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Cloud [144]
3 years ago
15

What is precautionary measure?​

Chemistry
1 answer:
vichka [17]3 years ago
4 0
Precautionary measures are rules often followed when working in labs and/or with chemicals.
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When a gas is heated, the gas particles have more kinetic energy. Complete the following sentence: This means they move about mo
Volgvan

Answer:

Contact

Explanation:

As the speed of the particles increases, they move around more, which means they hit the sides of the container more.

3 0
3 years ago
3. If you start with 8x1025 molecules of Cl, and 25 grams of KI, how many grams of KCl would
miskamm [114]

Answer:

Percent yield = 89.1%

Explanation:

Based on the equation:

Cl₂ + 2KI → 2KCl + I₂

<em>1 mole of Cl₂ reacts with 2 moles of KI to produce to moles of KCl</em>

<em />

To solve this quesiton we must find the moles of each reactant in order to find the limiting reactant. With the limiting reactant we can find the moles of KCl and the mass:

<em>Moles Cl₂:</em>

8x10²⁵ molecules * (1mol / 6.022x10²³ molecules) = 133 moles

<em>Moles KI -Molar mass: 166.0028g/mol-</em>

25g * (1mol / 166.0028g) = 0.15 moles

Here, clarely, the KI is the limiting reactant

As 2 moles of KI produce 2 moles of KCl, the moles of KCl produced are 0.15 moles. The theoretical mass is:

0.15 moles * (74.5513g / mol) =

11.2g KCl

Percent yield is: Actual yield (10.0g) / Theoretical yield (11.2g) * 100

<h3>Percent yield = 89.1%</h3>
3 0
3 years ago
What is the standard model? What can it explain?
Arturiano [62]

Answer:

The Standard Model of particle physics is the theory describing three of the four known fundamental forces (the electromagnetic, weak, and strong interactions, and not including the gravitational force) in the universe, as well as classifying all known elementary particles. It was developed in stages throughout the latter half of the 20th century, through the work of many scientists around the world, with the current formulation being finalized in the mid-1970s upon experimental confirmation of the existence of quarks.

7 0
3 years ago
Read 2 more answers
A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate o
Ostrovityanka [42]

Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

4 0
3 years ago
What is the molarity of 50.84 g of Na2CO3 dissolved in 0.400 L solution?
borishaifa [10]

Answer:

1.20 M

Explanation:

Convert grams of Na₂CO₃ to moles.  (50.84 g)/(105.99 g/mol) = 0.4797 mol

Molarity is (moles of solute)/(liters of solvent) = (0.4797 mol)/(0.400 L) = 1.20 M

4 0
3 years ago
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