All categories

3 years ago

Which of these substances are most likely crystalline solids? Select all that apply.

Chemistry

2 answers:

Lostsunrise [7]3 years ago

8 0

Answer:

you did not show the options

Explanation:

Svet_ta [14]3 years ago

5 0

I am unsure if this is the correct question, but I think these are the answer choices that are part of this question:

O diamond

O sugar

O rubber

O salt (NaCl)

O wood

O ice

O flour

The answers to this question are diamond, sugar, salt (NaCl), and ice (1st, 2nd, 4th, and 6th options).

You might be interested in

Ammonia and sulfuric acid react to form ammonium sulfate. Determine the starting mass of each reactant if 20.3 g of ammonium sul

strojnjashka [21]

Hope you are able to see

8 0

4 years ago

How many TOTAL atoms are contained in 10 molecules of TNT, C7H5N3O6?

nlexa [21]

Answer:7 atom of carbon+5atom of hydrogen +3atom of nitrogen +6 atom of oxygen =21

5 0

3 years ago

What determines the property of hardness in a mineral?

Rama09 [41]

Answer:

It's determined by the ability of one mineral to scratch another mineral.

5 0

3 years ago

Volume of 14.00g of nitrogen at 5.64atm and 315k

VikaD [51]

The volume of a 14.00g of nitrogen at 5.64atm and 315K is 4.59L.

<h3>How to calculate volume?</h3>

The volume of an ideal gas can be calculated using the following ideal gas equation formula;

PV = nRT

Where;

P = pressure (atm)
V = volume (L)
n = number of moles
R = gas law constant
T = temperature

An ideal gas is a hypothetical gas, whose molecules exhibit no interaction, and undergo elastic collision with each other and with the walls of the container.

The number of moles in 14g of nitrogen can be calculated as follows:

moles = 14g ÷ 14g/mol = 1mol

5.64 × V = 1 × 0.0821 × 315

5.64V = 25.86

V = 25.86 ÷ 5.64

V = 4.59L

Therefore, 4.59L is the volume of the gas

Learn more about volume at: brainly.com/question/12357202

#SPJ1

4 0

1 year ago

Zinc metal and aqueous silver nitrate react to give Zn(NO3)2(aq) plus silver metal. When 5.00 g of Zn(s) and solution containing

Fittoniya [83]

Answer : The percent yield is, 83.51 %

Solution : Given,

Mass of Zn = 5.00 g

Mass of $AgNO_3$ = 25.00 g

Molar mass of Zn = 65.38 g/mole

Molar mass of $AgNO_3$ = 168.97 g/mole

Molar mass of Ag = 107.87 g/mole

First we have to calculate the moles of Zn and $AgNO_3$ .

$\text{ Moles of }Zn=\frac{\text{ Mass of }Zn}{\text{ Molar mass of }Zn}=\frac{5.00g}{65.38g/mole}=0.0765moles$

$\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=\frac{25.00g}{168.97g/mole}=0.1479moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag$

From the balanced reaction we conclude that

As, 2 mole of $AgNO_3$ react with 1 mole of $Zn$

So, 0.1479 moles of $AgNO_3$ react with $\frac{0.1479}{2}=0.07395$ moles of $Zn$

From this we conclude that, $Zn$ is an excess reagent because the given moles are greater than the required moles and $AgNO_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Ag$