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2 years ago

A car with a mass of 1500 kg is moving at a rate of 4 m s . Determine its kinetic energy

Chemistry

2 answers:

Eva8 [605]2 years ago

6 0

Answer:- The Kinetic energy of the car is 12000 J.

Explanation:

Kinetic energy is the energy posessed by an object by virtue of its motion.

$K.E=\frac{1mv^2}{2}$

m= mass of object = 1500 kg

v= velocity of object = 4 m/s

Putting in the values in the equation:

$K.E=\frac{1500kg\times 4^2ms^{-1}}{2}$

$K.E=12000kgm^2s^{-2}=12000N$ $1kgm^2s^{-2}=1J$

Thus Kinetic energy of the car is 12000 J.

rodikova [14]2 years ago

4 0

the correct answer is b) 12000 J

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Calculate the atomic mass of gallium, symbol Ga. Gallium has two isotopes: Ga-69 and Ga-71. The relative abundance of Ga-69 is 6

amm1812

Answer:

The atomic mass of gallium (Ga) = 69.723 g/mol

Explanation:

Given: Two isotopes of Gallium (Ga) are Gallium-69 (⁶⁹Ga) and Gallium-71 (⁷¹Ga)

For ⁶⁹Ga: 

Relative abundance = 60.12% = 60.12 ÷ 100 = 0.6012; Atomic mass = 68.9257 g/mol

For ⁷¹Ga:

Relative abundance = 39.88% = 39.88 ÷ 100 = 0.3988; Atomic mass = 70.9249 g/mol

∴ The atomic mass of Ga = (Relative abundance of ⁶⁹Ga × Atomic mass of ⁶⁹Ga) + (Relative abundance of ⁷¹Ga × Atomic mass of ⁷¹Ga)

⇒ Atomic mass of Ga = (0.6012 × 68.9257 g/mol) + (0.3988 × 70.9249 g/mol) = 69.723 g/mol

Therefore, the atomic mass of gallium (Ga) = 69.723 g/mol

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3 years ago

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Answer:

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2 years ago

Why does every sample of water have the same boiling point?

Tamiku [17]

Answer:

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3 years ago

How many moles of aspartame are present in 1.00 mg of aspartame?

Diano4ka-milaya [45]

Answer:- There are $3.40*10^-^6$ moles.

Solution:- It is a unit conversion problem where we are asked to convert mg of aspartame to moles. Aspartame is $C_1_4H_1_8N_2O_5$ and it's molar mass is 294.31 grams per mole.

mg are converted to grams and then the grams are converted to moles as:

$1.00mg Aspartame(\frac{1g}{1000mg})(\frac{1mole}{294.31g})$

= $3.40*10^-^6$ moles of aspartame

So, there would be $3.40*10^-^6$ moles of aspartame in 1.00 mg of it.

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3 years ago

A laboratory analysis of a sample finds it is composed of 38.8% carbon, 16.2% hydrogen, and 45.1% nitrogen. What is its empirica

Sladkaya [172]

Answer: The empirical formula for the given compound is $CH_5N$

Explanation : Given,

Percentage of C = 38.8 %

Percentage of H = 16.2 %

Percentage of N = 45.1 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 38.8 g

Mass of H = 16.2 g

Mass of N = 45.4 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{38.8g}{12g/mole}=3.23moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{16.2g}{1g/mole}=16.2moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{45.4g}{14g/mole}=3.24moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.23 moles.

For Carbon = $\frac{3.23}{3.23}=1$

For Hydrogen = $\frac{16.2}{3.23}=5.01\approx 5$

For Oxygen = $\frac{3.24}{3.23}=1.00\approx 1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 1 : 5 : 1

Hence, the empirical formula for the given compound is $C_1H_5N_1=CH_5N$

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3 years ago