LOTS OF POINTS PLZ HELP!!!!! ...I got the first two for ya...

If 7.84 × 107 J of energy is released from a fusion reaction, what amount of mass in kilograms would be lost? Recall that c = 3

katrin [286]

m = 7.84x107/(3x108)2kg = 7.84x107/9x1016kg = 0.871x10-9 kg = 8.71x10-10 kg

3 0

3 years ago

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A chemist wants to make 6.5 L of a 0.340 M CaCl2 solution. Part A What mass of CaCl2 (in g) should the chemist use? Express your

katovenus [111]

Answer:

The answer to your question is 245 grams

Explanation:

Data

Volume 6.5 L

Molarity = 0.34

mass of CaCl₂ = ?

Process

1.- Calculate the molar mass of CaCl₂

molar mass = (1 x 40) + (2 x 35.5)

= 40 + 71

= 111 g

2.- Convert the grams to moles

111 g of CaCl₂ -------------- 1 mol

x ---------------0.34 mol

x = (0.34 x 111) / 1

x = 37.74 g

3.- Calculate the total mass

37.74 g ------------------ 1 L

x ------------------ 6.5 L

x = (6.5 x 37.74) / 1

x = 245.31

3 0

3 years ago

A 155 g sample of an unknown substance was heated from 25°c to 40°c. in the process, the substance absorbed 569 calories of ener

Tasya [4]

Energy= 2381 joules
heat= Mass(kg) *change in temperature(K) * Cp
2381=0.155*(15)*Cp
Cp=1024 J/kg K

5 0

3 years ago

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When 0.49 g of a molecular compound was dissolved in 20.00 g of cyclohexane, the freezing point of the solution was lowered by 3

IRISSAK [1]

Answer: The molecular mass of this compound is 131 g/mol

Explanation:

Depression in freezing point:

$\Delta T_f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$\Delta T_f$ = depression in freezing point = $3.9^oC$

$k_f$ = freezing point constant = $20.8^0C/m$

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

$w_2$ = mass of solute = 0.49 g

$w_1$ = mass of solvent (cyclohexane) = 20.00 g

$M_2$ = molar mass of solute = ?

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

$(3.9)^oC=1\times (20.8^oC/m)\times \frac{(0.49g)\times 1000}{M_2\times (20.00g)}$

$M_2=131g/mol$

Therefore, the molar mass of solute is 131 g/mol

7 0

2 years ago

Complete the sentences about heme. Some terms will not be used. The prosthetic group of hemoglobin and myoglobin is . The organi

nevsk [136]

Explanation:

Haemoglobin consists of heme unit which is comprised of an $Fe^{2+}$ and porphyrin ring. The ring has four pyrrole molecules which are linked to the iron ion. In oxyhaemoglobin, the iron has coordinates with four nitrogen atoms and one to the F8 histidine residue and the sixth one to the oxygen. In deoxyhaemoglobin, the ion is displaced out of the ring by 0.4 Å.

The prosthetic group of hemoglobin and myoglobin is - Heme

The organic ring component of heme is - Porphyrin

Under normal conditions, the central atom of heme is - $Fe^{2+}$

In deoxyhemoglobin , the central iron atom is displaced 0.4 Å out of the plane of the porphyrin ring system.

The central atom has six bonds: four to nitrogen atoms in the porphyrin, one to a histidine residue, and one to oxygen.

7 0

2 years ago