Pls help ASAP. I will give brainliest

The temperature of a sample of gas in steel container at 30.3 Atm is

vladimir1956 [14]

Answer:

74.09 atm

Explanation:

Using the gas laws ( Charles and Boyle's law). We have the formula ,

P1/T1 = P2/T2

Where P1 = 30.3atm

T1 = -100 degree Celsius

to kelvin = -100+ 273 = 173K

T2 = 150 degree Celsius

To Kelvin = 150 = 150+273 = 423K

Imputing values

P1/T1 = P2/T2

30.3/173 = P2/ 423

Cross multiply

173×P2 = 30.3 ×423

173P2 = 12816.9

Divide both sides by 173

P2 = 12816.9/173

P2 = 74.09 atm

I hope this was helpful, please mark as brainliest

6 0

2 years ago

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Copper (II) sulfate pentahydrate may be heated to drive off the water of hydration. If 5 g of water are produced, what was the o

____ [38]

Answer:

13.9g of CuSO4•5H2O

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CuSO4•5H2O → CuSO4 + 5H2O

Next, we shall determine the mass of CuSO4•5H2O heated and the mass of H2O produced from the balanced equation.

This is illustrated below

Molar mass of CuSO4• 5H2O = 63.5 + 32 + (16x4) + 5(2x1 + 16)

= 63.5 + 32 + 64 + 5(18) = 249.5g/mol

Mass of CuSO4•5H2O from the balanced equation = 1 x 249.5 = 249.5g

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O from the balanced equation = 5 x 18 = 90g

From the balanced equation above,

249.5g of CuSO4•5H2O produced 90g of H2O.

Now, we can determine the mass of CuSO4•5H2O needed to produce 5g of H2O. This can be achieved as shown below:

From the balanced equation above,

249.5g of CuSO4•5H2O produced 90g of H2O.

Therefore, Xg of CuSO4•5H2O will produce 5g of H2O i.e

Xg of CuSO4•5H2O = (249.5 x 5)/90

Xg of CuSO4•5H2O = 13.9g

Therefore, 13.9g of CuSO4•5H2O is needed to produce 5g of H2O.

6 0

3 years ago

What has a higher density Ice cubes or ice spheres pls pls pls help

Verizon [17]

The answer is Ice spheres

6 0

2 years ago

Which of the following are the starting substances in photosynthesis?

Misha Larkins [42]

A: water and carbon dioxide.

3 0

3 years ago

Read 2 more answers

calculate the mass of calcium phosphate and the mass of sodium chloride that could be formed when a solution containing 12.00g o

Leviafan [203]

Answer : The mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

Explanation : Given,

Mass of $Na_3PO_4$ = 12.00 g

Mass of $CaCl_2$ = 10.0 g

Molar mass of $Na_3PO_4$ = 164 g/mol

Molar mass of $CaCl_2$ = 111 g/mol

Molar mass of $NaCl$ = 58.5 g/mol

Molar mass of $Ca_3(PO_4)_2$ = 310 g/mol

First we have to calculate the moles of $Na_3PO_4$ and $CaCl_2$ .

$\text{Moles of }Na_3PO_4=\frac{\text{Given mass }Na_3PO_4}{\text{Molar mass }Na_3PO_4}$

$\text{Moles of }Na_3PO_4=\frac{12.00g}{164g/mol}=0.0732mol$

and,

$\text{Moles of }CaCl_2=\frac{\text{Given mass }CaCl_2}{\text{Molar mass }CaCl_2}$

$\text{Moles of }CaCl_2=\frac{10.0g}{111g/mol}=0.0901mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is:

$2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2$

From the balanced reaction we conclude that

As, 3 mole of $CaCl_2$ react with 2 mole of $Na_3PO_4$

So, 0.0901 moles of $CaCl_2$ react with $\frac{2}{3}\times 0.0901=0.0601$ moles of $Na_3PO_4$

From this we conclude that, $Na_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $CaCl_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$ and $Ca_3(PO_4)_2$

From the reaction, we conclude that

As, 3 mole of $CaCl_2$ react to give 6 mole of $NaCl$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{6}{3}\times 0.0901=0.1802$ mole of $NaCl$

and,

As, 3 mole of $CaCl_2$ react to give 1 mole of $Ca_3(PO_4)_2$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{1}{3}\times 0.0901=0.030$ mole of $Ca_3(PO_4)_2$

Now we have to calculate the mass of $NaCl$ and $Ca_3(PO_4)_2$

$\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl$

$\text{ Mass of }NaCl=(0.1802moles)\times (58.5g/mole)=10.5g$

and,

$\text{ Mass of }Ca_3(PO_4)_2=\text{ Moles of }Ca_3(PO_4)_2\times \text{ Molar mass of }Ca_3(PO_4)_2$

$\text{ Mass of }Ca_3(PO_4)_2=(0.030moles)\times (310g/mole)=9.3g$

Therefore, the mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

5 0

2 years ago