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wolverine [178]
3 years ago
9

What do elements in the same group on the periodic table have in common?

Chemistry
1 answer:
gtnhenbr [62]3 years ago
8 0
They all have the same number is valence electrons.
You might be interested in
How many molecules of CF₂Cl₂ are in 45.7 grams of CF₂Cl₂? (Show work)
posledela

Answer:

The answer to your question is: 2.20 x 10 ²³ molecules

Explanation:

Data

mass = 45.7 g

molecules of CF₂Cl₂ = ?

Process

1.- Calculate the mass number of CF₂Cl₂

   C = 12    F =2 x 19    Cl = 2 x 35.5

   total = 12 + 38 + 71

   total = 121 g

2.- Use the Avogradro's number to solve the problem

                       121 g -------------------  6.023 x 10²³ molecules of CF₂Cl₂

                        45.7 g ---------------     x

                        x = (45.7 x 6.023 x 10²³) / 121

                        x = 2.75 x 10²⁵ / 121

                        x = 2.20 x 10 ²³ molecules

4 0
3 years ago
What is the product of the unbalanced equation below?
kvasek [131]

Answer:

I think B

Explanation:

2Ca + O2 -> 2CaO

4 0
2 years ago
for the reaction 2Fe + O2 = 2FeO, how many grams of iron oxide are produced from 8.00 mol of iron? when o2 is an excess
Luba_88 [7]

Answer:

2Fe + O₂ -------------------> 2FeO

8 mol Fe produce

8 mol Fe * 2 mol FeO / 2 mol Fe = 8 mol FeO

Mass of FeO = 8 mol FeO * 71.85 g/mol = 574.8 grams FeO

Explanation:

Having 8 mol of Iron means 8 moles of iron oxide can be produced. Each mole of iron oxide has a molecular weight of 71.85 grams. Therefore, 8 moles of iron oxide should weight 574.8 grams.

7 0
3 years ago
What is the correct ionic charge for a barium atom
AVprozaik [17]
+2

Barium has a positive charge of 2
3 0
3 years ago
Biacetyl, the flavoring that makes margarine taste "just like butter," is extremely stable at room temperature, but at 200°C it
Sholpan [36]

Answer:

3.91 minutes

Explanation:

Given that:

Biacetyl breakdown with a half life of 9.0 min after undergoing first-order reaction;

As we known that the half-life for first order is:

t__{1/2}}= \frac{0.693}{k}

where;

k = constant

The formula can be re-written as:

k = \frac{0.693}{t__{1/2}}

k = \frac{0.693}{9.0 min}

k = 0.077 min^{-1}

Let the initial amount of butter flavor in the food be (N_0) = 100%

Also, the amount of butter flavor retained at 200°C (N_t)= 74%

The rate constant k = 0.077 min^{-1}

To determine how long can the food be heated at this temperature and retain 74% of its buttery flavor; we use the formula:

\frac{N_t}{N_0}= -kt

t = - (\frac{1}{k}*In\frac{N_t}{N_0}  )

Substituting our values; we have:

t = - (\frac{1}{0.077}*In\frac{74}{100}  )

t = 3.91 minutes

∵ The time needed for the food to be heated at this temperature and retain 74% of its buttery flavor is 3.91 minutes

4 0
3 years ago
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