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earnstyle [38]
2 years ago
13

Sodium (Na), lithium (Li), and potassium (K) are grouped together in one column on the periodic table. Which property do they sh

are?
Chemistry
1 answer:
tatuchka [14]2 years ago
8 0

Answer:

They´re pretty reactive whit wather forming hidroxydes, they have the same metal brigthness as well the same electrons valence

Explanation:

This can be for the ubications of the protons to the nucleoid of athom

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1. BaBr2(aq) + H2SO4(aq) →?<br> please balance the equation and predict the products
Andrew [12]

Answer:

BaBr2 (aq) + H2SO4 (aq) → BaSO4 (s) + 2 HBr (aq)

Explanation:

This is a precipitation reaction: BaSO4 is the formed precipitate.

4 0
3 years ago
Express the following number in scientific notation: 0.000000675.
tekilochka [14]
There are 7 digits from decimal to 1st digit, and it's coming from right, so exponent will be in negative 7

In short, Your Answer would be: Option C) <span>6.75 × 10-7
</span>
Hope this helps!
6 0
3 years ago
The protein lysozyme unfolds at a transition temperature of 75.5°C, and the standard enthalpy of transition is 509 kJ mol-1. Cal
spin [16.1K]

Answer:

0.4774 KJ/K.mol

Explanation:

We are told that the transition at 25.0°C occurs in three steps. Steps i, ii and iii.

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii

Now,

C_p,m(unfolded protein) = C_p,m(folded protein) + 6.28 kJ/K.mol

Now, for the first process, ΔS_i is given as;

ΔS_i = C_p,m × In(T2/T1)

We are given;

T1 = 25°C = 25 + 273.15K = 298.15 K

T2 = 75.5°C = 75.5 + 273.15 K=348.65 K

Thus;

ΔS_i = C_p,m × In(348.65/298.15)

Now, for the third process, ΔS_iii is given as;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(T1/T2)

Thus;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)

Now, we don't know C_pm. So, we have to find a way to eliminate it. We will do it by rewriting In(298.15/348.65) in such a way that when ΔS_iii is added to ΔS_i, C_p,m will cancel out. Thus;

In(298.15/348.65) can also be written as;

In(348.65/298.15)^(-1) or

- In(348.65/298.15)

Thus;

ΔS_iii = - [(C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)]

Now, let's add ΔS_iii to ΔS_i to get;

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] + [(-C_p,m - 6.28 kJ/K.mol) × In(348.65/298.15)]

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] - [C_p,m × In(348.65/298.15)] - [6.28In(348.65/298.15)]

First 2 terms will cancel out to give;

ΔS_i + ΔS_iii = -6.28In(348.65/298.15)

ΔS_i + ΔS_iii = -0.9826 KJ/K.mol

Now,for process ii;

ΔS_ii = standard enthalpy of transition/Transition Temperature

Thus;

ΔS_ii = (509 KJ/K.mol)/348.65

ΔS_ii = 1.46 KJ/K.mol

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii = -0.9826 + 1.46 = 0.4774 KJ/K.mol

5 0
3 years ago
Which statement is true at STP? (The atomic mass of Zn is 65.39 u.)
earnstyle [38]
Zinc is a metal. At STP, it exists as solid and is stable as it is. It is an important mineral and is used in many applications like in food, metal and drugs. Zinc can be found in the Earth's crust and also it is present in small amounts in some food.
7 0
3 years ago
Reaction 1: C3H8(g)+5O2(g)→3CO2(g)+4H2O(g), ΔH1=−2043 kJ
Leokris [45]
Given teh equation adn the heat of reaction, reaction 2's heat of reaction can be obtained by simply multiplying teh heat of reaction of 1 by 3. The final answer is -6129 kJ. 
6 0
3 years ago
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