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3 years ago

1. Identify whether each of the elements is a metal, non-metal, or metalloid:

Chemistry

2 answers:

lina2011 [118]3 years ago

8 0

Answer:

Ruthenium → metalloid
Neon → non-metal
Germanium → metalloid
Neptunium → metal
Iodine → non-metal
Strontium → metal
Caesium → metal
Europium → metal

Liono4ka [1.6K]3 years ago

3 0

Answer:

neon - non metal

Explanation:

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sodium reacts with chlorine gas to form sodium chloride. if you have 60 L of chlorine gas at STP and 30 g of sodium, how many gr

Stels [109]

Answer:

75.9 grams of salt

Explanation:

The reaction is the following:

2Na(s) + Cl₂(g) → 2NaCl(s) (1)

We have:

m(Na): the mass of sodium = 30 g

V(Cl₂): the volume of the chlorine gas at STP = 60 L

So, to find the mass of NaCl we need to calculate the number of moles of Na and Cl₂.

$n_{Na} = \frac{m}{A_{r}} = \frac{30 g}{22.99 g/mol} = 1.30 moles$

The number of moles of Cl₂ can be found by the Ideal gas law equation:

$PV = n_{Cl_{2}}RT$

Where:

P: is the pressure = 1 atm (at STP)

R: is the gas constant = 0.082 L*atm/(K*mol)

T: is the temperature = 273 K (at STP)

$n_{Cl_{2}} = \frac{PV}{RT} = \frac{1 atm*60 L}{0.082 L*atm/(K*mol)*273 K} = 2.68 moles$

Now we need to find the limiting reactant. From the stoichiometric relation between Na and Cl₂ (equation 1), we have that 2 moles of Na react with 1 mol of Cl₂, so:

$n_{Na} = \frac{2 moles Na}{1 mol Cl_{2}}*2.68 moles Cl_{2} = 5.36 moles$

Since we have 1.30 moles of Na, the limiting reactant is Na.

Finally, we can find the number of moles of NaCl and its mass.

$n_{NaCl} = n_{Na} = 1.30 moles$

$m_{NaCl} = n_{NaCl}*M = 1.30 moles*58.44 g/mol = 75.9 g$

Therefore, would be formed 75.9 grams of salt.

I hope it helps you!

3 0

2 years ago

How many electrons are there in second principal energy<br> level(n=2) of a phosphorus atom?

salantis [7]

Answer:

8 electrons

Explanation:

Phosphorous is the element of group 15 and third period. The atomic number of chlorine is 15 and the symbol of the element is P.

The electronic configuration of the element phosphorus is:-

$1s^22s^22p^63s^23p^3$

Thus, from the electronic configuration shown above,

In n = 2, the total number of electrons are 8 which are present in $2s^22p^6$

6 0

3 years ago

Pesticide concentrations in the Rhine River between Germany and France between 1969 and 1975 averaged 0.55 mg/L of hexachloroben

sertanlavr [38]

Answer:

1.93×10⁻³ mmoles/L of C₆Cl₆; 1.58×10⁻⁴ mmoles/L of C₁₂H₈Cl₆O; 3.51×10⁻³ mmoles/L of C₆H₆Cl₆

Explanation:

We have to find out the molar mass of each pesticide to calculate the moles, and then the milimoles

C₆Cl₆ → 12. 6 + 35.45 .6 = 284.7 g/m

C₁₂H₈Cl₆O → 12 . 12 + 8.1 + 35.45 .6 + 16 = 380.7 g/m

C₆H₆Cl₆ → 12.6 + 6.1 + 35.45 .6 = 290,7 g/m

Let's convert mg to g (/1000)

0.55 mg / 1000 = 5.5×10⁻⁴ g

0.060 mg / 1000 = 6×10⁻⁵ g

1.02 mg / 1000 = 1.02×10⁻³ g

Now we can know the moles (mass / molar mass)

5.5×10⁻⁴ g / 284.7 g/m = 1.93×10⁻⁶ moles of C₆Cl₆

6×10⁻⁵ g / 380.7 g/m = 1.58×10⁻⁷ moles of C₁₂H₈Cl₆O

1.02×10⁻³ g / 290,7 g/m = 3.51×10⁻⁶ moles of C₆H₆Cl₆

Milimoles = Mol . 1000

1.93×10⁻⁶ . 1000 = 1.93×10⁻³ mmoles of C₆Cl₆

1.58×10⁻⁷ . 1000 = 1.58×10⁻⁴ mmoles of C₁₂H₈Cl₆O

3.51×10⁻⁶ . 1000 = 3.51×10⁻³ mmoles of C₆H₆Cl₆

6 0

3 years ago

What is the relationship between wind and ocean waves

S_A_V [24]

With creates the waves. The stronger the wind the larger and more powerful the waves are.

8 0

3 years ago

Hat is the oxidation state of each element in the compound CaSO4? Include + or - in your answers as appropriate.

quester [9]

As per the rule, oxidation number of alkaline earth metals in their compounds is +2. Oxidation number of oxygen in it's compounds is -2(except peroxides) and the sum of oxidation numbers of all the elements of a neutral compound is zero.

7 0

2 years ago