Ask question

Ask question

All categories

3 years ago

10

How many molecules are in 0.943 grams of water

2 answers:

ipn [44]3 years ago

7 0

Answer:

3.16 x 10²² molecules H₂O in 0.943 gram

Explanation:

Given 0.943 g H₂O = (0.943 g / 18 g·mol⁻¹) = 0.05253 mol H₂O

0.05253 mol H₂O = 0.05253 mol H₂O x 6.023 x 10²³ molecules·mol⁻¹

= 3.16 x 10²³ molecules H₂O

zavuch27 [327]3 years ago

5 0

Answer: ther are 065.3

Explanation:

You might be interested in

The process of joining many small molecules into larger molecules is called(1) neutralization (3) saponification(2) polymerizati

yulyashka [42]

(2) polymerization. polymerization<span> is a process of reacting monomer molecules together in a chemical reaction to form polymer chains or three-dimensional networks.</span>

8 0

3 years ago

Read 2 more answers

The sun revolved around the earth why ? help me

Sergio [31]

Answer:

You are moving because the Earth and everything in our solar system is constantly moving. ... As the Earth rotates, it also moves, or revolves, around the Sun. The Earth's path around the Sun is called its orbit. It takes the Earth one year, or 365 1/4 days, to completely orbit the Sun.

6 0

3 years ago

Read 2 more answers

Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti

gayaneshka [121]

Answer : The value of $K$ for this reaction is, $2.6\times 10^{15}$

Explanation :

The given chemical reaction is:

$CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)$

Now we have to calculate value of $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]$

where,

$\Delta G^o$ = Gibbs free energy of reaction = ?

n = number of moles

$\Delta G^0_{(HCH_3CO_2(g))}$ = -389.8 kJ/mol

$\Delta G^0_{(CH_3OH(g))}$ = -161.96 kJ/mol

$\Delta G^0_{(CO(g))}$ = -137.2 kJ/mol

Now put all the given values in this expression, we get:

$\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]$

$\Delta G^o=-89.4kJ/mol$

The relation between the equilibrium constant and standard Gibbs, free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol

R = gas constant = 8.314 J/L.atm

T = temperature = $30.0^oC=273+30.0=303K$

$K$ = equilibrium constant = ?

Now put all the given values in this expression, we get:

$-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K$

$K=2.6\times 10^{15}$

Thus, the value of $K$ for this reaction is, $2.6\times 10^{15}$

4 0

3 years ago

What is the mass of HCL is consumed by the reaction of 1.5 mole of magnesium

photoshop1234 [79]

Answer: 109.5g

Explanation:

Mg + 2HCl —> MgCl2 + H2

From the equation,

1mole of Mg required 2moles of HCl

Therefore 1.5moles of Mg will require = 1.5 x 2 = 3 moles of HCl.

Molar Mass of HCl = 1+35.5 =36.5g/mol

Mass conc. Of HCl = 3 x 36.5 = 109.5g

3 0

3 years ago

Which is the strongest reducing agent in the periodic table?

ycow [4]

"Caesium" is the strongest reducing agent in Periodic table.

Hope this helps!

8 0

3 years ago

Read 2 more answers