What would make a conversion not possible?

Chemistry

2 answers:

MAVERICK [17]2 years ago

6 0

if it is already as low as it can go

EleoNora [17]2 years ago

5 0

Answer:

you generally can't convert between units of mass and volume (e.g., grams to gallons).

Explanation: Examples of Conversion Factors Remember, the two values must represent the same quantity as each other. For example, it's possible to convert between two units of mass (e.g., grams to pounds),

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(08.02 MC)

lesantik [10]

Answer:

d. 0.208 M NaOH

Explanation:

M[NaOH] = 23+16+1= 40g/mol

2.40L = 2.4dm3

M=m/Mv

M=20.0g/40g/mol×2.4dm3

M=20.0g ÷ 96

M= 0.208 M NaOH

3 0

2 years ago

Use bond energies to calculate δhrxn for the reaction. n2(g)+3cl2(g)→2ncl3(g)

Jobisdone [24]

ΔHrxn = ΣδΗ(bond breaking) - ΣδΗ(bond making)

Bond enthalpies,
N ≡ N ⇒ 945 kJ mol⁻¹
N - Cl ⇒ 192 kJ mol⁻¹
Cl - Cl⇒ 242 kJ mol⁻¹

According to the balanced equation,
ΣδΗ(bond breaking) = N ≡ N x 1 + Cl - Cl x 3
= 945 + 3(242)
= 1671 kJ mol⁻¹
ΣδΗ(bond making) = N - Cl x 3 x 2
= 192 x 6
= 1152 kJ mol⁻¹

δHrxn = ΣδΗ(bond breaking) - ΣδΗ(bond making)
= 1671 kJ mol⁻¹ - 1152 kJ mol⁻¹
= 519 kJ mol⁻¹

3 0

3 years ago

Read 2 more answers

GIVING BRAINLIEST One mole of hydrogen gas (H2), reacts with one mole of bromine Br2(g) to produce 2 moles of hydrogen bromide g

JulsSmile [24]

Answer:

The equation to show the the correct form to show the standard molar enthalpy of formation:

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

Explanation:

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

Given, that 1 mole of $H_2$ gas and 1 mole of $Br_2$ liquid gives 2 moles of HBr gas as a product.The reaction releases 72.58 kJ of heat.