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kari74 [83]
3 years ago
6

BRAINLIEST!!! ❣

Chemistry
2 answers:
SOVA2 [1]3 years ago
8 0
Q1. A.

Q2. Nitrous oxide

Hope this helped ♥︎
mafiozo [28]3 years ago
6 0

Explanation:

the first question's answer is a

the second question's answer is c

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in a typical person, the level of glucose is about 85 mg/ 100 mL of blood. if an average body contains about 11 pt of blood, how
Vanyuwa [196]

Let's note that 1 pint = 473.1765 mL, so 11 pints should be 5204.9415 mL.

We make a proportion out of the word problem

(85 mg glucose/ 100 mL) times (1 g/ 1000 mg) = 4.4242 grams of glucose

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Can someone please answer the question, <br>What is heat?​
Levart [38]

Answer:

hot

Explanation:

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What is the ph of an aqueous solution with the hydronium ion concentration : [h3o+] = 3 x 10-5 m ?
BaLLatris [955]

The concentration of hydrogen can be shown as:

[H+ ] = 3 * 10-5 M

pH can be determined as:

pH = - log [H+ ]

= - log (3 * 10-5)

= 4.53

Thus the pH of solution is 4.53


8 0
3 years ago
which of the following elements are in this unbalanced chemical reaction: li(s) + H2O(l) lioh(aq)+h2(g)​
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Answer:

Li and H

Explanation:

2Li(s)+2H2O(i)→2LiOH(aq)+H2(g) is full balanced

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2 years ago
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The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures
QveST [7]

<u>Answer:</u>

<u>For A:</u> The K_p for the given reaction is 4.0\times 10^1

<u>For B:</u> The K_c for the given reaction is 1642.

<u>Explanation:</u>

The given chemical reaction follows:

2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

  • <u>For A:</u>

The expression of K_p for the above reaction follows:

K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}

We are given:

p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm

Putting values in above equation, we get:

K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1

Hence, the K_p for the given reaction is 4.0\times 10^1

  • <u>For B:</u>

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

K_p = equilibrium constant in terms of partial pressure = 4.0\times 10^1

K_c = equilibrium constant in terms of concentration = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 500 K

\Delta ng = change in number of moles of gas particles = n_{products}-n_{reactants}=2-3=-1

Putting values in above equation, we get:

4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642

Hence, the K_c for the given reaction is 1642.

7 0
2 years ago
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