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3 years ago

Which mineral can be found in the rocks phyllite, sandstone, and granite?

Chemistry

2 answers:

shusha [124]3 years ago

8 0

Quartz

I would appreciate marking me Brainliest!

Dominik [7]3 years ago

6 0

Answer:

1 quartz

Explanation:

pls tell me if I'm wrong

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What is the balanced equation for H3PO4=H4P2O7+H2O

podryga [215]

Answer:

2H3PO4=H4P2O7+H2O

Explanation:

2H3PO4=H4P2O7+H2O

5 0

3 years ago

Ammonia is produced industrially via this exothermic reaction.

Cloud [144]

Answer:

high pressure of 200-300 atm.

low temp. of between 400-500 degrees celsius:this is for continuous development of ammonia since it decomposes at high temp fathermore the reaction is exothermic

a catalyst to speed up the rate of reaction:i guess it is finely divided iron impregnated in aluminium oxide

platinum can be used as a catalyst but it is easily poisoned

hope it helps

Explanation:

6 0

3 years ago

How many moles of oxygen are needed to react with 23.8 moles of aluminum?

antoniya [11.8K]

Answer:

17.85moles

Explanation:

Check attachment

7 0

3 years ago

A sucrose packet contains 4.0g sucrose C12H22O11, how many moles of sucrose does the packet contain?

SpyIntel [72]

Answer:

And we have to calculate the number of moles of sucrose present in a lb mass of sucrose: Moles of sucrose=454⋅g342.30⋅g⋅mol−1=1.33⋅mol .

Explanation:

<u>Brainliest</u><u> </u><u>Answer </u><u>Pls</u>

8 0

3 years ago

A volume of 105 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If

NemiM [27]

Answer:

<h3>25.0 grams is the mass of the steel bar.</h3>

Explanation:

Heat gained by steel bar will be equal to heat lost by the water

$Q_1=-Q_2$

Mass of steel= $m_1$

Specific heat capacity of steel = $c_1=0.452 J/g^oC$

Initial temperature of the steel = $T_1=2.00^oC$

Final temperature of the steel = $T_2=T=21.50^oC$

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2= 105 g$

Specific heat capacity of water= $c_2=4.18 J/g^oC$

Initial temperature of the water = $T_3=22.00^oC$

Final temperature of water = $T_2=T=21.50^oC$

$Q_2=m_2c_2\times (T-T_3)-Q_1=Q_2(m_1c_1\times (T-T_1))=-(m_2c_2\times (T-T_3))$

On substituting all values:

$(m_1\times 0.452 J/g^oC\times (21.50^o-2.00^oC))=-(105 g\times 4.18 J/g^oC\times (21.50^o-22.00^o))\\\\m_1*8.7914=241.395\\\\m_1=\frac{219.45}{8.7914} \\\\m_1=24.9\\\\ \approx25 \texttt {grams}$

<h3>25.0 grams is the mass of the steel bar.</h3>

3 0

3 years ago

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