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2 years ago

How much heat is required to convert 90. g of ice at -4.0 C to water vapor (steam) at 110 C

1 answer:

6 0

Answer: How much heat is required to convert 29 g of ice at -4.0 ˚C to water vapor at 105 ˚C (report your answer to three significant figures)? Answer all five parts, showing your detailed calculations: i) Heat to warm ice from -4.0 to 0 oC: ii) Heat to convert ice to liquid: iii) Heat to warm liquid to boiling: iv) Heat to convert liquid to gas

Explanation:

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When the following reaction comes to equilibrium, will the concentrations of the reactants or products be greater? Does the answ

Nutka1998 [239]

Answer:

At equilibrium, the concentration of the reactants will be greater than the concentration of the products. This does not depend on the initial concentrations of the reactants and products.

Explanation:

The value of Kc gives us an idea of the extent of the reaction. A big Kc (Kc > 1) means that in the equilibrium there are more products than reactants, and the opposite happens for a small Kc (Kc < 1). The equilibrium is reached no matter what the initial concentrations are.

The value of the equilibrium constant is relatively SMALL; therefore, the concentration of reactants will be GREATER THAN the concentration of products. This result is INDEPENDENT OF the initial concentration of the reactants and products.

4 0

3 years ago

Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe

leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as molarity, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

$c = \frac{n}{V}$

n = moles (unit mol)

V = volume (unit L)

Find the molar mass (M) of potassium hydroxide.

$M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}$

$M_{KOH} = 56.106 \frac{g}{mol}$

Calculate the moles of potassium hydroxide.

$n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}$

$n_{KOH} = 0.25941(9)mol$

Carry one insignificant figure (shown in brackets).

Convert the volume of water to litres.

$V = \frac{500.0mL}{1}*\frac{1L}{1000mL}$

$V = 0.5000L$

Here, carrying an insignificant figure doesn't change the value.

Calculate the concentration.

$c = \frac{n}{V}$

$c = \frac{0.25941(9)mol}{0.5000 L}$

$c = 0.5188(3) \frac{mol}{L}$ <= Keep an insignificant figure for rounding

$c = 0.5188 \frac{mol}{L}$ <= Rounded up

$c = 0.5188M$ <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

7 0

3 years ago

The compound FeCl3 is made of...

notka56 [123]

The compound FeCl3 is made of one atom and three atoms of chlorine

explanation

FeCl3 is made up of two elements.

that is iron with Fe chemical symbol and chlorine with Cl chemical symbol.

There are 3 atoms of chlorine in FeCl3 because the subscript in front of Cl in FeCl3 is 3.

3 0

3 years ago

Read 2 more answers

Why do astronauts on the moon seem like they're "Walking on springs" while on earth we are firmly attached to the ground

Sergio [31]

We have gravity here on earth while in space there’s none

8 0

3 years ago

Read 2 more answers

Calculate the concentration in % (m/m) of a solution containing 30.0g of mgcl2 dissolved in 270.0g of h20

qaws [65]

As the question tells you, you need to use the formula

% mass= mass of solute/ mass of solution x 100

mass solute= 30.0 g
mass of solution= 30.0 + 270.0= 300.0 g

% mass= 30.0/ 300.0 x 100= 10%

answer is B

4 0

3 years ago