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3 years ago

If the solvent in a solution is water, the solution is

Chemistry

1 answer:

BlackZzzverrR [31]3 years ago

7 0

Answer:

I don't know

Explanation:

Yet but I will find it for you

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3 years ago

What is the color would phenolphthalein be if placed in water

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It's would turn pink in a basic solution suck as water.

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4 years ago

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kow [346]

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3 years ago

Read 2 more answers

A sample of CaCO3(s) is introduced into a sealed container of volume 0.638 L and heated to 1000 K until equilibrium is reached.

JulsSmile [24]

Answer:

the mass of CaO present at equilibrium is, 0.01652g

Explanation:

$K_p = [CO_2]$ = 3.8×10⁻²

Now we have to calculate the moles of CO₂

Using ideal gas equation,

PV =nRT

P = pressure of gas = 3.8×10⁻²

T = temperature of gas = 1000 K

V = volume of gas = 0.638 L

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mole.k

$\frac{3.8 * 10^2 * 0.638}{0.0821 * 1000} \\= 2.95 * 10^-^4$

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mass = 2.95 * 10 ⁻⁴ × 56

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Therefore,

the mass of CaO present at equilibrium is, 0.01652g

7 0

3 years ago

A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at

lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

Explanation:

Step 1: Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

Step 2: Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

Step 3: Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25 = 8.2 °C

q = 1200 * 4.184 * 8.2 = 41170.56 J

Step 4: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

Step 5: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

Step 6: Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

8 0

3 years ago