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4 years ago

12

Two forces are acting on a wheelbarrow.One force is pushing to the right and an equal force is pushing to the left.What can you

say about the wheelbarrows movement?
A.it is slowing down
B.it is not accelerating
C.it is moving faster and faster
D.its speed is changing at a constant rate

2 answers:

andreev551 [17]4 years ago

7 0

B. It is not accelerating since equal and opposite forces cancel each other. Hope it helps

babymother [125]4 years ago

4 0

Answer: The correct answer is option B.

Explanation: We are given that two forces are acting on wheelbarrow. One of which is acting on right and the equal amount of force is acting on the left direction. This meas that the two forces are balanced forces.

Balanced forces are defined as the forces acting on the same object and are equal in magnitude but act in opposite direction. These forces cancel each other and not allow the object to move.

Hence, the object will not accelerate. So, the correct option is B.

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Sodium sulfide reacts with hydrochloric acid to produce hydrosulfuric acid and sodium chloride. (you need to write and balance t

VikaD [51]

Answer:

2.5 moles of NaCl

Explanation:

The balanced chemical reaction equation is shown in the image. Since it takes 2 moles of Hydrochloric acid to form two moles of sodium. Chloride, then 2.5 moles of hydrochloric acid should also form 2.5 moles of sodium chloride according to the balanced reaction equation.

4 0

3 years ago

What is the mole fraction of NaOH in an aqueous solution that

astraxan [27]

Answer:

The mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882

Explanation:

We are given that

Aqueous solution that contains 22.9% NaOH by mass means

22.9 g NaOH in 100 g solution.

Mass of NaOH(WB)=22.9 g

Mass of water =100-22.9=77.1

Na=23

O=16

H=1.01

Molar mass of NaOH(MB)=23+16+1.01=40.01

Number of moles = $\frac{Given\;mass}{Molar\;mass}$

Using the formula

Number of moles of NaOH $(n_B)=\frac{W_B}{M_B}=\frac{22.9}{40.01}$

$n_B=0.572moles$

Molar mass of water=16+2(1.01)=18.02g

Number of moles of water $(n_A)=\frac{77.1}{18.02}$

$n_A=4.279 moles$

Now, mole fraction of NaOH

= $\frac{n_B}{n_B+n_A}$

$=\frac{4.279}{0.572+4.279}$

=0.882

Hence, the mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882

4 0

3 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

1. What elements are carbohydrates generally composed of?

Llana [10]

The correct answer is hydrogen<span>, and </span>oxygen<span>. </span>

4 0

3 years ago

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What gives the gem amethyst its purplish color?

Semenov [28]

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5 0

3 years ago