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3 years ago

Which type of radiation is identical in mass and charge to a helium nucleus?

1 answer:

3 0

The correct answer is option B. Alpha radiation is identical in mass and charge to a helium nucleus. Alpha particle contains two protons and neutrons. It is the same as giving out a doubly-ionized helium atoms.

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Please help me to do this assignment

kirill [66]

Answer:

1. Objective

2. Objective

3. Opinion

4. Objective

5. Opinion

6. Opinion

7. Opinion (I think.)

8. Opinion (I think.)

Explanation:

7 0

3 years ago

Analysis of an ore of calcium shows that it contains 13.61 g calcium and 21.77 g oxygen in a 46.28-g sample. Calculate the perce

kondor19780726 [428]

Answer:

29.41% of Calcium and 47.04% of Oxygen

Explanation:

The percent composition of an atom in a molecule is defined as 100 times the ratio between the mass of the atom and the mass of the molecule.

The mass of the molecule of the problem (Ore) is 46.28g. That means the percent composition of Calcium is:

13.61g / 46.28g * 100 = 29.41% of Calcium

And percent composition of Oxygen is:

21.77g / 46.28g * 100 = 47.04% of Oxygen

5 0

3 years ago

How many atoms are in a sample of 1.83 moles of potassium (K) atoms? Please explain the conversions to me. Thank you!

IRINA_888 [86]

1 mole K ------------- 6.02x10²³ atoms
1.83 moles K ------ ?? atoms

1.83 x (6.02x10²³) / 1 =

1.101x10²⁴ atoms of K

hope this helps!

6 0

3 years ago

A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 8.10kg of water at 33.9 degree

lions [1.4K]

Answer:

The new temperature of the water bath 32.0°C.

Explanation:

Mass of water in water bath ,m= 8.10 kg = 8100 g ( 1kg = 1000g)

Initial temperature of the water = $T_1=33.9^oC=33.9+273K=306.9 K$

Final temperature of the water = $T_2$

Specific heat capacity of water under these conditions = c = 4.18 J/gK

Amount of energy lost by water = -Q = -69.0 kJ = -69.0 × 1000 J

( 1kJ=1000 J)

$Q=m\times c\times \Delta T=m\times c\times (T_2-T_1)$

$-69.0\times 1000 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)$

$-69,000.0 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)$

$T_2=304.86 K=304.86 -273^oC=31.86^oC\approx 32.0^oC$

The new temperature of the water bath 32.0°C.

5 0

3 years ago

Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 3.50 moles of magnesium

Firdavs [7]

Answer:

$n_{Mg}=3.50molMg\\\\ n_{Cl}=7.00molCl\\\\n_O=28.0molO$

Explanation:

Hello there!

In this case, according to the given information it turns out possible for us to realize that one mole of the given compound, Mg(ClO₄)₂, has one mole of Mg, two moles of Cl and eight moles of O; thus, we proceed as follows:

$n_{Mg}=3.50molMg(ClO_4)_2*\frac{1molMg}{1molMg(ClO_4)_2}=3.50molMg\\\\ n_{Cl}=3.50molMg(ClO_4)_2*\frac{2molCl}{1molMg(ClO_4)_2}=7.00molCl\\\\n_O=3.50molMg(ClO_4)_2*\frac{8molO}{1molMg(ClO_4)_2}=28.0molO$

Best regards!

3 0

3 years ago