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Schach [20]
3 years ago
10

which of the following elements has the lowest first ionization eneygy A, potassium B, sodium C, calcium D, Argon​

Chemistry
1 answer:
Jlenok [28]3 years ago
6 0

Answer:

Argon

Explanation:

Because, it is positioned down on the periodic table

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Each substance on the left side of the arrow in a chemical equation is a ____.
S_A_V [24]

Answer:

<em>Each substance on the left side of the arrow in a chemical equation is a </em><u>reactant.</u>

Explanation:

The chemical changes (chemical reactions) are represented by <em>chemical equations.</em>

The chemical equations show the starting substances and the final substances.

The starting substances of the reaction are the reactants.

A single <em>arrow</em> is used to indicate the direction of the reaction, the <em>reactants </em>are placed <em>on the left side</em> of the equation, and the products appear on the right side.

This sketch shows that:

                              reactants        yield          products                                        

general form              A + B             →               C + D

                     

example                     Na + Cl          →               NaCl

7 0
4 years ago
SC. 8.P.8.9
ELEN [110]

Answer:

C

Explanation:

8 0
3 years ago
Explain why we see the pattern between predator and prey​
nlexa [21]

Answer:

The food chain starts as prey to a preditor and then when the preditor dies the prey eat the preditor

Explanation:

Just that

6 0
3 years ago
What volume of a 1.08M KI solution would contain 0.642 moles of KI?
il63 [147K]

Answer: Volume is 0.594 litres

Explanation: Amount of moles = volume · concentration

n = Vc  and V = n/c = 0.642 mol / 1.08 mol/l = 0,59344..  l

8 0
3 years ago
There are 0.55 moles of carbon dioxide gas in a 15.0 L container. This container is at a temperature of 300 K. What is the press
sertanlavr [38]

Answer:

\large \boxed{\text{D.) 91 kPa}}

Explanation:

We can use the Ideal Gas Law — pV = nRT

Data:

V = 15.0 L

n = 0.55 mol

T = 300 K

Calculation:

\begin{array}{rcl}pV & =& nRT\\p \times \text{15.0 L} & = & \text{0.55 mol} \times \text{8.31 kPa$\cdot$ L$\cdot$K$^{-1}$mol$^{-1}\times$ 300 K}\\15.0p & = & \text{1370 kPa}\\p & = & \textbf{91 kPa}\end{array}\\\text{The pressure in the container is $\large \boxed{\textbf{91 kPa}}$}

6 0
3 years ago
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