The term biodiversity means the variety of organism (true or false)

A 1.000 g sample of an unknown hydrate of cobalt chloride is gently dehydrated. The resulting mass is 0.546 g. The cobalt is iso

ira [324]

Answer:

Explanation:

From the 1 g sample you have:

0.546 grams of cobalt chloride

1-0.546=0.454 grams of water

Now:

1) The salt

Of the 0.546 g, 0.248 g are cobalt (Mr=58.9) and the rest id Cl (Mr=35.45):

$n_{Co}=\frac{0.248 g}{58.9 g/mol}=4.21*10^{-3}mol$

$n_{Cl}=\frac{0.298 g}{35.45g/mol}=8.4*10^{-3}mol$

Dividing:

$\frac{n_{Cl}}{n_{Co}}=\frac{8.4*10^{-3}mol}{4.21*10^{-3}mol}=2$

So the molecular formula will be:

$CoCl_2$

2) The water

The water's molecular weight is M=18 :

$n_{w}=\frac{0.454g}{18g/mol}=0.025 mol$

Bonding with the Co:

$\frac{n_{w}}{n_{Co}}=\frac{0.025mol}{4.21*10^{-3}mol}=6$

The complete formula of the hydrate:

$CoCl_2*6 H_2O$

8 0

4 years ago

Answer these please ASAP need help no idea how to do these

STALIN [3.7K]

Answer:

Explanation:

Cu:

Number of moles = Mass / molar masa

2 mol = mass / 64 g/mol

Mass = 128 g

Mg:

Number of moles = Mass / molar masa

0.5 mol = mass / 24 g/mol

Mass = g

Cl₂:

Number of moles = Mass / molar masa

Number of moles = 35.5 g / 24 g/mol

Number of moles = 852 mol

H₂:

Number of moles = Mass / molar mass

8 mol = Mass / 2 g/mol

Mass = 16 g

P₄:

Number of moles = Mass / molar masa

2 mol = mass / 124 g/mol

Mass = 248 g

O₃:

Number of moles = Mass / molar masa

Number of moles = 1.6 g /48 g/mol

Number of moles = 0.033 mol

H₂O

Number of moles = Mass / molar masa

Number of moles = 54 g / 18 g/mol

Number of moles = 3 mol

CO₂

Number of moles = Mass / molar masa

2 mol = mass / 124 g/mol

Mass = 248 g

NH₃

Number of moles = Mass / molar masa

Number of moles = 8.5 g / 17 g/mol

Number of moles = 0.5 mol

CaCO₃

Number of moles = Mass / molar masa

Number of moles = 100 g / 100 g/mol

Number of moles = 1 mol

a)

Given data:

Mass of iron(III)oxide needed = ?

Mass of iron produced = 100 g

Solution:

Chemical equation:

F₂O₃ + 3CO → 2Fe + 3CO₂

Number of moles of iron:

Number of moles = mass/ molar mass

Number of moles = 100 g/ 56 g/mol

Number of moles = 1.78 mol

Now we compare the moles of iron with iron oxide.

Fe : F₂O₃

2 : 1

1.78 : 1/2×1.78 = 0.89 mol

Mass of F₂O₃:

Mass = number of moles × molar mass

Mass = 0.89 mol × 159.69 g/mol

Mass = 142.124 g

100 g of iron is 1.78 moles of Fe, so 0.89 moles of F₂O₃ are needed, or 142.124 g of iron(III) oxide.

b)

Given data:

Number of moles of Al = 0.05 mol

Mass of iodine = 26 g

Limiting reactant = ?

Solution:

Chemical equation:

2Al + 3I₂ → 2AlI₃

Number of moles of iodine = 26 g/ 254 g/mol

Number of moles of iodine = 0.1 mol

Now we will compare the moles of Al and I₂ with AlI₃.

Al : AlI₃

2 : 2

0.05 : 0.05

I₂ : AlI₃

3 : 2

0.1 : 2/3×0.1 = 0.067

Number of moles of AlI₃ produced by Al are less so it will limiting reactant.

Mass of AlI₃:

Mass = number of moles × molar mass

Mass = 0.05 mol × 408 g/mol

Mass = 20.4 g

26 g of iodine is 0.1 moles. From the equation, this will react with 2 moles of Al. So the limiting reactant is Al.

c)

Given data:

Mass of lead = 6.21 g

Mass of lead oxide = 6.85 g

Equation of reaction = ?

Solution:

Chemical equation:

2Pb + O₂ → 2PbO

Number of moles of lead = mass / molar mass

Number of moles = 6.21 g/ 207 g/mol

Number of moles = 0.03 mol

Number of moles of lead oxide = mass / molar mass

Number of moles = 6.85 g/ 223 g/mol

Number of moles = 0.031 mol

Now we will compare the moles of oxygen with lead and lead oxide.

Pb : O₂

2 : 1

0.03 : 1/2×0.03 = 0.015 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 0.015 mol × 32 g/mol

Mass = 0.48 g

The mass of oxygen that took part in equation was 0.48 g. which is 0.015 moles of oxygen. The number of moles of Pb in 6.21 g of lead is 0.03 moles. So the balance equation is

2Pb + O₂ → 2PbO

6 0

3 years ago

hich quantity describes how spread out a set of data is? O A. Standard deviation O B. Percent error O C. Line of best fit D. Mea

juin [17]

The answer to this is B.

6 0

4 years ago

A waste treatment pond is 50 m long and 25 m wide, and has an average depth of 2 m. The density of the waste is 75.3lbm/ft^3 .Ca

umka21 [38]

Answer:

$W = 6.65 \cdot 10^{6} lbf$

Explanation:

To find the weight (W) of the pond contents first we need to use the following equation:

$W = m\cdot g$ (1)

Where m the mass and g is the gravity

Also, we have that the mass is:

$m = \rho*V$ (2)

Where ρ is the density and V the volume

We cand calculate the volume as follows:

$V = L*w*d$ (3)

Where L is the length, w is the wide and d is the depth

By entering equation (2) and (3) into (1) we have:

$W = \rho*L*w*d*g$

$W = 75.3 lbm/ft^{3}*50 m*25 m*2 m*9.81 m/s^{2}$

$W = 75.3 lbm/ft^{3}*\frac{(1 ft)^{3}}{(0.3048 m)^{3}}*\frac{0.454 kg}{1 lbm}*50 m*25 m*2 m*9.81 m/s^{2} = 2.96 \cdot 10^{7} N}*\frac{0.2248 lbf}{1 N} = 6.65\cdot 10^{6} lbf$

Therefore, the weight of the pond is 6.65x10⁶ lbf.

I hope it helps you!

6 0

4 years ago

How many valence electrons are in the methyl ammonium ion CH3NH3+

Anika [276]

Answer:

15

Explanation:

8 0

3 years ago