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2 years ago

Describe the characteristics that identify a sample of matter as being a substance.

1 answer:

6 0

No matter how big or little a material is, its properties remain the same. Size, form, color, and mass are the qualities that distinguish a material from non-substances. All of these qualities, including size, shape, color, and mass, can be seen and measured. Some characteristics are physical, while others are chemical. Physical characteristics include mass, volume, density, and color. Viscosity and solubility are examples of chemical qualities.

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How does the neutron number in each compare

tester [92]

Answer:

Neutrons are all identical to each other, just as protons are. Atoms of a particular element must have the same number of protons but can have different numbers of neutrons.

Explanation:

Since the vast majority of an atom's mass is found its protons and neutrons, subtracting the number of protons (i.e. the atomic number) from the atomic mass will give you the calculated number of neutrons in the atom. In our example, this is: 14 (atomic mass) – 6 (number of protons) = 8 (number of neutrons).

5 0

3 years ago

An excess of mg(s) is added to 100.ml of 0.400 m hcl. at 0c and 1 atm pressure, what volume of h 2 (g) can be obtained?

ra1l [238]

The balanced equation for the reaction between Mg and HCl is as follows
Mg + 2HCl --> MgCl₂ + H₂
stoichiometry of HCl to H₂ is 2:1

number of HCl moles reacted - 0.400 mol/L x 0.100 L = 0.04 mol of HCl
since Mg is in excess HCl is the limiting reactant
number of H₂ moles formed - 0.04/2 = 0.02 mol of H₂

we can use ideal gas law equation to find the volume of H₂
PV = nRT
where
P - pressure - 1 atm x 101 325 Pa/atm = 101 325 Pa
V - volume
n - number of moles - 0.02 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 0 °C + 273 = 273 K
substituting these values in the equation

101 325 Pa x V = 0.02 mol x 8.314 Jmol⁻¹K⁻¹ x 273 K
V = 448 x 10⁻⁶ m³
V = 448 mL
therefore answer is
c. 448 mL

5 0

3 years ago

Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)

Brums [2.3K]

Answer:1) Volume of $AgNO_3$ required is 55.98 mL.

2) 0.62577 grams of $Ag_3PO_4$ is produced.

Explanation:

$3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)$

1) Molarity of $AgNO_3,M_1=0.225 M$

Volume of $AgNO_3.V_1=?$

Molarity of $Na_3PO_4,M_2=0.135 M$

Volume of $Na_3PO_4,V_2=31.1 mL=0.0311 L$

$Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}$

$\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles$

According to reaction, 1 mole of $Na_3PO_4$ reacts with 3 mole of $AgNO_3$ , then, 0.0041985 moles of $Na_3PO_4$ will react with:

$\frac{3}{1}\times 0.0041985$ moles of $AgNO_3$ that is 0.0125955 moles.

$M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}$

$V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL$

Volume of $AgNO_3$ required is 55.98 mL.

$Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}$

Number of moles of $AgNO_3=0.195\times 0.023 L=0.004485 moles$

According to reaction, 3 moles of $AgNO_3$ gives 1 mole of $Ag_3PO_4$ , then 0.004485 moles of $AgNO_3$ will give: $\frac{1}{3}\times 0.004485$ moles of $Ag_3PO_4$ that is 0.001495 moles.

Mass of $Ag_3PO_4$ =

Moles of $Ag_3PO_4$ × Molar Mass of $Ag_3PO_4$

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of $Ag_3PO_4$ is produced.

7 0

4 years ago

Bill Nye: Chemical Reactions

anygoal [31]

Dang don’t know I need my answer

7 0

3 years ago

Read 2 more answers

10g of hyrogen react with excess of oxygen gas according to the equation:

Degger [83]

Answer:

Volume of O₂ = 56 dm³

mass of water vapors (H₂O) = 90 g

Explanation:

Data Given:

mass of Oxygen = 10 g

Volume of Oxygen = ?

mass of the water vapor = ?

Reaction Given:

2H₂+O₂---->2H₂O

Solution:

First we have to look at the reaction for the information required

2H₂ + O₂ -------> 2H₂O

2 mol 1mol 2 mol

now convert moles to grams

molar mass of H₂ = 2(1) = 2 g/mol

molar mass of O₂ = 2(16) = 32 g/mol

molar mass of H₂0 = 2(1) + 16 = 18 g/mol

So the masses will be

2H₂ + O₂ -------> 2H₂O

2 mol (2 g/mol) 1mol (32 g/mol) 2 mol (18 g/mol)

4 g 32 g 36 g

So now we know that

4 g of hydrogen combine with 32 g of Oxygen and give 36 g of water vapors.

By using above information

First we find the volume of Oxygen:

For this first we find mass and then moles of Oxygen

As we know

if 4 g of hydrogen combine with 32 g of Oxygen then how much oxygen will react with 10 g of hydrogen

Apply unity formula

4 g of hydrogen H₂ ≅ 32 g of Oxygen O₂

10 g of hydrogen H₂ ≅ X g of Oxygen O₂

by doing Cross multiplication

g of Oxygen O₂ = 32 g x 10 g / 4 g

g of Oxygen O₂ = 80 g

So,

mass of oxygen = 80 g

now find moles of oxygen

formula used:

no. of moles = mass in grams/ molar mass . . . . . . (1)

Put values in above equation 1

no. of moles = 80 g / 32 g/mol

no. of moles = 2.5

Now to find volume of oxygen

Formula used

Volume of O₂ = no. of moles x molar volume (22.4 dm³/ mol) . . . . . . (2)

Put values in equation 2

Volume of O₂ = 2.5 moles x 22.4 dm³/mol

Volume of O₂ = 56 dm³

______________________

Now to find mass of water vapors

As we now

if 4 g of hydrogen produce 36 g of water vapors then how much water vapor will produce from 10 g of hydrogen

Apply unity formula

4 g of hydrogen H₂ ≅ 36 g of water vapors (H₂O)

10 g of hydrogen H₂ ≅ X g of water vapors (H₂O)

by doing Cross multiplication

g of water vapors (H₂O) = 36 g x 10 g / 4 g

g of water vapors (H₂O) = 90 g

So,

mass of water vapors (H₂O) = 90 g

5 0

3 years ago