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2 years ago

Calculate the number of ATOMS in 1.0 mole of O2.

Chemistry

2 answers:

podryga [215]2 years ago

8 0

<u>Answer:</u> The number of atoms in given amount of oxygen gas is $6.022\times 10^{23}$

<u>Explanation:</u>

According to the mole concept:

1 mole of a substance contains $6.022\times 10^{23}$ number of atoms.

We are given:

Moles of oxygen gas = 1 mole

So, the number of atoms that will be present in 1 mole of oxygen gas will be $1\times 6.022\times 10^{23}=6.022\times 10^{23}$ number of atoms.

Hence, the number of atoms in given amount of oxygen gas is $6.022\times 10^{23}$

Wewaii [24]2 years ago

7 0

There is 6.02*10^23 molecule per mole. And there is 2 atoms per oxygen molecule. So the answer is 1.204*10^24 atoms in 1.0 mole of O2.

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2 years ago

A sample of nitric acid has a mass of 8.2g. It is dissolved in 1L of water. A 25mL aliquot of this acid is titrated with NaOH. T

AveGali [126]

Answer:

18.075 mL of NaOH solution was added to achieve neutralization

Explanation:

First, let's formulate the chemical reaction between nitric acid and sodium hydroxide:

NaOH + HNO3 → NaNO3 + H2O

From this balanced equation we know that 1 mole of NaOH reacts with 1 mole of HNO3 to achieve neutralization. Let's calculate how many moles we have in the 25 mL aliquot to be titrated:

63.01 g of HNO3 ----- 1 mole

8.2 g of HNO3 ----- x = (8.2 g × 1 mole)/63.01 g = 0.13014 moles of HNO3

So far we added 8.2 grams of nitric acid (0.13014 moles) in 1 L of water.

1000 mL solution ---- 0.13014 moles of HNO3

25 mL (aliquot) ---- x = (25 mL× 0.13014 moles)/1000 mL = 0.0032535 moles

So, we now know that in the 25 mL aliquot to be titrated we have 0.0032535 moles of HNO3. As we stated before, 1 mole of NaOH will react with 1 mole of HNO3, hence 0.0032535 moles of HNO3 have to react with 0.0032535 moles of NaOH to achieve neutralization. Let's calculate then, in which volume of the given NaOH solution we have 0.0032535 moles:

0.18 moles of NaOH ----- 1000 mL Solution

0.0032535 moles---- x=(0.0032535moles×1000 mL)/0.18 moles = 18.075mL

As we can see, we need 18.075 mL of a 0.18 M NaOH solution to titrate a 25 mL aliquot of the prepared HNO3 solution.

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2 years ago

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2 years ago

Calculate: (a) the weight (in lbf) of a 30.0 lbm object. (b) the mass in kg of an object that weighs 44N. (c) the weight in dyne

belka [17]

Answer:

a) 965,1 lbf

b) 4,5 kg

c) 1,33 * 10^6 dynes

Explanation:

Mass of an object refers to the amount of mattter it cotains, it can be expressed it gr, kg, lbm, ton, etc.

Weight of an object refers to a force, and is the measurement of the pull of gravitiy on an object. It may be definide as the mass times the acceleration of gravity.

w=mg

In Planet Earth, the nominal "average" value for gravity is 9,8 m/s² (in the International System) or 32,17 ft/s² (in the FPS system).

To solve this problem we'll use the following conversion factors:

1 lbf = 1 lbm*ft/s²

1 N = 1 kg*m/s²

1 dyne = 1 gr*cm/s² and 1 N =10^5 dynes

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1 k = 1000 gr

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$w = 30 lbm * 32,17 ft/s^{2} = 965, 1 \frac{lmb*ft}{s^{2} } = 965,1 lbf$

b) w = 44 N

First, we clear m of the weight equation and then we replace our data.

$m = \frac{w}{g} = \frac{44 N}{9,8 \frac{m}{{s}^{2}} } = 4,5 kg$

c) m = 15 ton

$m=15 ton * \frac{907,18 kg}{1 ton} = 13607,7 kg \\ w = mg = 13607,7 kg * 9,8 m/s2 = 133355,5 N * \frac{10^{5} dynes }{1 N} = 1,33 * 10^{6}dynes$

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3 years ago

Calculate the pH of a 2.3 10-3M[OH'] solution.

disa [49]

Answer:

11.33

Explanation:

-log(2.3x10^-3) = 2.67

14-2.67

- Hope this helped! Let me know if you need a further explanation.

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3 years ago