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krok68 [10]
3 years ago
12

The seawater of Puget Sound off the coast of Oregon has a salinity of about 24 PPT. What weight of salt is there in 1,000g of se

awater?
Chemistry
1 answer:
Juli2301 [7.4K]3 years ago
8 0

The given salinity is 24ppt

The relation between salinity and grams is

salinity(ppt)=\frac{gramsofsolute}{massofsolventinkg}

it means there is 24 parts of salt per 1000 parts of water

ppt: parts per thousand

Or

there is 24 grams of salt per 1000g of water

thus the value 24 ppt signifies that there is 24g of salt dissolved in 1000g or 1kg of water

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If the same amount of heat is added to 25.0 g of each of the metals, which are all at the same initial temperature, which metal
AVprozaik [17]

Answer:

The bismuth sample.

Explanation:

The specific heat c of a substance (might not be a metal) is the amount of heat required for heating a unit mass of this substance by unit temperature (e.g., \rm 1\; ^{\circ}C.) The formula for specific heat is:

\displaystyle c = \frac{Q}{m \cdot \Delta T},

where

  • Q is the amount of heat supplied.
  • m is the mass of the sample.
  • \Delta T is the increase in temperature.

In this question, the value of Q (amount of heat supplied to the metal) and m (mass of the metal sample) are the same for all four metals. To find \Delta T (change in temperature,) rearrange the equation:

\displaystyle c \cdot \Delta T = \frac{Q}{m},

\displaystyle \Delta T = \frac{Q}{c \cdot m}.

In other words, the change in temperature of the sample, \Delta T can be expressed as a fraction. Additionally, the specific heat of sample, c, is in the denominator of that fraction. Hence, the value of the fraction would be the largest for sample with the smallest specific heat.

Make sure that all the specific heat values are in the same unit. Find the one with the smallest specific heat: bismuth (\rm 0.123 \; J \cdot g\cdot \,^{\circ}C^{-1}.) That sample would have the greatest increase in temperature. Since all six samples started at the same temperature, the bismuth sample would also have the highest final temperature.

3 0
2 years ago
What is the acceleration of the object’s motion? 0.5 m/s2 -0.5 m/s2 2 m/s2 -2 m/s2
djverab [1.8K]

Answer:  -2m/s2

Explanation:

Using the following equation ; acceleration = Change in velocity / time

i.e  a =  v - u / t

where  'a'  = acceleration

              v  = final velocity

              u  =  initial velocity

              t   =  time

Therefore;  from the graph we have acceleration to be, 0 - 6m/s / 3s  = -2m/s2

5 0
3 years ago
If the sample contained 2.0 moles of KClO3 at a temperature of 214.0 °C, determine the mass of the oxygen gas produced in grams
Westkost [7]

Answer : The mass of the oxygen gas produced in grams and the pressure exerted by the gas against the container walls is, 96 grams and 1.78 atm respectively.

Explanation : Given,

Moles of KCl_3 = 2.0 moles

Molar mass of O_2 = 32 g/mole

Now we have to calculate the moles of MgO

The balanced chemical reaction is,

2KClO_3\rightarrow 2KCl+3O_2

From the balanced reaction we conclude that

As, 2 mole of KClO_3 react to give 3 mole of O_2

So, 2.0 moles of KClO_3 react to give \frac{2.0}{2}\times 3=3.0 moles of O_2

Now we have to calculate the mass of O_2

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

\text{ Mass of }O_2=(3.0moles)\times (32g/mole)=96g

Therefore, the mass of oxygen gas produced is, 96 grams.

Now we have to determine the pressure exerted by the gas against the container walls.

Using ideal gas equation:

PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}

where,

P = pressure of oxygen gas = ?

V = volume of oxygen gas

T = temperature of oxygen gas = 214.0^oC=273+214.0=487K

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

w = mass of oxygen gas

\rho = density of oxygen gas = 1.429 g/L

M = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the ideal gas equation, we get:

P=1.429g/L\times \frac{(0.0821L.atm/mole.K)\times (487K)}{32g/mol}

P=1.78atm

Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.

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Pani-rosa [81]

Answer:

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3 years ago
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