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3 years ago

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Chemistry

1 answer:

Oduvanchick [21]3 years ago

6 0

Answer:

7.58

Explanation:

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Why might genetically identical twins have different phenotype

KIM [24]

Answer:

They may have different phenotypes because of differences in their environments, such as nutrition and healthcare. Why might genetically identical twins have different phenotypes? Any genes on the same chromosome could be linked genes, whereas only genes on sex chromosomes can be sex-linked genes.

Explanation:

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3 years ago

The percent yield for the reaction in which 15.6g of aluminum hydroxide is reacted in excess hydrogen chloride gas is 92.5%. Wha

Misha Larkins [42]

Answer:

24.6 g

Explanation:

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3 years ago

What is the concentration of the solution prepared by dissolving 2.35 g of KBr (M = 119 g/mol) in 250 mL of water?

dusya [7]

Answer:

The concentration of KBr is $C = 0.07899 \ mol L^{-1}$

Explanation:

From the question we are told that

The mass of KBr is $m_{KBr} = 2.35 \ g$

The molar mass of KBr is $M_{KBr} = 119 g/mol$

Volume of water is $V = 250 \ mL = 250 *10^{-3} = 0.250 \ L$

This implies that the volume of the solution is $V = 250 mL$

The number of moles of KBr is

$n = \frac{m_{KBr}}{M_{KBr}}$

Substituting values

$n = \frac{2.35}{119}$

$n = 0.01975 \ mol$

The concentration of KBr is mathematically represented as

$C = \frac{0.01975}{0.250}$

$C = 0.07899 \ mol L^{-1}$

3 0

3 years ago

An initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid container at a certain temperature by the reaction At

anygoal [31]

Answer: [N2]₀ = 10M and [H2]₀ = 11M

Explanation: To calculate the initial concentration, you would have to set up an ICE table, which is an organized way of tracking known quantities or the ones you want to find. ICE stands for:

I is initial amount;

C is change in concentration;

E is for equilibrium concentration;

For the mixture,

N2 3H2 2NH3

I [N2]₀ [H2]₀ 0

C - x -3x +2x

E [N2]₀ - x =8 [H2]₀ - 3x =5 2x =4

With the product, we can find "x":

2x=4

x=2M

With x=2, find the concentrations:

[N2]₀ - x = 8

[N2]₀ = 10M

[H2]₀ - 3x = 5

[H2]₀ = 11M

The initial concentrations of nitrogen gas [N2] is 10.0 M and of hydrogen gas [H2] is 11.0 M.

8 0

3 years ago

Read 2 more answers

It is proposed to use Liquid Petroleum Gas (LPG) to fuel spark-ignition engines. A typical sample of the fuel on a volume basis

Norma-Jean [14]

Answer:

The overall balanced combustion reaction is written as :

$0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2 \ + \ 3.8H_2O \ + \ 18.612N_2$

$(F/A)_{stoichiometric} = 0.0424$

$(A/F)_{stoichiometric} = 23.562$

the higher heating values $(HHV)_f$ per unit mass of LPG = 49.9876 MJ/kg

the lower heating values $(LHV)_f$ per unit mass of LPG = 46.4933 MJ/kg

Explanation:

The stoichiometric equation can be expressed as :

$0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> aCO_2 \ + \ bH_2O \ + \ cN_2$

Now, equating the coefficient of carbon; we have:

(0.7×3)+(0.05×4)+(0.25×3) = a

a = 3.05

Also, Equating the coefficient of hydrogen : we have:

(0.7 × 8) +(0.05 × 10) + ( 0.25 × 6) = 2 b

2b = 7.6

b = 3.8

Equating the coefficient of oxygen

2x = 2a + b

$x = \frac{2a+b}{2} \\ \\ x = \frac{2(3.05)+3.8}{2} \\ \\ x = 4.95$

Equating the coefficient of Nitrogen

$c = 3.76x \\ \\ c = 3.76 *4.95 \\ \\ c = 18.612$

Therefore, The overall balanced combustion reaction can now be written as :

$0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2 \ + \ 3.8H_2O \ + \ 18.612N_2$

Now; To determine the stoichiometric F/A and A/F ratios; we have:

$(F/A)_{stoichiometric} = \frac{n_f}{n_a } \\ \\ (F/A)_{stoichiometric} = \frac{1}{4.95*(1+3.76)} \\ \\ (F/A)_{stoichiometric} = 0.0424$

$(A/F)_{stoichiometric} = \frac{n_a}{n_f } \\ \\ (A/F)_{stoichiometric} = \frac{4.95*(1+3.76)}{1} \\ \\ (A/F)_{stoichiometric} = 23.562$

What are the higher and lower heating values per unit mass of LPG?

Let calculate the molecular mass of the fuel in order to determine their mass fraction of the fuel components.

Molecular mass of the fuel $M_f = (0.7*M_{C_3H_5} ) + (0.05 *M_{C_4H_{10}}) + (0.25*M _{C_3H_6})$

= 30.8 + 2.9 + 10.5

= 44.2 kg/mol

Mass fraction of the fuel components can now be calculated as :

$m_{C_3H_8} = \frac{30.8}{44.2} \\ \\ m_{C_3H_8} = 0.7 \\ \\ \\ m_{C_4H_{10}} = \frac[2.9}{44.2} \\ \\ m_{C_4H_{10}} = 0.06 \\ \\ \\ m_{C_3H_6} = \frac{10.5}{44.2} \\ \\ m_{C_3H_6} = 0.24$

Finally; calculating the higher heating values $(HHV)_f$ per unit mass of LPG; we have:

$(HHV)_f=(0.7 * HHV_{C_3H_8}) + (0.06 *HHV_{C_4H_{10}})+(0.24*HHV_{C_3H_6} \\ \\ (HHV)_f=(0.7*50.38)+(0.06*49.56)+(0.24*48.95) \\ \\ (HHV)_f=49.9876 \ MJ/kg$

calculating the lower heating values $(LHV)_f$ per unit mass of LPG; we have:

$(LHV)_f = (HHV)_f - \delta H_w \\ \\ (LHV)_f = (HHV)_f - [\frac{m_w}{m_f}h_{vap}] \\ \\ (LHV)_f = 49.9876 \ MJ/kg - [\frac{3.8*18}{44.2}*2.258 \ MJ/kg] \\ \\ (LHV)_f = 46.4933 \ M/kg$

7 0

3 years ago