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leonid [27]
3 years ago
9

How would you prepare 175 mL of a 0.350kmol/m^3 calcium nitrate solution

Chemistry
1 answer:
SVEN [57.7K]3 years ago
3 0
0,35 kmol/m³ = 0,35 mol/dm³ = 0,35 mol/L
175 mL = 0,175 L
*-*-*-*-*-*-*-*-*-*-*-*
C = n/V
n = 0,35×0,175
n = 0,06125 mol

mCa(NO₃)₂: 40+(14×2)+(16×6) = 164 g/mol

1 mol --------- 164g
0,06125 ---- X
X = 10,045g

To prepare 175 mL of 0,35M solution, add 10,045g of calcium nitrate and add water to a volume of 175ml.
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Convert 213.62 grams of xenon gas at STP to L.
jolli1 [7]

Answer: The coefficient is 3.645

The exponent is 1

There are 4 significant digits

The rightmost significant figure is 5

Explanation:

Scientific notation is defined as the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.

For example : 5000 is written as 5.0\times 10^3

According to avogadro's law, 1 mole of every gas contains avogadro's number 6.023\times 10^{23} of particles, occupy 22.4 L at STP and weighs equal to its molecular mass.

131.29 g of Xe occupy = 22.4 L at STP.

Thus 213.62 g of Xe occupy = \frac{22.4}{131.29}\times 213.62=36.45L at STP.

Scientific notation = 3.645\times 10^1L

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4 0
3 years ago
Calculate the mass in grams for 0.251 moles of Na2CO3
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Answer:

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3 years ago
metal weighing 6.98 g was heated to 91.29 °C and then put it into 114.84 mL of water (initially at 24.37 °C). The metal and wate
torisob [31]

Answer:

The specific heat of the metal is 10.93 J/g°C.

Explanation:

Given,

For Metal sample,

mass = 6.98 grams

T = 91.29°C

For Water sample,

volume = 114.84 mL

T = 24.37°C.

Final temperature of mixture = 33.54°C.

When the metal sample and water sample are mixed,

The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the  addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.

Since, heat lost by metal is equal to the heat gained by water,

Qlost = Qgain

However,

Q = (mass) (ΔT) (Cp)

(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)

After mixing both samples, their temperature changes to 27°C.

It implies that

water sample temperature changed from  24.37°C to 33.54°C and metal sample temperature changed from 91.29°C to 33.54°C.

We have all values, but, here mass of water is not given. It can be found by using the formula

Density = Mass/Volume

Since, density of water = 1 g/mL

we get, Mass = 114.84 grams.

Since specific heat of water is 4.184 J/g°C.

Now substituting all values in (mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)

(6.98)(91.29 - 33.54)(Cp) = (114.84)(33.54 - 24.37)(4.184)

solving, we get,

Cp = 10.93 J/g°C.

the specific heat of the metal is 10.93 J/g°C.

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What is the first thing you should do when any accident occurs during a science
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Answer:

Tell your teacher or learning coach.

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Hope this helps : )

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