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laila [671]
3 years ago
5

How many grams of Sn will be created when 114 grams of SnO2 react with excess H2? SnO2 + H2 ->Sn + H2O

Chemistry
1 answer:
Snezhnost [94]3 years ago
5 0

Explanation:

CNN vfbtgnynynynynyuybunujy

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How many grams in 11.9 moles of sulfur? Avogadro’s number: 6.02x1023 atoms = 1 mole Molar mass of sulfur: 32.06 g sulfur = 1 mol
sergiy2304 [10]

Answer:

Mass = 0.37 g

Explanation:

Given data:

Number of moles of sulfur = 11.9 mol

Mass of sulfur in 11.9 mol = ?

Molar mass of sulfur = 32.06 g

Solution:

Number of moles = mass/molar mass

by putting values,

11.9 mol = mass/ 32.06 g/mol

Mass = 11.9 mol × 32.06 g/mol

Mass = 0.37 g

4 0
3 years ago
How do sediments form?
Valentin [98]
Sedimentary<span> rocks are formed when </span>sediment<span> is deposited out of air, ice, wind, gravity, or water flows carrying the particles in suspension</span>
6 0
3 years ago
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In molecular oxygen (O=O) which atom is partially positive?
saveliy_v [14]

the oxygen atom

Explanation:

Water is a molecular compound consisting of polar molecules that have a bent shape. The oxygen atom acquires a partial negative charge while the hydrogen atom acquires a partial positive charge.

4 0
3 years ago
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A mixture of H2 and water vapor is present in a closed vessel at 20°C. The total pressure of the system is 755.0 mmHg.
Masja [62]

THE ANSWER IS: <u>737.5</u>

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7 0
3 years ago
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Acetylene burns in air according to the following equation: C2H2(g) + 5 2 O2(g) → 2 CO2(g) + H2O(g) ΔH o rxn = −1255.8 kJ Given
professor190 [17]

Answer:  -227 kJ

Explanation:

The balanced chemical reaction is,

C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(g)

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+ n_{H_2O}\times \Delta H_{H_2O})]-[(n_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]

where,

n = number of moles

\Delta H_{O_2}=0 (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]

-1255.8=[(-787)+(-241.8)]-[(1\times \Delta H_{C_2H_2})+(0)]

\Delta H_{C_2H_2}=-227kJ

Therefore, the enthalpy change for C_2H_2 is -227 kJ.

7 0
3 years ago
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