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3 years ago

What element is denoted by 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d1

Chemistry

2 answers:

Anestetic [448]3 years ago

7 0

Element 39....Yttrium

alexgriva [62]3 years ago

4 0

The answer is Zr (Zirconium)

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15.1 L N2 at 25 °C and 125 kPa and 44.3 L O2 at 25 °C and 125 kPa were transferred to a tank with a volume of 6.25 L. What is th

Y_Kistochka [10]

Answer:

The total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.

Explanation:

For N2,

Pressure(P₁)=125 kPa

Volume(V₁)=15·1 L

Temperature (T₁)=25°C=25+273 K=298 K

Similarly, for Oxygen,

Pressure(P₂)= 125 kPa

Volume(V₂)= 44.3 L

Temperature(T₂)=25°C= 298 K

Then, for the mixture,

Volumeof the mixture( V)= 6.25 L

Pressure(P)=?

Temperature (T)= 51°C = 51+273 K=324 K

Then, By Combined gas laws,

$\frac{P_{1} V_{1} }{T_{1} } +\frac{P_{2} V_{2} }{T_{2} } =\frac{PV}{T}$

or, $\frac{15.1*125}{298} +\frac{44.3*125}{298} =\frac{P*6.25}{324}$

or, $6.34+18.58=\frac{P*6.25}{324}$

or, $P=\frac{24.92*324}{6.25}$

∴P=1291.85 kPa

So the total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.

3 0

3 years ago

How do atoms form a new substance?

lubasha [3.4K]

By sharing electeons with each other
if they lose or gain electrons then they only form ions
they cannot lose neutrons as they are locked inside the nucleus

8 0

3 years ago

How many moles in 4.93 x 10E23 atoms of silver?

leva [86]

<h3>Answer:</h3>

0.819 mol Ag

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

Using Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

4.93 × 10²³ atoms Ag

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

Set up: $\displaystyle 4.93 \cdot 10^{23} \ atoms \ Ag(\frac{1 \ mol \ Ag}{6.022 \cdot 10^{23} \ atoms \ Ag})$
Divide: $\displaystyle 0.818665 \ mol \ Ag$