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3 years ago

Calculate the molarity of glucose in a solution that which contains 35 g of glucose in 0.16 kg of phenol

1 answer:

5 0

C(Molarity) = n of solute/V of solution (mol/L)

glucose(C6H12O6) = 180g/mol
glucose 35g = 35g/(180g/mol) = 0.1944mol

We need density of solution here, and I assume it as density of phenol, 1.07g/mL. (I don't know why the question doesn't contain it)

phenol 0.16kg = 0.16kg/(1.07kg/L) = 0.1495L

0.1944mol/0.1495L = 1.300M(mol/L)

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Explain how you would calculate the q for warming 100.0 grams of liquid water from 0°C to 100 °C.

mojhsa [17]

Answer:

mass = 100 g

T1 = 0°C

T2 = 100 °C

C = 1 cal/g°C

Q = mC(T2 -T1)

Q = 100(1)(100 - 0)

Q = 100(100)

Q = 10000 cal

Explanation:

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3 years ago

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2 years ago

2. 27.8 mL of an unknown were added to a 50.0-mL flask that weighs 464.7 g. The total mass of the flask and the liquid is 552.4

agasfer [191]

Answer:

$d=4.24x10^{-4}\frac{lb}{in^3}$

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary for us to set the equation for the calculation of density and mass divided by volume:

$d=\frac{m}{V}$

Thus, we can find the mass of the unknown by subtracting the total mass of the liquid to the mass of the flask and the liquid:

$m=552.4g-464.7g=87.7g$

So that we are now able to calculate the density in g/mL first:

$d=\frac{87.7g}{27.8mL}=3.15g/mL$

Now, we proceed to the conversion to lb/in³ by using the following setup:

$d=3.15\frac{g}{mL}*\frac{1lb}{453.6g}*\frac{1in^3}{16.3871mL}\\\\d=4.24x10^{-4}\frac{lb}{in^3}$

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2 years ago

A flexible container at an initial volume of 6.13 L contains 2.51 mol of gas. More gas is then added to the container until it r

aleksley [76]

Answer:

2.12 moles of gas were added.

Explanation:

We can solve this problem by using Avogadro's law, which states that at constant temperature and pressure:

V₁n₂=V₂n₁

Where in this case:

V₁ = 6.13 L

n₂ = ?

V₂ = 11.3 L

n₁ = 2.51 mol

We input the data:

6.13 L * n₂ = 11.3 L * 2.51 mol

n₂ = 4.63

As 4.63 moles is the final number of moles, the number of moles added is:

4.63 - 2.51 = 2.12 moles

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2 years ago