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3 years ago

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What is the acceleration of a ball traveling horizontally with an initial velocity of 20 meters/second and, 2.0 seconds later, a

velocity of 30 meters/second? A. 1.6 meters/second2 B. 3.0 meters/second2 C. 3.4 meters/second2 D. 5.0 meters/second2 Reset Next

1 answer:

Pepsi [2]3 years ago

6 0

Answer: <em>Acceleration of the ball in the given system is 5 meter per Second Square</em>

<em>The laws of motion are used to determine various aspects of an object in motion</em>.

Explanation:

Applying the first law of motion to calculate acceleration; if formula used in first law is given as $v=u+at$

Here we have a final velocity as 40 meter per second and initial velocity as 20 meter per second and time span is given as 2 second applying the given values in the given equation and finding the value of a

$30=20+a \times 2 = a=5 m/s^2$

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A chemistry student must write down in her lab notebook the concentration of a solution of potassium chloride. The concentration

DerKrebs [107]

Answer:

3.65 g / ml correct to 3 sig. fig.

Explanation:

The computation of the concentration required is shown below:

As we know that

[A] = mass of solute ÷ volume of solution

Before that first find the mass of solute

Given that

Initial weight = 5.55g

And,

Final weight = 92.7 g

So,

Mass of KCl is

= 92.7 - 5.55

= 87.15 g ~ 87.2 g

Now the KCi is fully dissolved, so the volume is 23.9 ml

So, concentration is

= 87.2 g ÷ 23.9 ml

= 3.65 g / ml correct to 3 sig. fig.

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3 years ago

3Co2 (aq) 6NO3(aq) 6Na (aq) 2PO43–(aq) → Co3(PO4)2(s) 6Na (aq) 6NO3(aq) Identify the net ionic equation for this reaction.

Burka [1]

Answer : The net ionic equation will be,

$3Co^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Co_3(PO_4)_2(s)$

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The given balanced ionic equation will be,

$3Co^{2+}(aq)+6NO_3^-(aq)+6Na^+(aq)+2PO_4^{3-}(aq)\rightarrow Co_3(PO_4)_2(s)+6Na^+(aq)+6NO_3^-(aq)$

In this equation, $NO_3^-(aq)\text{ and }Na^+(aq)$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

Thus, the net ionic equation will be,

$3Co^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Co_3(PO_4)_2(s)$

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3 years ago

What is the unit used for water vapor?

SpyIntel [72]

Answer:

relative humidity

Explanation:

Relative humidity can be measured with a hygrometer

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3 years ago

When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant so

zmey [24]

Answer: 54.4 kJ/mol

Explanation:

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=1.0M\times 0.05=0.05mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=1.0\times 0.05L=0.05mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.05 mole of HCl neutralizes by 0.05 mole of NaOH

Thus, the number of neutralized moles = 0.05 mole

Now we have to calculate the mass of water:

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $50ml+50ml=100ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 100ml=100g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 100 g

$T_{final}$ = final temperature of water = $27.5^0C$

$T_{initial}$ = initial temperature of metal = $21.0^0C$

Now put all the given values in the above formula, we get:

$q=100g\times 4.18J/g^oC\times (27.5-21.0)^0C$

$q=2719.6J=2.72kJ$

Thus, the heat released during the neutralization = 2.72 KJ

Now we have to calculate the enthalpy of neutralization per mole of $HCl$ :

0.05 moles of $HCl$ releases heat = 2.72 KJ

1 mole of $HCl$ releases heat = $\frac{2.72}{0.05}\times 1=54.4KJ$

Thus the enthalpy change for the reaction in kJ per mol of HCl is 54.4 kJ

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