1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
musickatia [10]
3 years ago
6

The density of methanol is 0.7918g/mL. Calculate the mass of 89.9mL of this liquid.

Chemistry
1 answer:
Novosadov [1.4K]3 years ago
4 0
The mass is 71.2 g.

Mass = 89.9 mL × \frac{0.7918 g}{1 mL} = 71.2 g
You might be interested in
How many grams of lead(II) sulfate (303 g/mol) are needed to react with sodium chromate (162 g/mol) in order to produce 0.162 kg
Afina-wow [57]

Answer : The mass of PbSO_4 needed are, 1.515 grams.

Explanation :

First we have to calculate the mole of PbCrO_4.

\text{Moles of }PbCrO_4=\frac{\text{Mass of }PbCrO_4}{\text{Molar mass of }PbCrO_4}=\frac{0.162g}{323g/mole}=0.005mole

Now we have to calculate the moles of PbSO_4.

The balanced chemical reaction will be,

PbSO_4+Na_2CrO_4\rightarrow PbCrO_4+Na_2SO_4[tex]From the balanced chemical reaction, we conclude thatAs, 1 mole of [tex]PbCrO_4 produced from 1 mole of PbSO_4

So, 0.005 mole of PbCrO_4 produced from 0.005 mole of PbSO_4

Now we have to calculate the mass of PbSO_4

\text{Mass of }PbSO_4=\text{Moles of }PbSO_4\times \text{Molar mass of }PbSO_4

\text{Mass of }PbSO_4=0.005mole\times 303g/mole=1.515g

Therefore, the mass of PbSO_4 needed are, 1.515 grams.

6 0
3 years ago
Which of the following would have the largest pKa?
garri49 [273]

Answer:

CH3CH2NH3+/CH3CH2NH2 would have the largest pKa

Explanation:

To answer this question we must know Kb of CH3CH2NH2 is 5.6x10⁻⁴, and for C6H5NH2 is 4.0x10⁻¹⁰. And the CH3CH2NH3+ and C6H5NH3+ are related with these substances because are their conjugate base. That means:

pKa of  CH3CH2NH3+ =  CH3CH2NH2;  C6H5NH3+ =  C6H5NH2

Also, Kw / Kb = Ka

Thus:

pKa of CH3CH2NH3+/CH3CH2NH2 is:

Kw / kb = Ka = 1.79x10⁻¹¹

-log Ka = pKa

pKa = 10.75

pKa of C6H5NH3+/ C6H5NH2 is:

Kw / kb = Ka = 2.5x10⁻⁵

-log Ka = pKa

pKa = 4.6

That means CH3CH2NH3+/CH3CH2NH2 would have the largest pKa

5 0
3 years ago
State and explain modern periodic table. ⊕⊕
Svetradugi [14.3K]

Explanation:

the physical and chemical properties of an element are periodic functions of their atomic number.

7 0
3 years ago
Read 2 more answers
Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia. N 2 ( g ) + 3 H 2 ( g )
Kisachek [45]

Answer:

After complete reaction, 0.280 moles of ammonia are produced

Explanation:

Step 1: Data given

Number of moles N2 = 0.140 moles

Number of moles H2 = 0.434 moles

Step 2: The balanced equation

N2(g) + 3H2 (g) ⟶ 2NH3 (g)

Step 3: Calculate the limiting reactant

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

N2 is the limiting reactant. It will completely be consumed (0.140 moles).

H2 is in excess. There will react 3*0.140 = 0.420 moles

There will remain 0.434 - 0.420 = 0.014 moles

Step 4: Calculate moles NH3

For 0.140 moles N2 we'll have 2*0.140 = 0.280 moles NH3

After complete reaction, 0.280 moles of ammonia are produced

5 0
3 years ago
or a particular isomer of C 8 H 18 , the combustion reaction produces 5104.1 kJ of heat per mole of C 8 H 18 ( g ) consumed, und
beks73 [17]

Answer:

The standard enthalpy of formation of this isomer of C_{8}H_{18} is -220.1 kJ/mol.

Explanation:

The given chemical reaction is as follows.

C_{8}H_{18}(g)+ \frac{25}{2}O_{2}(g) \rightarrow 8CO_{2}(g)+9H_{2}O(g)

\Delta H^{o}_{rxn}= -5104.1kJ/mol

The expression for the entropy change for the reaction is as follows.

\Delta H^{o}_{rxn}=[8\Delta H^{o}_{f}(CO_{2}) +9\Delta H^{o}_{f}(H_{2}O)]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}\Delta H^{o}_{f}(O_{2})]

\Delta H^{o}_{f}(H_{2}O)= -241.8kJ/mol

\Delta H^{o}_{f}(CO_{2})= -393.5kJ/mol

\Delta H^{o}_{f}(O_{2})= 0kJ/mol

Substitute the all values in the entropy change expression.

-5104.1kJ/mol=[8(-393.5)+9(-241.8)kJ/mol]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}(0)kJ/mol]

-5104.1kJ/mol=-5324.2kJ/mol -\Delta H^{o}_{f}(C_{8}H_{18})

\Delta H^{o}_{f}(C_{8}H_{18}) =-5324.2kJ/mol +5104.1kJ/mol

=-220.1kJ/mol

Therefore, The standard enthalpy of formation of this isomer of C_{8}H_{18} is -220.1 kJ/mol.

4 0
3 years ago
Other questions:
  • The oxidation number of As in Na3AsO2 is
    11·1 answer
  • Please Help! Let P and V represent the pressure and volume of Xe(g). If a piston is used to reduce the volume of the gas to V/2
    13·1 answer
  • A water solution whose pH is 4_______. a. is always neutral b. is always acidic c. is always basic d. might be neutral, basic or
    14·2 answers
  • What is the answer?
    8·2 answers
  • Which has a greater momentum, a car or a bike moving at the same speed​
    8·1 answer
  • What is the period of a wave with a frequancy of 3 Hz and an amplitude off 0.01
    9·1 answer
  • Which law states that the pressure and absolute temperature of a fixed quantity of gas are directly proportional under constant
    7·1 answer
  • What is the substance? cdH) (Η) (Η<br> (H)C<br> cco<br> Ή) (Η<br> Η<br> Η
    15·1 answer
  • How many grams of carbon are contained in one mole of C3H8?
    13·1 answer
  • What is the IUPAC name of the following compound a. (R)-3-chloro-1-methylcyclohexene b. (S)-3-chloro-1-methylcyclohexene c. (R)-
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!