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3 years ago

An atom with 13 protons and an unknown number of neutrons is called____?

Chemistry

2 answers:

MaRussiya [10]3 years ago

4 0

Aluminum contains 13 protons

Sergio [31]3 years ago

3 0

<h2>The ans is 13 protons.... </h2>

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Elements can be represented by chemical symbols. Which of the following is the correct way to write the chemical symbol for sodi

mixer [17]

hope this helps :)

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3 years ago

How many moles of kf are contained in 244 ml of 0.135 m kf solution? the density of the solution is 1.22 g/ml?

galben [10]

The Molarity of a solution = number of moles / volume.
Volume = 244ml = 0.244L
So it follows that number of moles = Molarity * volume
Number of moles = 0.135 * 0.244 = 0.03945.
Hence the number of moles = 0.03945

6 0

3 years ago

Read 2 more answers

How many moles of H are present in 3 moles of carbon trihydride?

PSYCHO15rus [73]

Answer:

answer is b.12

Explanation:

8 0

3 years ago

14.Sorbitol is used in sugar free gums as a sweetener. After analysis, a sample of

evablogger [386]

<h3>Answer:</h3>

C6H12O6

Explanation:

To find empirical formular you write out the element

Carbon Hydrogen Oxygen

3.96 769 5.28

then u althrough divide by their atomic numbers

3.96÷ 12 769÷1 5.28÷16

=0.33 =769 =0.33

also divide althrough by the lowest number

the lowest number is 0.33 so we use it to divide

0.33÷0.33. 769÷0.33. 0.33÷0.33

= 1 = 2 = 1

empirical formular= CH2O

To find the Molecular formular

Molecular number = molar mass

(CH2O)n = 182

(12+1×2+16)n = 182

30n = 182

n = 6

therefore the Molecular formular is C6H12O6

3 0

2 years ago

En un balneario necesitan calentarse 1 millón de litros de agua anuales, subiendo la temperatura desde 15 ºC a 50 ºC y para ello

MAXImum [283]

Answer:

a) $m_{CH_4}=2630kg$

b) 1657 €

Explanation:

Hola,

a) En este problema, vamos a considerar el millón de litros de agua anuales, ya que con ellos podemos calcular el calor requerido para dicho calentamiento, sabiendo que la densidad del agua es de 1 kg/L:

$Q_{H_2O}=m_{H_2O}Cp(T_2-T_1)=1x10^6LH_2O*\frac{1kgH_2O}{1LH_2O}*4.18\frac{kJ}{kg\°C}(50-15) \°C\\Q_{H_2O}=146.3x10^6kJ$

Luego, usamos la entalpía de combustión del metano para calcular su requerimiento en kilogramos, sabiendo que la energía ganada por el agua, es perdida por el metano:

$Q_{H_2O}=-Q_{CH_4}=-146.3x10^5kJ=m_{CH_4}\Delta _cH_{CH_4}$

$m_{CH_4}= \frac{Q_{CH_4}}{\Delta _cH_{CH_4}} =\frac{-146.3x10^5kJ}{-890kJ/molCH_4} *\frac{16gCH_4}{1molCH_4} \\\\m_{CH_4}=2630112.36g=2630kg$

b) En este caso, consideramos que a condiciones normales de 1 bar y 273 K, 1 metro cúbico de metano cuesta 0,45 €, con esto, calculamos las moles de metano a dichas condiciones:

$n_{CH_4}=\frac{PV}{RT}=\frac{1atm*1000L}{0.082\frac{atm*L}{mol*K}*273K} =44.67mol$

Con ello, los kilogramos de metano que cuestan 0,45 €:

$44.67molCH_4*\frac{16gCH_4}{1molCH_4}*\frac{1kg}{1000g} =0.715kgCH_4$

Luego, aplicamos la regla de tres:

0.715 kg ⇒ 0.45 €

2630 kg ⇒ X

X = (2630 kg x 0.45 €) / 0.715 kg

X = 1657 €

Regards.

3 0

3 years ago