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3 years ago

Question 1 (1 point)

Chemistry

1 answer:

Dima020 [189]3 years ago

4 0

Answer:

1. Both are made up of two substances that are chemically combined. 2. oxygen(O2) , 3. CARBON (C) , 4. AIR(N2 nixed with O2 AND CO2), 5. CANNOT BE SEPARATED BY PHYSICAL MEANS

Explanation:

<em>HOPE IT HELPS</em>

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A container was found in the home of the victim that contained 120 g of ethylene glycol in 550 g of liquid. How many drinks, eac

Andrei [34K]

Answer:

0.432 drinks are toxic

Explanation:

The toxic dose of ethylene glycol is 0.1 mL per kg body weight (mL/kg). In grams (Density ethylene glycol = 1.11g/mL):

1.11g/mL * (0.1mL / kg) = 0.111g/kg

If the victim weighs 85kg, its letal dose is:

85kg * (0.111g/kg) = 9.435g of ethylene glycol

Using the concentration of ethylene glycol in the liquid:

9.435g of ethylene glycol * (550g liquid / 120g ethylene glycol) = 43.2g of liquid are toxic.

The drinks are:

43.2g of liquid * (1 drink / 100 g) =

<h3>0.432 drinks are toxic</h3>

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3 years ago

What is the radius of a hydrogen atom whose electron is bound by 0.544 ev? express your answer with the appropriate units?

insens350 [35]

First, we need to calculate the principal quantum number n for this electron, using the equation:
E = (-13.60 eV) / (n x n)
where E is the energy that is used to bound the electron (here, E = - 0.544 eV).
- 0.544 eV = (-13.60 eV) / (n x n)
n x n = (- 13.60 eV) / (- 0.544 eV)
n x n = 25
n = 5

The orbital radius that is equal to the radius of a hydrogen atom is calculated using the equation:
r = 0.053 nm x n x n
r = 0.053 nm x 5 x 5
r = 0.053 nm x 25
r = 1.325 nm

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3 years ago

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Hi there!

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3 years ago

This for number 8 and 9

3241004551 [841]

Okay for 8 it is the maximum levels will be higher and for 9 is poor circulation leads to lack of nutrients and oxygen

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3 years ago

6. Cuando se oxidan en el aire 12,120 g de vapor de Zinc se obtienen 15,084 g del óxido. ¿Cuál es la fórmula empírica del óxido?

Eddi Din [679]

Answer:

12120 g + O2 = 15084 g

m Zn = 12.120 Kg

m óxido = 15.084 Kg

1. calcular la masa de cinc en gramos

g = 12,120 Kg x 1000 = 12120 g de cinc

g = 15.084 Kg x 100 = 15084 g de oxígeno

2. calcular gramos de Oxigeno

g O = 15084 g - 12120 g = 2964 g O2

3. calcular % de Zn y O

%m/m ( m soluto / m solc.) x 100

%m/m (Zn) = ( 1210 g / 15084 g ) x 100

% m/m (Zn) = 80.35 % = 80.35 g

%m/m (O) = ( 2964 g / 15084 g ) x 100

% m/m (Zn) = 19.65 % = 19.65 g

4. Calcular moles de cada elemento

Zn: 80.35 g / 65.38 g/mol = 1.228 mol

O: 19.65 g / 16 g/mol = 1.228 mol

5. dividir entre el menor de los elementos

Zn: 1.228 mol / 1.228 mol = 1

O: 1.228 mol / 1.228 mol = 1

6. Fórmula empírica: ZnO

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3 years ago