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sdas [7]
3 years ago
10

When two air masses meet what do they form?

Chemistry
2 answers:
Yanka [14]3 years ago
6 0

Answer:

ice. ......

Explanation:

because it cold

Dafna11 [192]3 years ago
3 0

Answer: a weather front

Explanation:one air mass is lifted above the other

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A compound is 52.14% C, 13.13% H, and 34.73% O. What is the empirical formula of the compound?
alina1380 [7]
% H = 100 - ( 52.14 + 34.73 )=13.13 % 

<span>assume 100 g of this compound </span>
<span>mass H = 13.13 g </span>
<span>moles H = 13.13 g / 1.008 g/mol=13 </span>

<span>mass C = 52.14 g </span>
<span>moles C = 52.14 g/ / 12.011 g/mol=4 </span>

<span>mass O = 34.73 g </span>
<span>moles O = 34.73 g/ 15.999 g/mol=2 </span>

<span>the empirical formula is C4H13O2</span>
7 0
3 years ago
Read 2 more answers
How many Cl atoms are in Zn(ClO3)2?<br> O A. 2<br> O B. 1<br> O c. 3<br> O D. 6
White raven [17]
The answer cl atoms are in zn(cio3)2 is A:2 because I looked it up and it gave me 2.
3 0
3 years ago
True or false elements in the same colum have simular properties
AlexFokin [52]
True since ithave same number of electrons and protons
3 0
3 years ago
To calculate the enthalpy change for the reaction, 2CO (g) + O2 (g) Imported Asset 2 CO2 (g), you can use ΔHf0 values for each r
svetoff [14.1K]

Answer:

ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]

Explanation:

Chemical equation:

CO + O₂   →  CO₂

Balanced chemical equation:

2CO + O₂   →  2CO₂

The standard enthalpy for the formation of CO = -110.5 kj/mol

The standard enthalpy for the formation of O₂  = 0  kj/mol

The standard enthalpy for the formation of CO₂  = -393.5 kj/mol

Now we will put the values in equation:

ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]

ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol + 0]

ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol]

ΔH0reaction = -283 kj/mol

7 0
3 years ago
A water sample has a pH of 8.2 and a bicarbonate concentration of 97 mg/L. What is the alkalinity of the sample in moles/L and i
goldfiish [28.3K]

Answer:6.94

Explanation:

Molar mass of CaCO3=40+12+16×3

=40+12+48=100g/mol

Moles=mass of substance/molar mass

=97mg/100g=0.097/100=0.00097moles/L.

PH=-log[CaCo3]=-log(0.00097)=6.94

P.s it's log to base e

3 0
3 years ago
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