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3 years ago

How many atoms are in 16.1 G Sr

Chemistry

1 answer:

marysya [2.9K]3 years ago

4 0

Answer: There are $1.11\times 10^{23}$ atoms of Sr

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{16.1g}{87.6g/mol}=0.184moles$

1 mole of Sr contains = $6.023\times 10^{23}$ atoms of Sr

Thus 0.184 moles of Sr contains = $\frac{6.023\times 10^{23}}{1}\times 0.184=1.11\times 10^{23}$ atoms of Sr

Thus there are $1.11\times 10^{23}$ atoms of Sr

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How much heat must be removed from 25.0g of steam at 118.0C in order to form ice at 15C

NemiM [27]

Answer:

-10778.95 J heat must be removed in order to form the ice at 15 °C.

Explanation:

Given data:

mass of steam = 25 g

Initial temperature = 118 °C

Final temperature = 15 °C

Heat released = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 15 °C - 118 °C

ΔT = -103 °C

now we will put the values in formula

q = m . c . ΔT

q = 25 g × 4.186 J/g.°C × -103 °C

q = -10778.95 J

so, -10778.95 J heat must be removed in order to form the ice at 15 °C.

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3 years ago

Why doesn't light get absorbed

ziro4ka [17]

It isn't a materialistic object.

6 0

3 years ago

To determine the freezing point depression of a LiCl solution, Toni adds 0.411 g of LiCl to the sample test tube along with 19.7

Cerrena [4.2K]

Answer:

LiCl = 0.492 m

Explanation:

Molal concentration is the one that indicates the moles of solute that are contained in 1kg of solvent.

Our solute is lithium chloride, LiCl.

Our solvent is distilled water.

We do not have the mass of water, but we know the volume, so we should apply density to determine mass.

Density = mass / volume

Density . volume = mass

1 g/mL . 19.7 mL = 19.7 g

We convert g to kg → 19.7 g . 1 kg / 1000g = 0.0197 kg

Let's determine the moles of LiCl

0.411 g . 1 mol / 42.394 g = 9.69×10⁻³ moles

Molal concentration (m) = 9.69×10⁻³ mol / 0.0197 kg → 0.492 m

7 0

3 years ago

Two identical 732.0 L tanks each contain 212.0 g of gas at 293 K, with neon in one tank and nitrogen in the other. Based on the

Vladimir [108]

Answer: (a) Neon, Nitrogen; (b) Neon, Nitrogen; (c) Neon is lower than Nitrogen; (d) It doesn't affect;

Explanation: The kinetic-molecular theory studies the behavior of particles under pre-determinated situation. In cases of gases, the particles moving around colliding with each other and the walls of the container, without loss of energy. In the case in question, all the parameters are the same (same temperature, volume and pressure), except for the gases, which has different molar masses. In this sense, Neon has lower average speed due to its molar mass being higher, which means, its particles moves slower for being heavier. Related to pressure, as velocity is lower, it collides less with the walls of the tank, and so pressure is lower. For density, it doesn't affect the behavior of the system nor the kinetic energy.

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3 years ago

What is the total volume of gaseous products formed when 116 liters of butane (C4H10) react completely according to the followin

Contact [7]

Answer: The total volume of the gaseous products is 1044.29 L

Explanation:

We are given:

Volume of butane = 116 L

At STP:

22.4 L of volume is occupied by 1 mole of a gas

So, 116 L of volume will be occupied by = $\frac{1}{22.4}\times 116=5.18mol$ of butane

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

For carbon dioxide:

By Stoichiometry of the reaction:

2 moles of butane produces 8 moles of carbon dioxide

So, 5.18 moles of butane will produce = $\frac{8}{2}\times 5.18=20.72mol$ of carbon dioxide

Volume of carbon dioxide at STP = (20.72 × 22.4) = 464.13 L

For water vapor:

By Stoichiometry of the reaction:

2 moles of butane produces 10 moles of water vapor

So, 5.18 moles of butane will produce = $\frac{10}{2}\times 5.18=25.9mol$ of water vapor

Volume of water vapor at STP = (25.9 × 22.4) = 580.16 L

Total volume of the gaseous products = [464.13 + 580.16] = 1044.29 L

Hence, the total volume of the gaseous products is 1044.29 L

3 0

3 years ago