All categories

3 years ago

What must models and theories do in order to remain valid

Chemistry

1 answer:

raketka [301]3 years ago

5 0

There must not be credible contrary evidence to the ideas suggested by the theory.

You might be interested in

Calculate the mass percent of titanium in the mineral ilmenite, FeTiO3. What mass of ilmenite (in grams) is required if you wish

muminat

We are given the mineral ilmenite with a molecular formula of FeTiO3. The molar mass of the compound is equal to 151.72 g/mol. The percent titanium thus is 31.55 percent.when 750 grams of titanium is present, the mass of ilmenite is equal to 2377.18 grams

4 0

3 years ago

How many moles of CoH12O6 are consumed<br> if 6 moles of O2 are consumed?

Rashid [163]

Answer:

Explanation:

There are 3 moles O2 in 1 mole CoH12O6 so 2 moles are consumed

7 0

3 years ago

Find the molecular mass of MgSO4

rewona [7]

120.38 g/mol is the answer

7 0

3 years ago

Read 2 more answers

What is the length of the pencil in cm?

Levart [38]

Answer:

16 cm

Explanation:

Because when you look at the ruler, it shows the length as 16 cm

7 0

2 years ago

A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$