How does the law of conservation of matter apply to chemical equations?

What is required in order to determine whether or not an object moves?

AysviL [449]

Answer:

A reference point.

Explanation:

Reference point:

A reference point is a point which is used to determine weather the object is in motion or not. The moving object is compared with the reference point.

An object is in the state of motion when its distance is changed relative to the other object or reference point.

In order to determine weather an object is moving or not a point of reference is needed.

Without reference point displacement can not be measured

3 0

3 years ago

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A 21.82 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 21.33 grams of CO2 and 4

morpeh [17]

Answer: The molecular formula for the given organic compound is $C_2H_2O_4$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=21.33g$

Mass of $H_2O=4.366g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 21.33 g of carbon dioxide, $\frac{12}{44}\times 21.33=5.82g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 4.366 g of water, $\frac{2}{18}\times 4.366=0.485g$ of hydrogen will be contained.

Mass of oxygen in the compound = (21.82) - (5.82 + 0.485) = 15.515 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5.82g}{12g/mole}=0.485moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.485g}{1g/mole}=0.485moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{15.515g}{16g/mole}=0.969moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.485 moles.

For Carbon = $\frac{0.485}{0.485}=1$

For Hydrogen = $\frac{0.485}{0.485}=1$

For Oxygen = $\frac{0.969}{0.485}=1.99\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is $C_1H_{1}O_2=CHO_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Thus, the molecular formula for the given organic compound is $C_2H_2O_4$

4 0

3 years ago

Earth has a low air pressure zone at the equator because what

SashulF [63]

Rising air in the tropics is called the Intertropical convergence zone and is a region of rising hot air. As it rises, the moisture in the air eventually reaches its saturation point and the moisture falls out as rain. This is why the tropics are know for their very heavy rainfall periods

7 0

3 years ago

Hydrogen gas, iodine vapor, hydrogen iodine are mixed in a flask and heated to 696°C. H2(g) + I2(g) ⇋ 2 HI(g) Kc = 52 at 696°C I

Anettt [7]

Answer: The equilibrium concentration of hydrogen gas is 0.0269 M

Explanation:

The chemical reaction follows the equation:

$H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$

At t = 0 0.044M 0.044M 0.177M

At $t=t_{eq}$ (0.044-x)M (0.044-x)M (0.177+x)M

The expression for $K_c$ for the given reaction follows:

$K_c=\frac{[HI]^2}{[H_2]\times [I_2]}$

We are given:

$K_c=52$

Putting values in above equation, we get:

$52=\frac{(0.177+x)^2}{(0.044-x)^2}$

$x=0.0171M$

Hence, the equilibrium concentration of hydrogen gas is (0.044-x) M =(0.044-0.0171) M= 0.0269 M

7 0

3 years ago

What is the mass in grams of 9.45*10^24 molecules of methanol (CH3OH)?

Angelina_Jolie [31]

Number of moles:

1 mole ---------- 6.02x10²³ molecules
? moles --------- 9.45x10²⁴ molecules

1 x ( 9.45x10²⁴) / 6.02x10²³ =

9.45x10²⁴ / 6.02x10²³ => 15.69 moles of CH3OH

Therefore:

Molar mass CH3OH = 32.04 g/mol

1 mole ------------ 32.04 g
15.69 moles ----- mass methanol

Mass methanol = 15.69 x 32.04 / 1 => 502.7076 g

6 0

3 years ago