AN ARRANGEMENT OF ELEMENTS INTO ROWS and columns according to similiarties in their properties

Which element has the smallest atomic weight? Which element has the greatest atomic weight?

Aleks [24]

The lightest is hydrogen and the heaviest is hassium.

6 0

3 years ago

Not enough food to eat is _____<br><br> A: Abiotic<br><br> B: Biotic

elena-14-01-66 [18.8K]

Not enough food to eat is A: Abiotic

8 0

3 years ago

Read 2 more answers

When the following equation is balanced using the smallest possible integers, what is the number in front of the substance in bo

weeeeeb [17]

The balanced equation :

8Al + 3Fe₃O₄→ 4Al₂O₃ + 9Fe

You just have to look for which element is in bold because the question is not clear

<h3>Further explanation</h3>

Equalization of chemical reaction equations can be done using variables. Steps in equalizing the reaction equation:

1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.

2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product

3. Select the coefficient of the substance with the most complex chemical formula equal to 1

Reaction

Al + Fe₃O₄→ Al₂O₃ + Fe

give a coefficient

aAl + Fe₃O₄→ bAl₂O₃ + cFe

O, left=4, right=3b⇒3b=4⇒b=4/3

Al, left=a, right=2b⇒a=2b⇒a=2.4/3⇒a=8/3

Fe, left=3, right=c⇒c=3

The equation becomes :

8/3Al + Fe₃O₄→ 4/3Al₂O₃ + 3Fe x3

8Al + 3Fe₃O₄→ 4Al₂O₃ + 9Fe

7 0

3 years ago

How many moles and numbers of ions of each type are present in the following aqueous solution? 63.1 mL of 1.85 M magnesium chlor

Alexxandr [17]

Answer:

We have 0.117 moles Mg^2+ and 0.234 moles Cl-

The number of Mg^2+ ions = 7.04 *10^22 ions

The number of Cl- ions = 1.4*10^23 ions

Explanation:

Step 1: Data given

Volume of magnesium chloride = 63.1 mL

Concentration of solution = 1.85 M

Step 2: Calculate moles MgCl2

Moles MgCl2 = concentration * volume

Moles MgCl2 = 1.85 M * 0.0631 L

Moles MgCl2 = 0.117 moles

Step 3: Calculate moles of ions

MgCl2 → Mg^2+ + 2Cl-

For 1 mol MgCl2 we have 1 mol Mg^2+ and 2 moles Cl-

For 0.117 moles MgCl2 we have 0.117 moles Mg^2+ and 2*0.117 = 0.234 moles Cl-

The number of Mg^2+ ions = 0.117 * 6.022 *10^23 = 7.04 *10^22 ions

The number of Cl- ions = 1.4*10^23 ions

4 0

4 years ago

Suppose you heat a metal object with a mass of 31.7 g to 96.5 oC and transfer it to a calorimeter containing 100.0 g of water at

OverLord2011 [107]

Answer: The specific heat of the metal in $1.34J/g^0C$

Explanation:

$Q_{absorbed}=Q_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]$

where,

$m_1$ = mass of metal = 31.7 g

$m_2$ = mass of water = 100.0 g

$T_{final}$ = final temperature = $24.6^oC$

$T_1$ = temperature of metal = $96.5^oC$

$T_2$ = temperature of water = $17.3^oC$

$c_1$ = specific heat of metal= ?

$c_2$ = specific heat of water = $4.184J/g^0C$

Now put all the given values in equation (1), we get

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$

$-(31.7\times c_1\times (24.6-96.5)^0C)=(100.0\times 4.184\times (24.6-17.3)]$

$c_1=1.34J/g^0C$

Therefore, the specific heat of the metal in $1.34J/g^0C$

4 0

3 years ago