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AveGali [126]
3 years ago
12

. For the reaction 2HNO3 + Mg(OH), → Mg(NO3)2 + 2H20, how many grams of magnesium nitrate are produced from 8.00 mol of nitric a

cid, HNO,?​
Chemistry
1 answer:
GREYUIT [131]3 years ago
6 0

Mass of Magnesium nitrate produced : 593.2 g

<h3>Further explanation</h3>

The reaction equation is the chemical formula of reagents and product substances

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

Reaction

2HNO₃ + Mg(OH)₂ → Mg(NO₃)₂ + 2H₂0

mol HNO₃ = 8

From the equation, mol ratio of HNO₃ : Mg(NO₃)₂ = 2 : 1, so mol Mg(NO₃)₂ :

\tt \dfrac{1}{2}\times 8=4~moles

Mass Mg(NO₃)₂(MW=148,3 g/mol) :

\tt mass=mol\times MW\\\\mass=4\times 148,3 g/mol\\\\mass=593.2~g

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probably because most of our food comes from animals-

Explanation:

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Question 8 of 8&gt; 0 Attempt 2 A chemist needs to determine the concentration of a sulfuric acid solution by titration with a s
Nesterboy [21]

Answer:

12.29 M

Explanation:

  • The reaction that takes place is:

H₂SO₄ + 2NaOH → 2Na⁺ + SO₄⁻² + 2H₂O

  • Now let's calculate the <u>moles of H₂SO₄ that were titrated</u>:

0.1284 M * 0.1915L*\frac{1molAcid}{2molNaOH} = 0.01229 mol H₂SO₄.

  • Thus, the <u>concentration of the diluted solution is</u>:

0.01229 mol H₂SO₄ / 0.010 L = 1.229 M

  • Finally, the <u>concentration of the original acid solution is:</u>

1.229 M*\frac{250mL}{25mL} = 12.29 M

5 0
3 years ago
An ideal gas occupies a volume V at an absolute temperature T. If the volume is halved and the pressure kept constant, what will
Kruka [31]

Answer:

It will be halve of T

Explanation:

V1 = V

T1 = T

V2 = ½V

T2 = x

V1/T1 = V2/T2

V/T = ½V/x

Vx = ½VT

2Vx = VT

2x = T

x = ½T

6 0
3 years ago
What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig
Mashutka [201]

Answer:

P_H =2.86

c=1.4\times 10^{-4}

Explanation:

first write the equilibrium equaion ,

C_3H_6O_3  ⇄ C_3H_5O_3^{-}  +H^{+}

assuming degree of dissociation \alpha =1/10;

and initial concentraion of C_3H_6O_3 =c;

At equlibrium ;

concentration of C_3H_6O_3 = c-c\alpha

[C_3H_5O_3^{-}  ]= c\alpha

[H^{+}] = c\alpha

K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}

\alpha is very small so 1-\alpha can be neglected

and equation is;

K_a = {c\alpha \times \alpha}

[H^{+}] = c\alpha = \frac{K_a}{\alpha}

P_H =- log[H^{+} ]

P_H =-logK_a + log\alpha

K_a =1.38\times10^{-4}

\alpha = \frac{1}{10}

P_H= 3.86-1

P_H =2.86

composiion ;

c=\frac{1}{\alpha} \times [H^{+}]

[H^{+}] =antilog(-P_H)

[H^{+} ] =0.0014

c=0.0014\times \frac{1}{10}

c=1.4\times 10^{-4}

6 0
3 years ago
A volume of 80.0 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. I
murzikaleks [220]

Answer:

The mass of the steel bar is 26.833 grams

Explanation:

<u>Step 1: </u>data given

ΣQ gained = ΣQ lost

Q=m*C*ΔT

with m = mass in grams

with C= specific heat capacity ( in J/(g°C))

with ΔT = change in temperature = T2-T1

Qsteel = Qwater

msteel * Csteel * (T2steel - T1steel) = mwater * Cwater * (T2water - T1water)

Mass of steel = TO BE  DETERMINED

mass of water =⇒ since 1mL = 1g : 80 mL = 80g

Csteel =0.452 J/(g °C

Cwater = 4.18 J/(g °C

initial temperature steel T1 : 2 °C

final temperature steel T2 = 21.3 °C

initial temperature water T1 =22 °C

final temperature water T2 = 21.3 °C

<u>Step 2:</u> Calculate mass of steel

msteel * Csteel * (T2steel - T1steel) = mwater * Cwater * (T2water - T1water)

msteel * 0.452 *(21.3-2) = 80 * 4.18 * (21.3-22)

msteel = (80 * 4.18 * (-0.7)) / (0.452 * 19.3)

msteel = -234.08 / 8.7236

msteel = -26.833 g

Since mass can't be negative we should take the absolute value of it = 26.833g

The mass of the steel bar is 26.833 grams

6 0
3 years ago
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