What is the visual differences between elements compounds and mixtures

Answer:

After 3 half lives 10 g of radio active material left.

Explanation:

Given data:

Total amount of radio active material = 80 g

Amount left after 3 half lives = ?

Solution:

At time zero = 80 g

At first half life = 80 g/2 = 40 g

At 2nd half life = 40 g/2 = 20 g

At 3rd half life = 20 g/2 = 10 g

Thus, after 3 half lives 10 g of radio active material left.

The [H₃0⁺] of the solution : 1.778  x 10⁻¹² M

pH is the degree of acidity of a solution that depends on the concentration of H⁺ ions. The greater the value the more acidic the solution and the smaller the pH.  

pH = - log [H⁺]  

So that the two quantities between pH and [H⁺] are inversely proportional because they are associated with negative values.  

A solution whose value is different by n has a difference in the concentration of H⁺ ion of 10ⁿ.  

pH of solution = 11.75

the [H₃0⁺] of the solution :

[tex]\tt pH=-log[H_3O^+]\\\\11.75=-log[H_3O^+]\\\\(H_3O^+]=10^{-11.75}=1.778\times 10^{-12}[/tex]

Answer:

1. They will react.

2. 2HCI + Zn —> ZnCl2 + H2

Explanation:

1. From the question given above,

We can see that Zn is higher than H in the activity series.

NOTE: Elements higher (i.e at the top) in the activity series will displace those lower (i.e at the bottom) in the series. Thus, Zn is higher in the series than H. Therefore, Zn will displace H from solution. Hence, they will react.

2. The product obtained from the reaction can be seen as follow:

HCI + Zn —> ZnCl2 + H2

The above equation can be balance as follow:

There are 2 atoms of H on the right side and 1 atom on the left side. It can be balance by putting 2 in front of HCl as shown below:

2HCI + Zn —> ZnCl2 + H2

Answer:

3.68 g/cm3

Explanation:

D = M / V

350 / 95

Answer:

bottom right corner with arrows going up

thermal contact

When heat flows from one object or substance to another it is in contact with, the objects or substances are said to be in thermal contact. The state where objects in thermal contact with each other reach the same temperature, no heat flows between them.

Answer:

i do k12 too!

Explanation:

lol

The characteristic that could distinguish a crystalline solid from an amorphous solid is option b. The way it breaks.

Crystalline solids contained edges and faces, that should be diffracted x-rays, and have sharp melting points.

While on the other hand, amorphous solids should contain irregular does not provide well-resolved x-ray diffraction patterns.

hence, the option b is correct.

And, the rest of the options are wrong.

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Answer: The empirical formula is [tex]NaBO_2[/tex]

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given:

Mass of Na= 34.5 g

Mass of B= 16.4 g

Mass of O = 48.6 g

Step 1 : convert given masses into moles.

Moles of Na =[tex]\frac{\text{ given mass of Na}}{\text{ molar mass of Na}}= \frac{34.5g}{23g/mole}=1.5moles[/tex]

Moles of B =[tex]\frac{\text{ given mass of B}}{\text{ molar mass of B}}= \frac{16.4g}{11g/mole}=1.5moles[/tex]

Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{48.6g}{16g/mole}=3moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Na = [tex]\frac{1.5}{1.5}=1[/tex]

For B = [tex]\frac{1.5}{1.5}=1[/tex]

For O =[tex]\frac{3}{1.5}=2[/tex]

The ratio of Na: B: O= 1: 1: 2

Hence the empirical formula is [tex]NaBO_2[/tex]

Answer:

dunno

Explanation:

Answer:

35.453 amu.

Explanation:

From the question given above, the following data were obtained:

Isotope A (Cl–35):

Abundance (A%) = 75.77%

Mass of A = 34.969 amu

Isotope B (Cl–37):

Abundance (B%) = 24.23%

Mass of B = 36.966 amu

AVERAGE atomic mass of Cl =?

The average atomic mass of Cl can be obtained as follow:

Average = [(Mass of A × A%)/100] + [(Mass of B × B%)/100]

Average = [(34.969 × 75.77)/100] + [(36.966 × 24.23)/100]

Average = 26.496 + 8.957

Average atomic mass of Cl = 35.453 amu.

Answer:

The answer is c, Clouds fill with moisture and get too hravy

Answer:

Molarity = 0.3 M

Explanation:

Given data:

Moles of NaOH = 0.720 mol

Volume of water = 2.40 L

Molarity = ?

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

Molarity = 0.720 mol / 2.40 L

Molarity = 0.3 mol/L

Molarity = 0.3 M

Answer:

What's the rest of the question

Explanation:

Answer:

843.75

Explanation:

ok so the moon radius is 1,350,000meters right so meters to miles

1 mile=1600meters

1,350,000/1600=843.75

843.75 miles

Hope that helps!

Merry Christmas!

Answer: In the case of water, we know that the O-H covalent bond is polar, due to the different electronegativities of hydrogen and oxygen. Since there are two O-H bonds in water, their bond dipoles will interact and may result in a molecular dipole which can be measured.

Explanation:

Answer:

carbon mono oxide CO

Explanation:

as both are in 1:1 ratio so CO is formed

Answer:

P4O10

Explanation:

Answer:

C. are found in the nucleus of an atom

Explanation:

Protons and Neutrons are always located at the center of the atom also called as the nuclei of the atom and constitute nearly all the mass of the atom, while the electrons revolve around them in orbits called shells. Only protons have a positive charge and electrons have negative charges, due to which only protons and electrons attract each other, while neutrons lack charges or have 'Neutral Charges'.

Answer:

i not sure but i searched it up and it said Q=104525J

Explanation:

Answer:

0.9612 g

Explanation:

First we calculate how many moles are there in 3.00 g of CCl₃F, using its molar mass:

Now, we need to calculate how many grams of N₂O would have that same number of molecules, or in other words, the same amount of moles.

Thus we calculate how many grams would 0.0218 moles of N₂O weigh, using the molar mass of N₂O :

The  mass of dinitrogen monoxide is 0.9612 g

The number of  moles are there in 3.00 g of CCl₃F, using its molar mass:

= 3.00 g CCl₃F ÷ 137.37 g/mol

= 0.0218 mol CCl₃F

Now the mass should be

=[ 0.0218 mol N₂O ×  44.013 g/mol

= 0.9612 g N₂O

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Answer:

True. Hope this helps. :)

Hi there! :)

[tex]\large\boxed{x = 2.49 L}[/tex]

Use the proportion for Charles' Law where:

[tex]\frac{v_{1}}{t_{1}}= \frac{v_{2}}{t_{2}}[/tex]

v1 = initial volume

t1 = initial temperature

v2 = final volume

v2 = final temperature

Substitute in the given values into the proportion:

v1 = 4.39 L

t1 = 44° C

t2 = 25°C

v2 = x L

Set up the proportion:

[tex]\frac{4.39}{44} = \frac{x}{25}[/tex]

Cross multiply:

[tex]25 * 4.39 = 44x\\\\109.75 = 44x\\\\x = 2.49 L[/tex]

Answer:

The new volume will be approximately 2.49 L.

Explanation:

We are given that a gas sample occupies 4.39 L at 44° C.

We are also given that we are going decrease the temperature to 25° C.

Charles' Law has a formula which shows the relationship between the volumes of a samples and the temperatures of the samples.

The proportion is shown as:

[tex]\displaystyle \bullet \ \ \ \frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex],

where V₁ and V₂ are the initial and final volume respectively and T₁ and T₂ are the initial and final temperature respectively.

This proportion can be used to determine what our unknown is. We need to know the final volume.

[tex]\displaystyle \frac{4.39 \ \text{L}}{44 \ \text{C}} = \frac{\text{x}}{25 \ \text{C}}\\\\\\\frac{4.39}{44} = \frac{x}{25}\\\\\\4.39 \times 25 = 44 \times x\\\\\\44x = 109.75\\\\\\\frac{44x}{44}=\frac{109.75}{44}\\\\\\x = 2.49432 \approx \boxed{2.49 \ \text{L}}[/tex]

Therefore, if we reduce the temperature to 25 °C, we will be reducing the volume of the substance to 2.49 L.

Answer:

18.84 g of silver.

Explanation:

We'll begin by calculating the number atoms present in 5.59 g of sulphur. This can be obtained as follow:

From Avogadro's hypothesis,

1 mole of sulphur contains 6.02×10²³ atoms.

1 mole of sulphur = 32 g

Thus,

32 g of sulphur contains 6.02×10²³ atoms.

Therefore, 5.59 g of sulphur will contain = (5.59 × 6.02×10²³) / 32 = 1.05×10²³ atoms.

From the calculations made above, 5.59 g of sulphur contains 1.05×10²³ atoms.

Finally, we shall determine the mass of silver that contains 1.05×10²³ atoms.

This is illustrated below:

1 mole of silver = 6.02×10²³ atoms.

1 mole of silver = 108 g

108 g of silver contains 6.02×10²³ atoms.

Therefore, Xg of silver will contain 1.05×10²³ atoms i.e

Xg of silver = (108 × 1.05×10²³)/6.02×10²³

Xg of silver = 18.84 g

Thus, 18.84 g of silver contains the same number of atoms (i.e 1.05×10²³ atoms) as 5.59 g of sulfur

Protons,neutrons,electrons