of the ball in kg?

Tasya [4]

Answer:

0.725 kg

Explanation:

Step 1: Given data

Kinetic energy of the softball (K): 145 J

Speed of the softball (v): 20.0 m/s

Mass of the softball (m): ?

Step 2: Calculate the mass of the softball

We will use the following expression.

K = 1/2 × m × v²

m = 2 K / v²

m = 2 × 145 J / (20.0 m/s)²

m = 0.725 kg

The mass of the softball is 0.725 kg.

3 0

2 years ago

calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w

kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is $-22Jmole^-^1k^-^1$

(b) The normal boiling point of water (J·K−1·mol−1) is $-109Jmole^-^1K^-^1$

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is 109J/mole

Explanation:

Lets calculate

(a) - General equation -

$(\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p$ = $-5_m(\beta )+5_m(\alpha )$ = $-\frac{\Delta H}{T}$

$\alpha ,\beta$ → phases

ΔH → enthalpy of transition

T → temperature transition

$(\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p$ = $= -\frac{\Delta_fH}{T_f}$

= $\frac{-6.008kJ/mole}{273.15K}$ ( $\Delta_fH$ is the enthalpy of fusion of water)

= $-22Jmole^-^1k^-^1$

(b) $(\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}$

= $\frac{40.656kJ/mole}{373.15K}$ ( $\Delta_v_a_p_o_u_rH$ is the enthalpy of vaporization)

= $-109Jmole^-^1K^-^1$

(c) $\Delta\mu =\Delta\mu(l)-\Delta\mu(s)$ = $-S_m\DeltaT$

$[\mu(l-5$ ° $C)-\mu(l,0$ ° $C)]$ = $[\mu(s-5$ ° $C)-\mu(s,0$ ° $C)]$ $=-S_m$ ΔT

$\mu(l,-5$ ° $C)-\mu(s,-5$ ° $C)=-Sm\DeltaT [\mu(l,0$

$\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)$

= 109J/mole

6 0

2 years ago

What effect does a quadrupling of the mass have upon the acceleration of the object

netineya [11]

The state in which all of the external forces acting upon an object are balanced; there is no acceleration. friction ..... quadrupling. doubling distance and quadrupling mass has the overall effect of the force

8 0

3 years ago

Yesterday we combined Hydrochloric Acid HCl with Sodium Hydroxide NaOH in a violent reaction that resulted in water H2O and comm

IceJOKER [234]

Reaction:
HCl + NaOH ---> NaCl + H2O

1 mole of HCl = 36,5 g
1 mole of NaOH = 40g

so, according to the reaction:
1 mol HCl = 1 mol NaOH
so, we need > 36,5 g HCl (hydrochloric acid)

answer: 36,5 g HCl (hydrochloric acid)
 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
next question.

1 mole of NaCl = 58,5 g
1 mole of H2O = 18g

so, according to the reaction:
1 mole of HCl (36,5 g) ----------------- - 1 mole of NaCl (58,5 g)
(the same for NaOH)
i
1 mole of HCl (36,5 g) ------------------ 1 mole of H2O (18 g)
(the same for NaOH)

so, this reaction is stechiometric

answer: 58,5 g NaCl i 18g H2O

7 0

3 years ago

Read 2 more answers

How many moles of carbon dioxide gas should be produced when 10.0 g of C2H6 are combusted at STP?

Finger [1]

Answer:

0.665 moles of CO₂

Explanation:

The balance chemical equation for the combustion of Ethane is as follow:

2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

Step 1: Calculate moles of C₂H₆ as;

Moles = Mass / M.Mass

Putting values,

Moles = 10.0 g / 30.07 g/mol

Moles = 0.3325 moles

Step 2: Calculate Moles of CO₂ as;

According to balance chemical equation,

2 moles of C₂H₆ produced = 4 moles of CO₂

So,

0.3325 moles of C₂H₆ will produce = X moles of CO₂

Solving for X,

X = 0.3325 mol × 4 mol ÷ 2 mol

X = 0.665 moles of CO₂

3 0

2 years ago