All categories

3 years ago

Which element was first discovered on the sun? hydrogen carbon oxygen helium

2 answers:

3 0

The element first discovered in the sun is the element Helium. This noble gas is the second most abundant element in the universe and was discovered on the sun before it was discovered on the earth. Helium is the result of the nuclear fusion of hydrogen creating thus the mass of energy in the sun.

Anon25 [30]3 years ago

3 0

Answer:

Helium

Explanation:

You might be interested in

6.45 g of C6H12O6 is burned in a bomb calorimeter containing 950. g of water and the temperature goes from 35.0∘C to 42.3∘C. If

emmasim [6.3K]

Answer:

$q_{rxn}=-488.86\frac{kJ}{mol}$

Explanation:

Hello!

In this case, since this calorimetry problems can by analyzed by figuring out that the heat lost due to the combustion of C6H12O6 is gained by the water and the calorimeter which undergo the mentioned temperature rise, we can write:

$Q_{rxn}=-(m_wC_w+C_c)(T_2-T_1)$

Thus, by plugging in the given data, we obtain:

$Q_{rxn}=-(950.0g*4.184\frac{J}{g\°C} +933\frac{J}{\°C})(42.3\°C-35.0\°C)\\\\Q_{rxn}=-17.5kJ$

Next, we compute the moles of C6H12O6 by using its molar mass (180.18 g/mol) as shown below:

$n=6.45g*\frac{1mol}{180.18g} =0.0358mol$

Thus, the value for q of the reaction turns out:

$q_{rxn}=\frac{Q_{rxn}}{n} \\\\q_{rxn}=\frac{17.5kJ}{0.0358mol} \\\\q_{rxn}=-488.86\frac{kJ}{mol}$

Best regards!

8 0

3 years ago

Read 2 more answers

Tincture of iodine has antiseptic properties. This solution is made by dissolving.

ser-zykov [4K]

Tincture of iodine has antiseptic properties and is made by dissolving iodine crystals in alcohol.

<h3>What are antiseptics?</h3>

Antiseptics are substance which prevent or stops the growth of microorganisms.

Microorganisms have been known to cause diseases, therefore the application of antiseptics are useful to prevent diseases.

An example of an antiseptic is tincture of iodine.

Tincture of iodine is prepared by dissolving iodine crystals in alcohol.

In conclusion, antiseptics are useful in the prevention of diseases by microorganism. An example of an antiseptic is tincture of iodine.

Learn more about tincture of iodine at: brainly.com/question/3209701

#SPJ1

7 0

2 years ago

Using table 9.4, calculate an approximate enthalpy (in kj) for the reaction of 1.02 g gaseous methanol (ch3oh) in excess molecul

Tanya [424]

Given:

Mass of methanol = 1.02 g

To determine:

Enthalpy for the reaction of 1.02 g of methanol with excess O2

Explanation:

Balanced equation-

2CH3OH(g) + 3O2(g) → 2CO2(g) + 4H2O(g)

The reaction enthalpy is given as:

ΔHrxn = ∑nH°f(products) - ∑nH°f(reactants)

where n = number of moles

H°f = standard enthalpy of formation.

ΔHrxn = [2H°f(CO2(g)) + 4H°f(H2O(g))] - [2H°f(CH3OH(g)) + 3H°f(O2(g))]

= [2(-393.5) + 4(-241.8)]-[2(-201.5) + 3(0)] = -1351.2 kJ

Now, 1 mole of CH3OH = 32 g

The calculated ΔHrxn corresponds to 2 moles of CH3OH. i.e.

The enthalpy change for 64 g of Ch3OH = -1351.2 kJ

Therefore, for 1.02g gaseous methanol we have:

ΔH = 1.02 * -1351.2/64 = -21.5 kJ

Ans: The enthalpy for the given reaction is -21.5 kJ

6 0

3 years ago

What is the [oh−] of a solution if its ph = 8.78? answer in units of m?

dexar [7]

PH = -log([H+])
[H+] = 10^(-pH)

[H+] = 10^(-8.78) = 1.65*10^-9

[H+][OH-] = Kw
Kw = 1.0*10^-14 at 25 degrees celsius.

[OH-] = Kw/[H+] = (1.0*10^-14)/(1.65*10^-9) = 6.06*10^-6

The concentration of OH- ions is 6.1*10^-6 M.

7 0

3 years ago

A solution is made by dissolving 23.0 g KOH in 1.60 L H2O. What is the molality of the solution? The density of water is 1.00 g/

artcher [175]

The molarity of a solution is defined as the moles of solute dissolved per kilogram of solvent. Therefore, we first compute the moles of KOH using:

Moles = mass / Mr
Moles = 23 / 56
Moles = 0.41

The volume of solvent is 1.6 liters
The density is 1 gram/cm³ = 1 kg/L
Mass of solvent = density * volume
Mass of solvent = 1 * 1.6
Mass of solvent = 1.6 kg

Molality = moles / kilogram
Molarity = 0.41 / 1.6
Molarity = 0.26

The molality of the solution is 0.26 molal.

4 0

3 years ago

Read 2 more answers