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2 years ago

The most common carbonate rock is A) dolomite. B) halite. 35) C) limestone. D) calcite.

Chemistry

1 answer:

babunello [35]2 years ago

6 0

Answer:

The correct option is : C) limestone.

Explanation:

Carbonate rocks are a type of sedimentary rocks. The carbonate rocks are composed of carbonate minerals. The carbonate minerals are the minerals containing carbonate ion (CO₃²⁻).

The most common type of carbonate rock is limestone. Limestone is composed of the minerals calcite and aragonite, which have a different crystal form of calcium carbonate.

Therefore, Limestone is the most common type of carbonate rock.

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Copper is commonly mined as an ore with a variable percent composition of copper (II) sulfide. This ore is also sometimes referr

pantera1 [17]

Answer:

$m_{CuO}=93.6gCuO$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$CuS+\frac{3}{2} O_2\rightarrow CuO+SO_2$

Thus, given the 1.00-kg of 12.5% ore, we can compute the theoretical yield of copper (II) oxide via stoichiometry:

$m_{CuO}^{theoretical}=1.00kgCuS*\frac{1000gCuS}{1kgCuS} *\frac{12.5gCuS}{100gCuS} *\frac{1molCuS}{95.6gCuS} *\frac{1molCuO}{1molCuS} *\frac{79.5gCuO}{1molCuO} \\\\m_{CuO}^{theoretical}=103.95gCuO$

Whereas the third factor accounts for the percent purity of the covellite. Then, given the percent yield, we can compute the actual yield by:

$m_{CuO}=103.95gCuO*0.9\\\\m_{CuO}=93.6gCuO$

Regards.

6 0

3 years ago

La valencia que utiliza el hidrógeno al formar un hidruro metálico es

Serjik [45]

Answer: Esta tendencia es tan regular que el poder de combinación, o valencia, de un elemento se definió una vez como el número de átomos de hidrógeno unidos al elemento en su hidruro. El hidrógeno es el único elemento que forma compuestos en los que los electrones de valencia están en la capa n = 1.

Explanation:

¡Espero que esto ayude!

4 0

2 years ago

A compound distributes between benzene (solvent 1) and water (solvent 2) with a distribution coefficient, K = 2.7. If 1.0g of th

mote1985 [20]

Explanation:

The given data is as follows.

Solvent 1 = benzene, Solvent 2 = water

$K_{p}$ = 2.7, $V_{S_{2}}$ = 100 mL

$V_{S_{1}}$ = 10 mL, weight of compound = 1 g

Extract = 3

Therefore, calculate the fraction remaining as follows.

$f_{n} = [1 + K_{p}(\frac{V_{S_{2}}}{V_{S_{1}}})]^{-n}$

= $[1 + 2.7(\frac{100}{10})]^{-3}$

= $(28)^{-3}$

= $4.55 \times 10^{-5}$

Hence, weight of compound to be extracted = weight of compound - fraction remaining

= 1 - $4.55 \times 10^{-5}$

= 0.00001

or, = $1 \times 10^{-5}$

Thus, we can conclude that weight of compound that could be extracted is $1 \times 10^{-5}$ .

7 0

2 years ago

Read 2 more answers

If 16.4 grams of calcium nitrate is heated as shown in the reaction:

vfiekz [6]

Answer:

7.2

Explanation:

you first have to find the number of moles of nitrogen dioxide by using the number of moles for calcium nitrate and the mole to mole ratios

number of moles of calcium nitrate=mass/mm

=16.4/102

=0.16g/mol

then you use the mole to mole ratios

2 : 4

0.16: x

2x/2=0.64/2

x=0.32g/moles of nitrogen dioxide

then you use the formula for the volume

v=22.4n

=22.4×0.32

=7.2

I hope this helps

4 0

2 years ago

Which process is characterized by the movement of particles?.

Serjik [45]

Answer: Diffusion is the movement of particles from a high to low particle concentration, while osmosis is the movement of water from a high to a low water concentration.

Explanation:

6 0

2 years ago

Read 2 more answers