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3 years ago

2CO +0, 200, what is the mole ratio of carbon monoxide to carbon

Chemistry

1 answer:

Nastasia [14]3 years ago

6 0

Answer:

B. 2.2

Explanation:

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A strong acid, HA, is added to water. Which statement about the solution is true?

Arturiano [62]

My answer to this question is C

3 0

3 years ago

A sample of gas occurs 9.0mL at a pressure of 500mmHg. A new volume of the same sample is at 750mmHg. Use two significant figure

Anit [1.1K]

Answer:

V₂ = 6.0 mL

Explanation:

Given data:

Initial volume = 9.0 mL

Initial pressure = 500 mmHg

Final volume = ?

Final pressure = 750 mmHg

Solution:

According to Boyle's Law

P₁V₁ = P₂V₂

V₂ = P₁V₁ / P₂

V₂ = 500 mmHg × 9.0 mL / 750 mmHg

V₂ = 4500 mmHg .mL / 750 mmHg

V₂ = 6.0 mL

5 0

3 years ago

Identity the anode and cathode

Anna007 [38]

Answer:

Pb(s) ---> Pb+2 + 2e- is the anode

Cu+2(aq) + 2e- ---> Cu(s) is the cathode

7 0

3 years ago

Determine the boiling point of a 3.70 m solution of phenol in benzene. Benzene has a boiling point of 80.1°C and a boiling point

xeze [42]

Answer: The boiling point of a 3.70 m solution of phenol in benzene is $89.5^0C$

Explanation:

Elevation in boiling point:

$\Delta T_b=i\times k_b\times m$

where,

$\Delta T_b$ = change in boiling point

i= vant hoff factor = 1 (for benzene which is a non electrolyte )

$k_b$ = boiling point constant = $2.53^0C/kgmol$

m = molality = 3.70

$T_{solution}-T_{solvent}=i\times k_b\times m$

$T_{solution}-80.1^0C=1\times 2.53\times 3.70$

$T_{solution}=89.5^0C$

Thus the boiling point of a 3.70 m solution of phenol in benzene is $89.5^0C$

8 0

3 years ago

Which missing item would complete this beta decay reaction?<br> PLATO

goldfiish [28.3K]

Answer: $_{-1}^0\beta$

Explanation:

General representation of an element is given as: $_Z^A\textrm{X}$

where,

Z represents Atomic number

A represents Mass number

X represents the symbol of an element

Beta decay : In this process, a neutron gets converted into a proton and an electron releasing a beta-particle. The beta particle released carries a charge of -1 units.

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta$

The given reaction would be represented as:

$_{43}^{98}\textrm{Tc}\rightarrow_{44}^{98}\textrm{Pb}+_{-1}^0\beta$

7 0

3 years ago

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