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ziro4ka [17]
3 years ago
13

35 Before the discovery of oxygen some scientists had a different theory about burning.

Chemistry
2 answers:
almond37 [142]3 years ago
7 0
C phlogiston hope this helps
skad [1K]3 years ago
3 0

Answer:

<h2>B i have to search it like 23 hours to find it but Thank you for sharing your question :)</h2>
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In the laboratory you are asked to make a 0.303 m cobalt(II) sulfate solution using 275 grams of water. How many grams of cobalt
valentinak56 [21]

Answer: 12.92g of CoSO4

Explanation:

Molar Mass of CoSO4 = 59 + 32 + (16x4) = 59 + 32 +64 = 155g/mol

Molarity of CoSO4 = 0.303mol/L

Mass conc. In g/L = Molarity x molar Mass

= 0.303x155 = 46.965g/L

275 grams of water = 0.275L of water

46.965g of CoSO4 dissolves in 1L

Therefore Xg of CoSO4 will dissolve in 0.275L i.e

Xg of CoSO4 = 46.965x0.275 = 12.92g

Therefore 12.92g of CoSO4 is needed

4 0
3 years ago
Answer the question below based on the periodic table entry for bromine.
Shalnov [3]
<span>The number of neutrons bromine will have are equal to
= protons + neutrons
so,
80-35=45</span>
4 0
3 years ago
Read 2 more answers
Are the bases on the interior or the exterior of the double helix? Are they randomly arranged or neatly stacked?
Lemur [1.5K]

the double helix is hydrogen bonded through the bases only so the bases are inside the helix only

as adenine combines with thymine and guanine with cytosine

phosphate are in the exterior of it

sugar groups constitute the double helix.

3 0
3 years ago
Read 2 more answers
The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor p
saul85 [17]

Answer : The vapor pressure of bromine at 10.0^oC is 0.1448 atm.

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of bromine at 10.0^oC = ?

P_2 = vapor pressure of propane at normal boiling point = 1 atm

T_1 = temperature of propane = 10.0^oC=273+10.0=283.0K

T_2 = normal boiling point of bromine = 58.8^oC=273+58.8=331.8K

\Delta H_{vap} = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})

P_1=0.1448atm

Hence, the vapor pressure of bromine at 10.0^oC is 0.1448 atm.

4 0
3 years ago
Nitrogen ion (Nitride) is ... Cation/Anion/Neither? # of protons? # of electrons? Charge (1-, 2-, 3-, 1+, 2+, 3+, or 0) Number o
OLga [1]

Answer:

3- is the charge and 8 dots on its Lewis dot structure.

Explanation:

Hello there!

In this case, since nitrogen is an element with five valence electrons (electrons on its outer shell), we infer that it needs three bonds to complete the octet, for which its charge, when forming nitride ions is 3-, which means it has received three electrons. Thus, when drawing the Lewis dot structure, it is evident that is will have 5+3 = 8 dots due to the electron reception.

Regards!

8 0
2 years ago
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