Charles Lyell believed that:

Which of the following equations is balanced for charge and mass?

Nataliya [291]

<span>B. S⁰(s) + 2H⁺ + 2e⁻ –--> H2S⁰(g)

by mass: 1 S and 2 H ----> 2 H and 1S True.
by charge : 0 +(2*(+1)) + 2*(-1) = 0, 0+2-2=0, 0 = 0 True.</span>

8 0

3 years ago

Read 2 more answers

Once the torch is lit why must the acetylene flow be increased until the flame stops smoking

mojhsa [17]

Once the torch is lit, the acetylene flow must be increased until the flame stops smoking <span>before the oxygen is turned on for adjustment in order to keep the tip of the torch cool.

You should also note that while lighting the torch, you should keep the spark lighter near the tip but not covering it.</span>

5 0

3 years ago

How do you determine the correct subscripts in a chemical formula

VMariaS [17]

BY CHECKING THE REACTIVITY OF AN ELEMENT WHICH IS MOST REACTIVE OR NOT AND YOU STUDY TYPE OF CHEMICAL REACTION IN 1 CH AND YOU CHECK THE REACTIVITY OF ELEMENTS IN 3 CH METALS AND NON METALS PAGE NO 45 IN NCERT BOOK

4 0

3 years ago

What is the mass of the solid NH4Cl formed when 75.5 g of NH3 is mixed with an equal mass of HCl? What is the volume of the gas

Gekata [30.6K]

Answer : The volume of the gas remaining is 56.5 liters.

The gas is hydrochloric acid and the formula of the gas is HCl.

The mass of $NH_4Cl$ produced is, 110.7 grams.

Explanation :

The balanced chemical reaction will be:

$NH_3+HCl\rightarrow NH_4Cl$

First we have to calculate the moles of $NH_3$ and HCl

$\text{Moles of }NH_3=\frac{\text{Mass of }NH_3}{\text{Molar mass of }NH_3}$

Molar mass of NH_3 = 17 g/mole

$\text{Moles of }NH_3=\frac{75.5g}{17g/mole}=4.44mole$

and,

$\text{Moles of }HCl=\frac{\text{Mass of }HCl}{\text{Molar mass of }HCl}$

Molar mass of HCl = 36.5 g/mole

$\text{Moles of }HCl=\frac{75.5g}{36.5g/mole}=2.07mole$

Now we have to calculate the limiting and excess reagent.

From the balanced reaction we conclude that

As, 1 mole of HCl react with 1 mole of $NH_3$

So, 2.07 mole of HCl react with 2.07 mole of $NH_3$

From this we conclude that, $NH_3$ is an excess reagent because the given moles are greater than the required moles and HCl is a limiting reagent and it limits the formation of product.

The remaining moles of HCl gas = 4.44 - 2.07 = 2.37 moles

Now we have to calculate the volume of the gas remaining.

Using ideal gas equation :

PV = nRT

where,

P = Pressure of gas = 752 mmHg = 0.989 atm (1 atm = 760 mmHg)

V = Volume of gas = ?

n = number of moles of gas = 2.37 moles

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

T = Temperature of gas = $14.0^oC=273+14.0=287K$

Putting values in above equation, we get:

$0.989atm\times V=2.37mole\times (0.0821L.atm/mol.K)\times 287K$

V = 56.5 L

Now we have to calculate the moles of $NH_4Cl$

As, 1 mole of HCl react with 1 mole of $NH_4Cl$

So, 2.07 mole of HCl react with 2.07 mole of $NH_4Cl$

Now we have to calculate the mass of $NH_4Cl$

$\text{ Mass of }NH_4Cl=\text{ Moles of }NH_4Cl\times \text{ Molar mass of }NH_4Cl$

Molar mass of $NH_4Cl$ = 53.5 g/mole

$\text{ Mass of }NH_4Cl=(2.07moles)\times (53.5g/mole)=110.7g$

Thus, the volume of the gas remaining is 56.5 liters.

The gas is hydrochloric acid and the formula of the gas is HCl.

The mass of $NH_4Cl$ produced is, 110.7 grams.

3 0

3 years ago

How are the shape of sand dunes determined?using one word

SVEN [57.7K]

"windweathered" because the use of wind weathering would bring the sand all around and it would be wind weathered

8 0

3 years ago