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Galina-37 [17]
3 years ago
14

If K3PO4= 0.250M, how many grams of K3PO4 are in 750.0ml of solution? Remember that M is the same as mol/L. Answer: 39.8g

Chemistry
1 answer:
sweet-ann [11.9K]3 years ago
4 0

Answer:

See below

Explanation:

Molarity = moles/Volume in Liters = (grams/formula wt)/Vol in Liters

=> Grams of solute = Molarity x Vol in Liters x formula wt

= (0.250M)(0.750L)(212.3g/mol)

= 39.8 grams

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PART A

Explanation:

3. A cylinder of compressed gas has a pressure of 4.882 atm on one day. The next

day, the same cylinder of gas has a pressure of 4.690 atm, and its temperature is

8°C. What was the temperature on the previous day in °C? Ans: 20°C.

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Can animals detect radio waves
Svetach [21]

Answer:

No.

Explanation:

No organism can detect X-rays or radio waves

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How many protons, electrons, and neutrons does silver have
Vera_Pavlovna [14]

Answer:

Heat of vaporization: 250.580 kJ/mol. Number of Protons/Electrons: 47. Number of neutrons: 61. Classification: Transition Metal.

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Explanation:

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3 years ago
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Determine a massa, em kg, de um material que está contida em um volume de 18L. Sabe-se que a densidade do material é de 0,9 g/cm
suter [353]
The question in English is "<span>Determine the mass, in kg, of a material that is contained in a volume of 18L. It is known that the material density is 0.9 g/cm 3"

Answer: 
</span>
We can use a simple equation to solve this problem. <span>
    d = m/v</span><span>

<span>Where </span>d <span>is the density, </span>m <span>is the mass and </span>v is the volume.

d = </span>0.9<span> g/cm³
m = ?
v = </span>18 L = 18 x 10³ cm³<span>

By applying the equation,
<span>    0.9 g/cm³ = m / </span></span>18 x 10³ cm³<span>
                  m = 0.9 g/cm³ x </span>18 x 10³ cm³<span> 
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Hence, the mass of 18 L of material is 16.2 kg.</span>
6 0
3 years ago
It is found that, when a dilute gas expands quasistatically from 0.40 to 5.0 L, it does 210 J of work. Assuming that the gas tem
fenix001 [56]

Answer:

n=0.033mole

Explanation:

From the question we are told that:

Initial volume V_1=0.40L

Final VolumeV_2=5.0L

Work W=210J

Temperature T=300k

Generally the equation for Ideal gas is mathematically given by

W=nRTIn\frac{V_2}{V_1}

n=\frac{W}{RTIn\frac{V_2}{V_1}}

n=\frac{210}{8.32*300In\frac{5.0}{0.4}}

n=0.033mole

6 0
3 years ago
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