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3 years ago

The kinetic energy of a ball with a mass of 0.5kg and a velocity of 10 m/s is ?

Chemistry

1 answer:

KATRIN_1 [288]3 years ago

3 0

Answer:

K.E = 25 J

Explanation:

Given data:

Mass of ball = 0.5 g

Velocity of ball = 10 m/s

Kinetic energy = ?

Solution:

Formula:

K.E = 1/2 mv²

m = mass

v = velocity

Now we will put the values in formula.

K.E = 1/2 mv²

K.E = 1/2× 0.5kg × (10m/s)²

K.E = 1/2 ×0.5kg × 100m²/s²

K.E = 25 Kg.m²/s²

Kg.m²/s² = J

K.E = 25 J

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Which of the following is NOT an indication of a chemical reaction?

PolarNik [594]

A. The substance dissolved. Just because a substance dissolved in a solvent doesn’t mean that a chemical action occurred. For example, if you put sugar in water as stir, it will dissolve, but the chemical make up of the sugar and the water don’t change. A chemical reaction can be indicated by a drastic change in color or temperature or if a substance precipitate.

Hope this helped :)

6 0

3 years ago

Read 2 more answers

Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = +

qwelly [4]

Answer: The $\Delta H^o_{rxn}$ for the reaction is -1052.8 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $X(s)+\frac{1}{2}O_2(g)\rightarrow XO(s)$ $\Delta H_1=-668.5kJ$

(2) $XCO_3(s)\rightarrow XO(s)+CO_2$ $\Delta H_2=+384.3kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-668.5))+(1\times (-384.3))=-1052.8kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is -1052.8 kJ.

7 0

3 years ago

If 10.0 mL of a .600 M of HNO3 reacts with 31.0 mL of .700M Ba(OH)2 solution, what is the molarity of Ba(OH)2 after the reaction

Tasya [4]

Answer:

0.456M

Explanation:

1. Balanced molecular equation

$2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_2+2H_2O$

2. Mole ratio

$\dfrac{2molHNO_3}{1molBa(OH)_2}$

3. Moles of HNO₃

Number of moles = Molarity × Volume in liters
n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃

4. Moles Ba(OH)₂

n = 0.700M × 0.0310 liter = 0.0217 mol

5. Limiting reactant

Actual ratio:

$\dfrac{0.0600molHNO_3}{0.0217molBa(OH)_2}\approx0.28$

Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.

Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.

6. Final molarity of Ba(OH)₂

Molarity = number of moles / volume in liters
Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M

5 0

3 years ago

HNO3 + H2SO4 + NO

77julia77 [94]

Explanation:

The two half equations are;

3e + HNO3 → NO

S→ H2SO4 + 6e

When balancing half equations, we have to make sure the number of electrons gained is equal to the number of electrons lost.

Which factor will you use for the top equation?

We multiply by 2 to make the number of electrons = 6e

Which factor will you use for the bottom equation?

We multiply by 1 to make the number of electrons = 6e

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3 years ago

What is a radioactive tracer? Give an example of how one might be used.

Norma-Jean [14]

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4 years ago