All categories

3 years ago

HELP!

Chemistry

2 answers:

ki77a [65]3 years ago

4 0

Answer:

(1)There are 1.5 moles of water in a 27 gram sample of water. The molar mass of water is 18.02 gmol g m o l .(2)

AnswersChemistryGCSEArticle

What is the mass (g) of 0.25mols of NaCl?

What you need for these equations are a calculator, periodic table and the following equation:

Mass (g) = Mr x Moles (important equation to remember)

In this case we already know the moles as it's in the question, 0.25 moles.

to find the Mr, you need to look at your periodic table. Find the relative atomic mass of Na and Cl and add the two numbers together.

Na = 22.99

Cl = 35.45

NaCl = 58.4

Now just put all of the numbers into the equation.

0.25 x 58.4 = 14.6g

photoshop1234 [79]3 years ago

4 0

2+16=18 gr 1mol = 18 gr ——-> 27gr/18gr = 1.5 mol
35.5 + 23 =58.5gr ——>0.25mol • 58.5 =14.625 gram

You might be interested in

Reaction of acetylene with silver nitrate

laila [671]

Answer:

When acetylene reacts with silver nitrate, its acidic hydrogen reacts with silver nitrate to form a white precipitate. So here acetylene shows its acidic nature.

Explanation:

this is what i found

3 0

3 years ago

Consider the following intermediate chemical equations.

geniusboy [140]

Answer: The enthalpy of the overall chemical equation is -205.7 kJ

Explanation:

Hess's law is defined as the law which states that the total amount of heat absorbed or released for a given chemical reaction will remain the same whether the process occurs in one step or several steps.

In this law, a chemical equation is treated as an ordinary algebraic expression that can be added or subtracted to yield the required overall equation.

The enthalpy change or heat of the overall reaction is the summation of the enthalpies or heats of the intermediate reactions involved in the process.

The given overall chemical equation follows:

$CH_4(g)+4Cl_2(g)\rightarrow CCl_4(g)+4HCl(g);\Delta H^o_{rxn}$

The intermediate equations for the above reaction are:

(1) $CH_4(g)\rightarrow C(s)+2H_2(g);\Delta H_1=74.6 kJ$

(2) $CCl_4(g)\rightarrow C(s)+2Cl_2(g);\Delta H_2=95.7kJ$

(3) $H_2(g)+Cl_2(g)\rightarrow 2HCl(g);\Delta H_3=-92.3kJ$

According to Hess law, the enthalpy of the reaction becomes:

$\Delta H^o_{rxn}=[1\times (\Delta H_1)] + [1\times (-\Delta H_2)] + [2\times (\Delta H_3)]$

Putting values in above expression, we get:

$\Delta H^o_{rxn}=[1\times (74.6)] + [1\times (-95.7)] + [2\times (-92.3)]\\\\\Delta H^o_{rxn}=-205.7kJ$

Hence, the enthalpy of the overall chemical equation is -205.7 kJ

8 0

3 years ago

Please answer the following question in the picture answer the question

raketka [301]

I think the answer is warm and high salinity

6 0

3 years ago

Read 2 more answers

Strong acids have a ______ value of Ka and the equlibrium for the reaction with water lies far to the _______.

nika2105 [10]

Answer:

All strong acids have a higher value of $K_{a}$ and the equilibrium for the reaction with water lies far to the right.

Explanation:

All strong acids dissociate completely in the solution. Higher the value of dissociation constant of the acid, higher will be the dissociation of the acid.

The reaction of the acid with water will be favored in the forward direction for acids having higher dissociation constant value ( $K_{a}$ ).

The dissociation of a strong acid say HA in water is shown below $\textrm{HA}\left ( aq \right )+\textrm{H}_{2}\textrm{O}\left ( l \right )\rightleftharpoons \textrm{A}^{-}\left ( aq \right )+\textrm{H}_{3}\textrm{O}^{+}\left ( aq \right )$

Higher the value of $K_{a}$ , more will be the dissociation of the acid in water. The reaction will move far to the right side.

4 0

3 years ago

Pls help me 20 points ty

qaws [65]

Answer:

what do you need help with i cant see it

Explanation:

6 0

3 years ago

Read 2 more answers