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3 years ago

Where are the metals located on the periodic table? the nonmetals?

Chemistry

1 answer:

vovikov84 [41]3 years ago

6 0

Answer:

The metals are located on the left and the non-metals are located on the right.

Explanation:

Hope this helps :)

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The number of protons present in ATOM 1 is______

Ira Lisetskai [31]

Three protons and three electrons. The protons are the plus signs because protons are positively charged, and the electrons are the minus signs because electrons are negatively charged. Also, the neutrons are the circles. Neutrons are negatively charged

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3 years ago

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How many grams of a solution that is 50% by weight NaOH

kkurt [141]

Answer:

12 g

Explanation:

We want to prepare 1 L of a 0.15 M NaOH solution. The moles of NaOH required are:

0.15 mol/L × 1 L = 0.15 mol

The molar mass of NaOH is 40.00 g/mol. The mass corresponding to 0.15 mol is:

0.15 mol × (40.00 g/mol) = 6.0 g

The starting solution is 50% by weight. The mass of solution that contains 6.0 g of NaOH is:

6.0 g NaOH × (100 g solution/50 g NaOH) = 12 g solution

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4 years ago

How many moles are in 60.0 grams of water?

Mashutka [201]

<h3>Answer:</h3>

3.33 mol H₂O

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

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

60.0 g H₂O (Water)

<u>Step 2: Identify Conversions</u>

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 60.0 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})$
Multiply: $\displaystyle 3.32963 \ mol \ H_2O$

<u>Step 4: Convert</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

3.32963 mol H₂O ≈ 3.33 mol H₂O

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3 years ago

Gold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver-gold

Sonja [21]

Answer:

The Number of gold atoms are = $N_{Au}=1.81718*10^{22}\ atoms/cm^3$

Explanation:

The formula we are going to use is:

$N_{Au}=\frac{N_A*C_{Au}}{{\frac{C_{Au}A_{Au} }{\rho_{Au}}}+\frac{A_{Au}}{\rho_{Ag}}(100-C_{Au})}$

Where:

$N_{Au}$ are number of gold atoms.

$N_A$ is Avogadro Number.

$C_{Au}$ is the amount of gold.

$A_{Au}$ is the atomic weight of gold.

$\rho_{Au}$ is the density of gold.

$\rho_{Ag}$ is the density of silver.

$C_{Ag}$ is the amount of silver.

$N_{Au}=\frac{6.023*10^{23}*45\%wt}{\frac{45\%wt*196.97}{19.32}+\frac{196.97}{10.49}(100-45\%wt)}\\ N_{Au}=1.81718*10^{22}\ atoms/cm^3$

The Number of gold atoms are = $N_{Au}=1.81718*10^{22}\ atoms/cm^3$

5 0

4 years ago

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3 years ago

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