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3 years ago

Help plsss quickly!!!

Chemistry

1 answer:

Luba_88 [7]3 years ago

6 0

It would be a continuous straight line

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Magnesium oxide does not readily decompose into magnesium and oxygen. The reaction is shown below. MgO(s) + 601.7 kJ mc017-1.jpg

galben [10]

TEMPERATURE IS THE ANSWER

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3 years ago

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WILL GIVE BRAINIEST Watered down koolaid is an example of

vovangra [49]

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3 years ago

When using Charles's law, temperature must be expressed in degrees Kelvin.

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The correct answer would more than likely have to be true, i’ve had this question before not to long ago.

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3 years ago

A student mixes a 10.0 mL sample of 1.0 M NaOH with a 10.0 mL sample of 1.0 M HCl in a polystyrene container. The temperature of

ElenaW [278]

Answer:

The experimental value of ΔH is -50 kJ/mol

Explanation:

Step 1: Data given

Volume of 1.0 M NaOH = 10.0 mL = 0.01 L

Volume of 1.0 M HCl = 10.0 mL = 0.01 L

Temperature before mixing = 20 °C

Final temperature = 26 °C

Specific heat of solution = 4.2 J/g°C

Density = 1g/mL

Step 2: Calculate q

q = m*c*ΔT

⇒ with m = the mass

⇒ 20.0 mL * 1g/mL = 20 grams

⇒ c = specific heat of solution = 4.2 J/g°C

⇒ ΔT = T2 -T1 = 26 -20 = 6 °C

q = 20g * 4.2 J/g°C * 6°C

q = 504 J

ΔHrxn = -q ( because it's an exothermic reaction)

ΔHrxn = -504 J

Step 3: Calculate number of moles

Moles = Molarity * volume

Moles = 1M *0.01 L = 0.01 moles

Step 4: Calculate the experimental value of ΔH

ΔHrxn = -504 / 0.01 mol = -50400 J/mol = -50.4 kJ/mol

The experimental value of ΔH is -50 kJ/mol

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4 years ago

Neon is compressed from 100 kPa and 16°C to 500 kPa in an isothermal compressor. Determine the change in the specific volume and

Viktor [21]

Answer:

$\Delta v=-0.952m^3/kg$

$\Delta h=0$

Explanation:

In this case, since neon could be considered as an ideal gas, the specific volumes at the first and second state are respectively:

$v_1=\frac{Rg*T_1}{p_1}=\frac{0.4119kJ/kg*K*289.15}{100kPa} =1.19m^3/kg\\$

$v_2=\frac{Rg*T_2}{p_2}=\frac{0.4119kJ/kg*K*289.15}{500kPa} =0.238m^3/kg\\$

Thus, the change in the specific volume turns out:

$\Delta v=v_2-v_1=0.238m^3/kg-1.19m^3/kg\\\Delta v=-0.952m^3/kg$

Which value has sense for the compression.

In addition, the specific enthalpy change just depend on the temperature as it is an ideal gas, therefore, since the process es isothermic:

$\Delta h=Cp\Delta T=0$

Best regards.

3 0

4 years ago