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4 years ago

Why is potassium placed in the periodic table where it is?

1 answer:

6 0

Potassium is placed where it is based on its properties and it's reactivity. It's also placed there based on it's atomic number.

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2 FONS

stiv31 [10]

Answer:

C. It decreases by a factor of 4

Explanation:

F1 = kq1*q2/r²

F2 = kq1*q2/(2r)² = kq1*q2/(4r²) = kq1*q2/(r²*4) = F1/4

7 0

3 years ago

The following data was collected when a reaction was performed experimentally in the laboratory. Determine the ammount of NaNO3

zepelin [54]

equalize the coefficients:

Al(NO3)3+3NaCl=3NaNO3+AlCl3

according to the equation 1 mol Al(NO3)3 reacted with 3 moles NaCl. hence, 4 moles Al(NO3)3 should react with 12 moles NaCl, and 9 moles reacted NaCl. So NaCl is limiting, we decide further on this reagent.

\frac{9}{3}=\frac{x}{3}

x=9

9 moles NaNO3

6 0

3 years ago

A 5.0 g sample of aluminum with a specific heat of 0.90 J/(g °C) was heated from 22.1°C to 32.1°C. How much heat, to the nearest

mina [271]

Heat gained in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:

Heat = mC(T2-T1)
Heat = 5(0.90)(32.1 - 22.1)
Heat = 45 J energy gained

7 0

3 years ago

Nicotine, a component of tobacco, is composed of c, h, and n. A 2.625-mg sample of nicotine was combusted, producing 7.1210 mg o

Vikki [24]

The mass sample of nicotine combusted = 2.625 mg (given)

Mass of $CO_2$ produced = 7.1210 mg (given)

Mass of $H_2O$ produced = 2.042 mg (given)

Molar mass of $CO_2$ = 44 g/mol

Molar mass of $H_2O$ = 18 g/mol

Percentage of Carbon = $\frac{12 g}{44 g/mol}\times \frac{7.120 mg of CO_2}{2.625 mg sample}\times 100 = 74.04$ %

Percentage of hydrogen = $\frac{2 g}{18 g/mol}\times \frac{2.04 mg of H_2O}{2.625 mg sample}\times 100 = 8.62$ %

Now, for percentage of nitrogen = $100 - (74.04+8.62) = 17.34$ %

Calculating the moles of each element:

$Number of moles = \frac{given mass}{Molar mass}$

For $C$

$\frac{74.04 g}{12 g/mol} = 6.17 mol$

For $H$

$\frac{8.62 g}{1 g/mol} = 8.62 mol$

For $N$

$\frac{17.34 g}{14 g/mol} = 1.24 mol$

Dividing with the smallest mole value to calculate the molar ratio of each element:

$C_{\frac{6.17}{1.24}} = C_{4.9} = C_5$

$H_{\frac{8.62}{1.24}} = H_{6.9} = H_7$

$N_{\frac{1.24}{1.24}} = N_1$

Hence, the empirical formula for nicotine is $C_5H_7N$ .

8 0

3 years ago

"Select True or False: The following data for the reaction: A(g) + 2B(s) === AB2(g) provides evidence that the reaction is exoth

galina1969 [7]

The reaction as shown is known to be an exothermic reaction.

<h2>Endothermic and exothermic reaction:</h2>

Depending on the temperature of the reaction vessel after reaction, the reaction could be classified as;

Endothermic
Exothermic

In an endothermic reaction, heat is absorbed and the rate of reaction increasing with increase in temperature while in an exothermic reaction, the rate of reaction decreases with increase in temperature.

Hence, the correct statements are;

The reaction is an exothermic reaction.
The value of K decreases as the value of T increases.

Learn more about exothermic reaction: brainly.com/question/14969584

7 0

3 years ago