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2 years ago

3

What is the mass of a glucose solution that fills a 1.00 l intravenous bottle if the density of the glucose solution is 1.15 g/m

l?

1 answer:

Finger [1]2 years ago

6 0

The mass is 1.15 g/ml because d=m/v.

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Cyclohexane has a freezing point of 6.50 ∘C and a Kf of 20.0 ∘C/m. What is the freezing point of a solution made by dissolving 0

ExtremeBDS [4]

Answer:

The freezing point will be $2.9^{0}C$

Explanation:

The depression in freezing point is a colligative property.

It is related to molality as:

$Depressioninfreezing point=K_{f}Xmolality$

Where

Kf= $20\frac{^{0}C}{m}$

the molality is calculated as:

$molality=\frac{moles_{solute}}{mass_{solvent}}$

$moles=\frac{mass}{molarmass}=\frac{0.694}{154}=0.0045mol$

$massofcyclohexane=25g=0.025Kg$

$molarity=\frac{0.0045}{0.025}=0.18m$

Depression in freezing point = $20X0.18=3.6^{0}C$

The new freezing point = $6.5^{0}C-3.6^{0}C=2.9^{0}C$

5 0

2 years ago

Arkeisha and Rodney are having a discussion about the correct use of the term "battery acid." Arkeisha insists that an alkaline

mote1985 [20]

Arkeisha is correct because the fluid in an alkaline battery has a ph between 7.1 and 14.0

8 0

2 years ago

4.82 g of an unknown metal is heated to 115.0∘C and then placed in 35 mL of water at 28.7∘C, which then heats up to 34.5∘C. What

nikitadnepr [17]

<u>Answer:</u> The specific heat of metal is 2.34 J/g°C

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 35 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{35mL}\\\\\text{Mass of water}=(1g/mL\times 35mL)=35g$

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of metal = 4.82 g

$m_2$ = mass of water = 35 g

$T_{final}$ = final temperature = 34.5°C

$T_1$ = initial temperature of metal = 115°C

$T_2$ = initial temperature of water = 28.7°C

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$4.82\times c_1\times (34.5-110)=-[35\times 4.186\times (34.5-28.7)]$

$c_1=2.34J/g^oC$

Hence, the specific heat of metal is 2.34 J/g°C

4 0

2 years ago

What is the specific heat (J/g°C) of a metal object whose temperature increases by 3.0°C when 17.5 g of metal was heated with 38

Verizon [17]

Answer:

a. 0.73

Explanation:

Given data

Added heat (Q): 38.5 J

Change in the temperature (ΔT): 3.0°C

Mass of the metal (m): 17.5 g

Specific heat of the metal (c): ?

We can determine the specific heat of the metal using the following expression.

Q = c × m × ΔT

c = Q / m × ΔT

c = 38.5 J / 17.5 g × 3.0°C

c = 0.73 J/g.°C

6 0

2 years ago

If it takes one second to count two rice grains,calculate the time in years required to count a mole of rice grainrg.

Anuta_ua [19.1K]

Mole is 6,022*10^23

6,022 * 10^23 /2 (here are seconds) then /60 (minutes), then /60 (hours) /24 (days), then /365 (year) equals 9547818366311516,9965 years

every 4 years there's one day more in a year, so 9547818366311516,9965 /4 ( here are days) then / 365 = 6539601620761,31 years less.

the answer is 9547818366311516,9965 - 6539601620761,31 = 9541278764690755,6865 years to calculate the rice

5 0

2 years ago