876 millimeters to meters

Explain the states of solid, liquid and gas in terms of molecular movement and molecular packing?*

lora16 [44]

Solid- molecules vibrate in place and tightly packed
liquid-molecules fur shape of container and can slide past each other
gas-molecules also fit shape of container and have the most room

7 0

3 years ago

How many grams are in 11.9 moles of chromirum

faust18 [17]

Answer:

Explanation:

Method 1 proportion

1 mole of chromium is 52 grams

11.9 moles = x grams

1/11.9 = 52/x Cross multiply

x = 11.9 * 52

x = 618.8 grams

Now I have used an approximate mass for Chromium. The answer you get here is expected to reflect the weigth given on your periodic table Use that to get your answer. You should give a number very close to mine. Round to 3 places as in 619.

Method Two Formula

mols = given mass / molecular mass

11.9 = given mass / 51.9961 Multiply both sides by 51.9961

11.9 *51.9961 = given mass

given mass = 618.75

given mass = 619

3 0

3 years ago

The rate-determining step in the acid-catalyzed dehydration of 2-methylcyclohexanol results in the formation of what type of int

Svetach [21]

Answer:

The rate determining step depends on the

Explanation:

The rate determining step depends on the mechanism of the reaction. If it proceeds by E1 reaction mechanism, it produces 1-methylcyclohexene while the E3 reaction mechanism gives 3-methylcyclohexene. The rate determining step for E1 is an intermediate that is formed when protonation of the hydroxyl group occurs to form H3O+ followed by electrophilic attack on the carbocation formed to absorb the hydrogen making the carbon atom -CH2 and forming a double bond between.

The E3 mechanism is similar only that the attack on the carbon atom is done at the -C3 instead of -C1.

5 0

4 years ago

Read 2 more answers

A 14.60g sample of an unknown compound, composed only of carbon, hydrogen, and oxygen, produced 28.6g of CO2 and 14.6g of H2O in

o-na [289]

Answer: The empirical formula for the given compound is $C_{2}H_{5}O$

Explanation:

The chemical equation for the combustion of compound 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 $CO_2=28.6g$

Mass of $H_2O=14.6g$

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

So, in 28.6 g of carbon dioxide, $\frac{12}{44}\times 28.6=7.8g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 14.6 g of water, $\frac{2}{18}\times 14.6=1.6$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = (14.60) - (7.8 + 1.6) = 5.2 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{7.8g}{12g/mole}=0.65moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.6g}{1g/mole}=1.6moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{5.2g}{16g/mole}=0.32moles$

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 0.32 moles.

For Carbon = $\frac{0.65}{0.32}=2.03\approx 2$

For Hydrogen = $\frac{1.6}{0.32}=5$

For Oxygen = $\frac{0.32}{0.32}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 5 : 1

Hence, the empirical formula for the given compound is $C_{2}H_{5}O_{1}=C_{2}H_{5}O$

8 0

3 years ago

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What is the specific heat of a 44 g piece of metal if 202 Joules of heat are required to

Ray Of Light [21]

Answer:

≈ 395,8 J/(kg * °C)

Explanation:

m = 44 g = 0,044 kg

$t_{1}$ = 22 °C

$t_{2}$ = 33,6 °C

Q = 202 J

The formula is: Q = c * m * ( $t_{2} - t_{1}$ )

c = $\frac{Q}{m * (t_{2} - t_{1} )}$

Calculating:

c = 202 J / 0,044 kg * (33,6 °C - 22 °C) ≈ 395,8 J/(kg * °C)

4 0

3 years ago