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3 years ago

How many moles are in 2.00 g of NaCl?7

Chemistry

1 answer:

coldgirl [10]3 years ago

7 0

Answer:

0.0342 moles

Explanation:

Moles and Mass are related as,

Moles = Mass / M.Mass

M.Mass of NaCl = 58.44 g/mol

Putting values,

Moles = 2.00 g / 58.44 g/mol

Moles = 0.0342 moles

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Consider the chemical equation in equilibrium. A(g) 2B(g) Double headed arrow. C(g) D(g) heat What will happen to the equilibriu

Mazyrski [523]

If the temperature is increased then reaction will shift to the left because heat is absorbed.

<h3>What is equilibrium state?</h3>

Equilibrium of any reaction is that state in which concentration of reactant and concentration of product will be constant.

Given chemical reaction is:

A(g) + 2B(g) ⇄ C(g) + D(g)

From the equilibrium state reaction will move only that side which will contribute to maintain the stable state. In the forward reaction heat is released as mention in the question. So, when the temperature of reaction is increased then it shifts towards the left side by absorbing the heat and maintain the stability.

Hence, option (2) is correct, i.e. It will shift to the left because heat is absorbed.

To know more about equilibrium, visit the below link:

brainly.com/question/14297698

6 0

2 years ago

A piece of unknown metal with mass 30 g is heated to 110.0 °C and dropped into 100.0 g of water at 20.0 °C. The final temperatur

Ymorist [56]

<u>Answer:</u> The specific heat of metal is 0.821 J/g°C

<u>Explanation:</u>

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of metal = 30 g

$m_2$ = mass of water = 100 g

$T_{final}$ = final temperature = 25°C

$T_1$ = initial temperature of metal = 110°C

$T_2$ = initial temperature of water = 20.0°C

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$30\times c_1\times (25-110)=-[100\times 4.186\times (25-20)]$

$c_1=0.821J/g^oC$

Hence, the specific heat of metal is 0.821 J/g°C

8 0

3 years ago

Choose all the answers that apply.

iVinArrow [24]

Answer:it all above

Explanation:

it all above because all the answer are truth so it all above

8 0

3 years ago

Read 2 more answers

A colorless liquid has a molar mass of 60.01 g/mol. When the liquid was analyzed, it was 46.7% nitrogen and 53.3% oxygen. What i

MAXImum [283]

first we have to find the empirical formula of the compound

empirical formula is the simplest ratio of whole numbers of components making up a compound

for 100 g of the compound

N O

mass 46.7 g 53.3 g

number of 46.7 g/ 14 g/mol 53.3 g/ 16 g/mol

moles = 3.34 mol = 3.33 mol

divide by the least number of moles

3.34/3.33 = 1.00 3.33/ 3,33 = 1.00

therefore number of atoms are

N - 1

O - 1

empirical formula is - NO

mass of empirical unit - 14 g/mol + 16 g/mol = 30 g

molecular formula is actual composition of elements in the compound

molecular mass - 60.01 g/mol

number of empirical units = molecular mass / empirical unit mass

= 60.01 g/mol / 30 g = 2

there are 2 empirical units

2(NO)

molecular formula = N₂O₂

6 0

3 years ago

A catalyst decreases the activation energy of a particular exothermic reaction by 34 kJ/mol, to 57 kJ/mol. Assuming that the mec

vagabundo [1.1K]

Answer:

the activation energy for the uncatalyzed reverse reaction is 83kJ/mol

Explanation:

see the attached file

4 0

3 years ago