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Dmitry_Shevchenko [17]
2 years ago
15

Balance the following chemical equation by providing the correct coefficients:

Chemistry
1 answer:
Mandarinka [93]2 years ago
7 0

Answer:

CO + 3H2 -> H20 + CH4 :)) !!

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CEYSDINO IS A FILE SPAMMER SHE OR HE USES DIS FILE TO GET YOUR INFORMATION...

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Indirect costs incurred in a manufacturing environment that cannot be traced directly to a product are treated as a.product cost
Alisiya [41]

Answer:

C. product costs and expensed when the goods are sold

Explanation:

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3 years ago
Osmotic pressure Π is given by the relation:Π = iMRTwhere i is the van’t Hoff factor, M is the concentration of solute, R is the
lions [1.4K]

<u>Answer:</u> The concentration of solute is 0.503 mol/L

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=icRT

where,

\pi = osmotic pressure of the solution = 24 atm

i = Van't hoff factor = 2 (for NaCl)

c = concentration of solute = ?

R = Gas constant = 0.08\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[273+25]=298K

Putting values in above equation, we get:

24atm=2\times c\times 0.08\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\c=0.503mol/L

Hence, the concentration of solute is 0.503 mol/L

5 0
3 years ago
(MARK BRAINLEST) PLEASE HELP ASAP: Network covalent solids have an unusual structure. Describe how they are formed, and give two
Amiraneli [1.4K]

Answer:

Covalent network solids are formed by networks or chains of atoms or molecules held together by covalent bonds. Consists of sp3 hybridized carbon atoms, each bonded to four other carbon atoms in a tetrahedral array to create a giant network. Examples of network covalent solids include diamond and graphite (both allotropes of carbon), and the chemical compounds silicon carbide and boron-carbide.

4 0
3 years ago
1. Calculate the energy change (q) of the surroundings (water) using the enthalpy equation
Rasek [7]

Answer:

Q1: 728.6 J.

Q2:

a) 668.8 J.

b) 0.3495 J/g°C.

Explanation:

<em>Q1: Calculate the energy change (q) of the surroundings (water) using the enthalpy equation:</em>

  • The amount of heat absorbed by water = Q = m.c.ΔT.

where, m is the mass of water (m = d x V = (1.0 g/mL)(24.9 mL) = 24.9 g).

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 32.2°C - 25.2°C = 7.0°C).

<em>∴ The amount of heat absorbed by water = Q = m.c.ΔT</em> = (24.9 g)(4.18 J/g°C)(7.0°C) = 728.6 J.

<em>Q2:  Calculate the energy change (q) of the surroundings (water) using the enthalpy equation </em>

<em>qwater = m × c × ΔT.  </em>

<em>We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL. calculate the specific heat of the metal. Use the data from your experiment for the unknown metal in your calculation.</em>

<em></em>

a) First part: the energy change (q) of the surroundings (water):

  • The amount of heat absorbed by water = Q = m.c.ΔT.

where, m is the mass of water (m = d x V = (1.0 g/mL)(25 mL) = 25 g).

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 31.6°C - 25.2°C = 6.4°C).

<em>∴ The amount of heat absorbed by water = Q = m.c.ΔT</em> = (25 g)(4.18 J/g°C)(6.4°C) = <em>668.8 J.</em>

<em>b) second part:</em>

<em>Q water = Q unknown metal. </em>

<em>Q unknown metal =  - </em>668.8 J. (negative sign due to the heat is released from the metal to the surrounding water).

<em>Q unknown metal =  - </em>668.8 J = m.c.ΔT.

m = 27.776 g, c = ??? J/g°C, ΔT = (final T - initial T = 31.6°C - 100.5°C = - 68.9°C).

<em>- </em>668.8 J = m.c.ΔT = (27.776 g)(c)( - 68.9°C) = - 1914 c.

∴ c = (<em>- </em>668.8)/(- 1914) = 0.3495 J/g°C.

<em></em>

3 0
3 years ago
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