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AleksAgata [21]
3 years ago
6

Mummification frequently preserves fine detail and internal organs true or false

Chemistry
1 answer:
mariarad [96]3 years ago
6 0
<span>I believe thats false. think about the mummys youve seen? do you see fine detail and intact organs</span>
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A- south<br> B- north <br> C-west<br> D- east <br> Help ASAP no links or I’m reporting
Agata [3.3K]

Answer:

east

Explanation:

5 0
3 years ago
What does an atom become if it gains or loses electrons?
Tju [1.3M]
I think a is the answerr
3 0
3 years ago
72.40% iron,27.60% Oxygen Empirical formula
VARVARA [1.3K]

Explanation:

Fe. O

72.40/ 56 27.60/16

____________________

1.29/1.29 1.725/1.29

_____________________

1. :. 1

<h3>Emperical formula = FeO</h3>
6 0
3 years ago
A chemist measures the enthalpy change ?H during the following reaction: Fe (s) + 2HCl (g) ? FeCl2 (s) + H2 (g) =?H?157.kJ Use t
erik [133]

Correct Question:

A chemist measures the enthalpy change ΔH during the following reaction: Fe(s) + 2HCl(g)-->FeCl2(s) + H2 ΔH=-157.0 kJ. Use this information to complete the table below. Round each of your answers to the nearest kJ/mol

Answer:

-314 kJ

+628 kJ

+157 kJ

Explanation:

The enthalpy change of a reaction measures the amount of heat that is lost or gained by it. If ΔH >0 the heat is gained, and the reaction is called endothermic, if ΔH<0, the heat is lost, and the reaction is called exothermic.

If the reaction is inverted, the value of ΔH is inverted too (the opposite endothermic reaction is exothermic), and if the reaction is multiplied by a constant, ΔH will be multiplied by it too.

1) 2Fe(s) + 4HCl --> 2FeCl2(s) + 2H2(g)

This reaction is the product of the given reaction by 2, so

ΔH = 2*(-157) = -314 kJ

2) 4FeCl2(s) + 4H2(g) --> 4Fe(s) + 8HCl(g)

This reaction is the inverted reaction given multiplied by 4, so

ΔH = 4*(157) = +628 kJ

3) FeCl2(s) + H2(g) --> Fe(s) + 2HCl

This reaction is the inverted reaction given, so

ΔH = +157 kJ

4 0
3 years ago
A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base
tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ =  0.00375  mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

HN_3 + OH- ---> N^-_{3} + H_2O

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction =  0.025 +  0.0133 = 0.0383  L

Concentration of HN_3 = \dfrac{0.001755}{0.0383} = 0.0458 M

Concentration of N^{-}_3 = \dfrac{ 0.001995 }{0.0383} = 0.0521 M

GIven that :

Ka = 1.9 x 10^{-5}

Thus; it's pKa = 4.72

pH =4.72 +  log(\dfrac{ \ 0.0521}{0.0458})

pH =4.72 + log(1.1376)

pH =4.72 + 0.05598

pH =4.77598

pH ≅ 4.80

3 0
3 years ago
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