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3 years ago

To a 0.0001 m solution of mg(no3)2, naoh was added to a final concentration of 0.001m did a precipitate form?

1 answer:

8 0

I looked on a solubility chart to answer this question, and hydroxides are generally insoluble (with some exceptions of course). However, it says to consider $Mg(OH)_{2}$ as an insoluble substance, though it may be moderately soluble.

The answer that you are most likely looking for is: Yes, a precipitate does form - this is due to the double placement reaction:

$Mg(NO_{3})_{2}_{(aq)}$ + $2NaOH_{(aq)}$ → $Mg(OH)_{2} {(s)}$ + $2NaNO_{3}_{(aq)}$

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1. What do microorganisms share in common?

alexgriva [62]

Answer:

Bacterial species are typified by their diversity. There are three notable common traits of bacteria, 1) lack of membrane-bound organelles, 2) unicellular and 3) small (usually microscopic) size.

Explanation:

What Are the Characteristics Common to All Bacteria?

Single-Celled. Perhaps the most straightforward characteristic of bacteria is their existence as single-celled organisms. ...

Absent Organelles. ...

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DNA.

4 0

3 years ago

Suppose that 0.48 g of water at 25∘C∘C condenses on the surface of a 55-gg block of aluminum that is initially at 25∘C∘C. If the

GarryVolchara [31]

Answer : The final temperature of the metal block is, $25^oC$

Explanation :

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminum = 55 g

$m_2$ = mass of water = 0.48 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminum = $25^oC$

$T_2$ = temperature of water = $25^oC$

$c_1$ = specific heat of aluminum = $0.900J/g^oC$

$c_2$ = specific heat of water= $4.184J/g^oC$

Now put all the given values in equation (1), we get

$55g\times 0.900J/g^oC\times (T_{final}-25)^oC=-[0.48g\times 4.184J/g^oC\times (T_{final}-25)^oC]$

$T_{final}=25^oC$

Thus, the final temperature of the metal block is, $25^oC$

6 0

3 years ago

Characteristics of matter that cannot be changed are called

olchik [2.2K]

It is called Mass.

Among the physical characteristics of a matter which includes Weight, Mass, Volume, and Density the only characteristic that does not change is the mass. The mass of an object never change from place to place. All the objects have a mass and that is why there is a weight but it changes depends on the gravity. Mass don't.

5 0

3 years ago

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

How many grams of H, are needed to react with 2.75 g of N,?

Elena-2011 [213]

Answer:

0.6 grams of hydrogen are needed to react with 2.75 g of nitrogen.

Explanation:

When hydrogen and nitrogen react they form ammonia.

Chemical equation:

N₂ + 3H₂ → 2NH₃

Given mass of nitrogen = 2.75 g

Number of moles of nitrogen:

Number of moles = mass/ molar mass

Number of moles = 2.75 g / 28 g/mol

Number of moles = 0.098 mol

Now we will compare the moles of nitrogen with hydrogen from balance chemical equation:

N₂ : H₂

1 : 3

0.098 : 3×0.098 = 0.3 mol

Mass of hydrogen:

Mass = number of moles × molar mass

Mass = 0.3 mol × 2 g/mol

Mass = 0.6 g

6 0

3 years ago