All categories

2 years ago

PLZ HELP ASAP PLATO!!! BRAINLIEST TO WHOEVER ANSWERS CORRECTLY!!!

Chemistry

1 answer:

katen-ka-za [31]2 years ago

3 0

Explanation:

a. 0.0093

Number of significant figures = 2

All zero’s preceding the first integers are never significant

b. 120.9

Number of significant figures = 4

All zero’s between integers are always significant.

c. 1,000

Number of significant figures = 1

All zeroes used solely for spacing the decimal point are not significant.

d. 1.008

Number of significant figures = 4

All zero’s between integers are always significant.

All zero’s after the decimal point are always significant.

e. 670

Number of significant figures = 2

All zeroes used solely for spacing the decimal point are not significant.

f. 0.184

Number of significant figures = 3

All zero’s after the decimal point are always significant.

g. 1.30

Number of significant figures = 3

All zero’s after the decimal point are always significant.

You might be interested in

2 M n O 2 + 4 K O H + O 2 + C l 2 → 2 K M n O 4 + 2 K C l + 2 H 2 O , there are 100.0 g of each reactant available. Which reacta

Sliva [168]

Answer:

The limiting reactant is KOH.

Explanation:

To find the limiting reactant we need to calculate the number of moles of each one:

$\eta = \frac{m}{M}$

<u>Where</u>:

η: is the number of moles

m: is the mass

M: is the molar mass

$\eta_{MnO_{2}} = \frac{100.0 g}{86.9368 g/mol} = 1.15 moles$

$\eta_{KOH} = \frac{100.0 g}{56.1056 g/mol} = 1.78 moles$

$\eta_{O_{2}} = \frac{100.0 g}{31.998 g/mol} = 3.13 moles$

$\eta_{Cl_{2}} = \frac{100.0 g}{70.9 g/mol} = 1.41 moles$

Now, we can find the limiting reactant using the stoichiometric relation between the reactants in the reaction:

$\eta_{MnO_{2}} = \frac{\eta_{MnO_{2}}}{\eta_{KOH}}*\eta_{KOH} = \frac{2}{4}*1.78 moles = 0.89 moles$

We have that between MnO₂ and KOH, the limiting reactant is KOH.

$\eta_{O_{2}} = \frac{\eta_{O_{2}}}{\eta_{Cl_{2}}}*\eta_{Cl_{2}} = \frac{1}{1}*1.41 moles = 1.41 moles$

Similarly, we have that between O₂ and Cl₂, the limiting reactant is Cl₂.

Now, the limiting reactant between KOH and Cl₂ is:

$\eta_{KOH} = \frac{\eta_{KOH}}{\eta_{Cl_{2}}}*\eta_{Cl_{2}} = \frac{4}{1}*1.41 moles = 5.64 moles$

Therefore, the limiting reactant is KOH.

I hope it helps you!

6 0

2 years ago

Read 2 more answers

Which of the following best describes electrolytes

adoni [48]

Answer:

a, b

Explanation:

Electrolytes dissociate to make ions, because of it they conduct electricity.

4 0

2 years ago

Explain why you cannot just take away a proton to make a halogen negative.

Ilia_Sergeevich [38]

The force that holds protons and neutrons together is too strong to overcome.

<h3>Explanation</h3>

Consider the location of the particles in an atom.

Electrons are found outside the nucleus.
Protons and neutrons are found within the nucleus.

Protons carry positive charges and repel each other. The nucleus will break apart without the strong force that holds the protons and neutrons together. This force is much stronger than the attraction between the nucleus and the electrons. X-rays are energetic enough for removing electrons from an atom. However, you'll need a collider to remove protons from a stable nucleus. You could well have ionized the atom with all that energy.

Also, changing the number of protons per nucleus will convert the halogen atom to an atom of a different element. Rather than making the halogen negative, removing a proton will convert the halogen atom to the negative ion of a different element.

6 0

2 years ago

When heating a flammable or volatile solvent for a recrystallization, which of these statements are correct? More than one answe

otez555 [7]

Explanation:

A volatile substance is defined as the substance which can easily evaporate into the atmosphere due to weak intermolecular forces present within its molecules.

Whereas a flammable substance is defined as a substance which is able to catch fire easily when it comes in contact with flame.

Hence, when we heat a flammable or volatile solvent for a recrystallization then it should be kept in mind that should heat the solvent in a stoppered flask to keep vapor away from any open flames so that it won't catch fire.

And, you should ensure that no one else is using an open flame near your experiment.

Thus, we can conclude that following statements are correct:

You should heat the solvent in a stoppered flask to keep vapor away from any open flames.

You should ensure that no one else is using an open flame near your experiment.

3 0

2 years ago

For the Bradford assay, the instructor will make a Bradford reagent dye by mixing 50 ml of 95% v/v ethanol with 100 mg of Coomas

jolli1 [7]

Answer:

4,25% v/v H3PO4

Explanation:

The concentration of phosphoric acid (H3PO4) is expressed as a volume / volume percentage, which means:

%v/v H3PO4 = (mL of pure H3PO4/mL of solution)*100%

In other words, <u>we are only interested in the final volume of the solution to which the phosphoric acid was diluted, regardless of its composition</u>. Which in this case is 1 L (1000 mL).

We can then apply the following equation, commonly used to calculate the initial or final concentration (or volume) of a substance when it is diluted:

Ci*Vi=Cf*Vf

<u>Where</u>:

Ci, is the initial concentration of the substance.

Vi, the initial volume of the substance

Cf, the final concentration reached after dilution

Vf, the final volume of the solution at which the substance was diluted

In this case, the incognite would be the final concentration of H3PO4 reached after dilution, that is, Cf. Therefore, we proceed to clear Cf from the previous equation and replace our data:

Cf = (Ci*Vi)/Vf = (85% v/v * 50 mL)/1000 mL = 4,25 % v/v

Note that being up and down in the division, the mL unit is canceled to result in% v / v.

7 0

2 years ago