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3 years ago

When should you read the label on a chemical container?

Chemistry

1 answer:

atroni [7]3 years ago

6 0

Always. You never know what kind of chemical you’re dealing with and how powerful it is.

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What is the compound name for the ion chemical formula AgBr?

Gnom [1K]

Answer:

Silver bromide.

Explanation:

4 0

3 years ago

4. How many milligrams are in 5.25 x 10-13 kg? the “-13” is an exponent

rusak2 [61]

5. 25 x 10⁻⁷mg

Explanation:

This is mass conversion from mg to kg;

The kg is a quantity of mass used to measure the amount of matter in a substance.

Given mass = 5.25 x 10⁻¹³kg

The kilo- is a prefix that denotes 10³

therefore;

1000g = 1kilogram

the milli- is a prefix that denotes 10⁻⁻³

1000mg = 1g

Now that we know this, we can convert:

5.25 x 10⁻¹³kg x $\frac{1000g}{1kg} x \frac{1000mg}{1g}$ = 5. 25 x 10⁻¹³ x 10⁶mg

= 5. 25 x 10⁻⁷mg

learn more:

Conversion brainly.com/question/1548911

#learnwithBrainly

8 0

3 years ago

If the human body were a car, glucose would be:

NeTakaya

If the human body were a car, glucose would be the gasoline.
Glucose gives humans energy, we basically run on glucose, among other things, the same way a car would run on gas.

8 0

3 years ago

There are 9.88x1023 molecules of O2 available.

blondinia [14]

Answer:

1.64 moles O₂

Explanation:

Part A:

Remember 1 mole of particles = 6.02 x 10²³ particles

So, the question becomes, how many '6.02 x 10²³'s are there in 9.88 x 10²³ molecules of O₂?

This implies a division of given number of particles by 6.02 x 10²³ particles/mole.

∴moles O₂ = 9.88 x 10²³ molecules O₂ / 6.02 x 10²³ molecules O₂ · mole⁻¹ = 1.64 mole O₂

_______________

Part B needs an equation (usually a combustion of a hydrocarbon).

7 0

2 years ago

If 3.31 moles of argon gas occupies a volume of 100 L what volume does 13.15 moles of argon occupy under the same temperature an

kumpel [21]

Answer:

397 L

Explanation:

Recall the ideal gas law:

$\displaystyle PV = nRT$

If temperature and pressure stays constant, we can rearrange all constant variables onto one side of the equation:

$\displaystyle \frac{P}{RT} = \frac{n}{V}$

The left-hand side is simply some constant. Hence, we can write that:

$\displaystyle \frac{n_1}{V_1} = \frac{n_2}{V_2}$

Substitute in known values:

$\displaystyle \frac{(3.31 \text{ mol})}{(100 \text{ L})} = \frac{(13.15\text{ mol })}{V_2}$

Solving for V₂ yields:

$\displaystyle V_2 = \frac{(100 \text{ L})(13.15)}{3.31} = 397 \text{ L}$

In conclusion, 13.15 moles of argon will occupy 397* L under the same temperature and pressure.

(Assuming 100 L has three significant figures.)

3 0

2 years ago