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3 years ago

How many grams of NH4OH are present in 37.0 grams of 6.00 M NH4OH which has a density of 0.960 g/mL?

Chemistry

1 answer:

inn [45]3 years ago

6 0

Answer:

The mass of NH₄OH present in the solution is 8.085 g

Explanation:

Given;

molarity of NH₄OH = 6 M

density of NH₄OH = 0.960 g/mL

mass of NH₄OH = 37.0 grams

volume of NH₄OH = mass /density

= 37/0.96

= 38.54 mL

molarity = moles of solute / Liters of solution

moles of solute = molarity x Liters of solution

moles of solute = 6 x 38.54 x 10⁻³

moles of solute = 0.231 moles

Reacting mass = number of moles x molecular mass

Molecular mass of NH₄OH = 35 g/mol

Reacting mass = 0.231 mol x 35 g/mol

Reacting mass = 8.085 g

Therefore, the mass of NH₄OH present in the solution is 8.085 g

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2 years ago

An airplane travels with a constant velocity of 210 m/s and in the upper atmosphere where the plane is traveling there is a wind

kolezko [41]

Answer:

hello your question is incomplete the options are missing

Determine the resultant velocity for the plane when it is travelling

i) To the east

ii) To the west

answer :i) 270 i

ii) -150 i

Explanation:

velocity of Airplane = 210 m/s

wind velocity = 60 m/s to the east

The resultant velocity for the plane when it is travelling

let the velocity of the wind = V2

velocity of the plane = v1

i) The resultant velocity for the plane when travelling to the east

Vr = V2 i + V1 i

Vr= 60i + 210i = 270i

ii) resultant velocity when the plane is travelling to the west

Vr = - V1 i + V2i

= -210i + 60 i = -150 i

4 0

3 years ago

What is the standard enthalpy of a reaction for which the equilibrium constant is (a) doubled, (b) halved when the temperature i

Alexxandr [17]

Answer:

a) 48KJ

b) -48KJ

Explanation:

Given that;

ln(K2/K1) = ΔH°/R(1/T2 - 1/T1)

K2= equilibrium constant at T2

K1 = equilibrum constant at T1

R = gas constant

T1 = initial temperature

T2 = final temperature

When we double the equilibrium constant K1; K2 = 2K1

T1 = 310 K

T2 = 310 + 15 = 325 K

ln(2K1/K1) =- ΔH°/R(1/T2 - 1/T1)

ln2 = -ΔH°/8.314(1/325 - 1/310)

0.693 = -ΔH°/8.314(3.08 * 10^-3 - 3.2 * 10^-3)

0.693 = -ΔH°/8.314 (-0.00012)

0.693 = 0.00012ΔH°/8.314

0.693 * 8.314 = 0.00012ΔH°

ΔH° = 0.693 * 8.314/0.00012

ΔH° = 48KJ

b) K2 =K1/2

ln(K1/2/K1) =- ΔH°/R(1/T2 - 1/T1)

ln (1/2) = -ΔH°/8.314 (1/325 - 1/310)

-0.693 = -ΔH°/8.314 (-0.00012)

-0.693 = 0.00012ΔH°/8.314

-0.693 * 8.314 = 0.00012ΔH°

ΔH°= -0.693 * 8.314/0.00012

ΔH°= -48KJ

6 0

3 years ago

When methane ( CH4 ) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reac

slega [8]

<h2>Answer:</h2> $1.33*10^{-2}grams$

<h2>Explanations</h2>

The complete balanced equation for the given reaction is expressed as;

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

Given the following parameters

Mass of CH4 = 5.90×10^−3 g = 0.0059grams

Determine the moles of methane

$\begin{gathered} moles\text{ of CH}_4=\frac{mass}{molar\text{ mass}} \\ moles\text{ of CH}_4=\frac{0.0059}{16.04} \\ moles\text{ of CH}_4=0.000368moles \end{gathered}$

According to stoichimetry, 1 mole of methane produces 2 moles of water, hence the moles of water required will be:

$\begin{gathered} moles\text{ of H}_2O=\frac{2}{1}\times0.000368 \\ moles\text{ of H}_2O=0.000736moles \end{gathered}$

Determine the mass of water produced

$\begin{gathered} Mass\text{ of H}_2O=moles\times molar\text{ mass} \\ Mass\text{ of H}_2O=0.000736\times18.02 \\ Mass\text{ of H}_2O=0.0133grams=1.33\times10^{-2}grams \end{gathered}$

Therefore the mass of water produced from the complete combustion of 5.90×10−3 g of methane is 1.33 * 10^-2grams

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1 year ago

Is a cube of salt being crushed before being stirred into water a physical or chemical change

Olin [163]

By crushing the salt, you are performing a physical change because you aren't altering the chemical makeup of the salt, just the physical form. Hope this helps! :)

6 0

2 years ago

Read 2 more answers