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3 years ago

PLEASE HELP!!!!! WILL MARK AS BRAINLIEST!!!!! What is the volume of 15.0 grams of H2S at STP?

Chemistry

1 answer:

ICE Princess25 [194]3 years ago

6 0

Answer:

The volume of 15.0 grams of H2S at STP is 9.85 L.

Explanation:

1 unit of any gas under STP has volume 22.4 L.

Unit's mass M(H2S) = 2M(H) + M(S)= 2*1.0 +32.1=34.1 g/U

15g * 1U/34.1g= 0.440 U

0.440 U*1L/22.4L= 9.85 L

Hope this helped you :)

Good Luck :D

BTW, If you understand what I am saying and want the more technical answer, take all the (U) and replace it with (Mol) which is an abbreviation for Mole

:D have a good rest of your night

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Question 6 (1 point)

Mnenie [13.5K]

Answer:

The correct option is;

d 4400

Explanation:

The given parameters are;

The mass of the ice = 55 g

The Heat of Fusion = 80 cal/g

The Heat of Vaporization = 540 cal/g

The specific heat capacity of water = 1 cal/g

The heat required to melt a given mass of ice = The Heat of Fusion × The mass of the ice

The heat required to melt the 55 g mass of ice = 540 cal/g × 55 g = 29700 cal

The heat required to raise the temperature of a given mass ice (water) = The mass of the ice (water) × The specific heat capacity of the ice (water) × The temperature change

The heat required to raise the temperature of the ice from 0°C to 100°C = 55 × 1 × (100 - 0) = 5,500 cal

The heat required to vaporize a given mass of ice = The Heat of Vaporization × The mass of the ice

The heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal

The total heat required to boil 55 g of ice = 29700 cal + 5,500 cal + 4,400 cal = 39,600 cal

However, we note that the heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal.

The heat required to vaporize the 55 g mass of ice at 100°C = 4,400 cal

3 0

3 years ago

2.46 grams of Cu(NO3)2 would contain how many formula units

NikAS [45]

Answer:

it woul be 263

Explanation:

becaue it adds up like that

7 0

3 years ago

CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is

BaLLatris [955]

Answer : The moles of $O_2$ left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of $CH_4$ .

Using ideal gas equation:

$PV=nRT$

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = $27^oC=273+27=300K$

n = number of moles of gas = ?

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

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

$(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)$

$n=0.406mole$

Now we have to calculate the moles of $O_2$ .

The balanced chemical reaction will be:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that,

As, 1 mole of $CH_4$ react with 2 moles of $O_2$

So, 0.406 mole of $CH_4$ react with $2\times 0.406=0.812$ moles of $O_2$

Now we have to calculate the excess moles of $O_2$ .

$O_2$ is 20 % excess. That means,

Excess moles of $O_2$ = $\frac{(100 + 20)}{100}$ × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of $O_2$ left in the products.

Moles of $O_2$ left in the products = Excess moles of $O_2$ - Required moles of $O_2$

Moles of $O_2$ left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of $O_2$ left in the products are 0.16 moles.

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3 years ago

How many atoms of magnesium metal are needed to produce 3.87 moles of magnesium oxide

Ket [755]

Answer:

48.42g of Mg

Explanation:

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3 years ago

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In what way do the atoms in the same row of the Periodic Table increase from left to right?

zheka24 [161]

Answer:

As you move from left to right, the nucleus gains protons. This increases the positive charge of the nucleus and its attractive force on the electrons. At the same time, electrons are added to the atoms as you move from left to right across a period.

Explanation:

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3 years ago