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3 years ago

Equal but opposite forces are said to result in a(n)_____force of zero

Chemistry

2 answers:

Zina [86]3 years ago

5 0

Answer:

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Alenkinab [10]3 years ago

3 0

The answer for this question is Net

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Hydrogen cyanide, HCN, can be made by a two-step process. First, ammonia reacts with O2 to give nitric oxide, NO.

stealth61 [152]

Answer:

The mass of HCN is 79.65 g.

The mass of reactant which remain at the end of both reactions is 88.5 g.

Explanation:

Given that,

Mass of ammonia = 50.2 g

Mass of methane = 48.4 g

Hydrogen cyanide, HCN, can be made by a two-step process

Ammonia reacts with O₂ to give nitric oxide NO.

The reaction is,

$4NH_{3}+5O_{2}\Rightarrow 4NO+6H_{2}O$

We need to calculate the mole of NO

Using given data,

2.25 g NH_{3}=\dfrac{50.2}{17}= 2.95\ mole\ NH_{3} [/tex]

$4\ mole NH_{3}\ glose 4\ mol NO$

2.95 mol NH₃ will produced 2.95 mol NO

Then nitric oxide reacts with methane,

The reaction is,

$2NO+2CH_{4}\Rightarrow 2HCN+2H_{2}O+H_{2}$

We need to calculate the mole of methane

Using given data,

$mole\ of\ methane=\dfrac{48.4}{16}$

$mole\ of\ methane = 3.03\ moles$

2 mole NO produced 2 mole HCN

2.95 mol NO will produced $\dfrac{2.95\times3.03}{3.03}$ = 2.95 mol HCN

We need to calculate the mass of HCN

Using formula of mass

$m=N\times M$

Where, N = number of mole

M = molecular mass

Put the value into the formula

$m=2.95\times27$

$m= 79.65\ g$

The mass of HCN is 79.65 g.

We need to calculate the mass of NO

Using formula of mass

$m=N\times M$

Where, N = number of mole

M = molecular mass

Put the value into the formula

$m=2.95\times30$

$m= 88.5\ g$

Hence, The mass of HCN is 79.65 g.

The mass of reactant which remain at the end of both reactions is 88.5 g.

4 0

3 years ago

(Yield Problem)

alex41 [277]

Answer:

Percent Yield Fe = 82.5%

Explanation:

The actual yield is the value produced after an experiment is conducted. The theoretical yield is the value calculated using the balanced chemical equation and atomic/molar masses.

To find the percent yield of iron (Fe), you need to (1) convert grams Al to moles Al (via atomic mass), then (2) convert moles Al to moles Fe (via mole-to-mole ratio from equation coefficients), then (3) convert moles Fe to grams Fe (via atomic mass), and then (4) calculate the percent yield. It is important to arrange the ratios in a way that allows for the cancellation of units. The final answer should have 3 sig figs to reflect the sig figs of the given values.

Atomic Mass (Mg): 24.305 g/mol

Atomic Mass (Fe): 55.845 g/mol

3 Mg + 2 FeCl₃ -----> 2 Fe + 3 MgCl₂

20.5 g Mg 1 mole 2 moles Fe 55.845 g
----------------- x ----------------- x ---------------------- x ----------------- =
24.305 g 3 moles Mg 1 mole

= 31.4 g Fe

Actual Yield
Percent Yield = ---------------------------------- x 100%
Theoretical Yield

25.9 g Fe
Percent Yield = -------------------- x 100%
31.4 g Fe

Percent Yield = 82.5%

5 0

2 years ago

When an aqueous solution of silver nitrate is

Andre45 [30]