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3 years ago

There pls help it’s due in 5 minutes!! Thanks if you do!

Chemistry

1 answer:

sergey [27]3 years ago

7 0

Explanation:

well when you heat ice the individual molecules gain kinetic energy. but until the temperature reaches it's melting point they don't have the energy to break bonds and hold their crystal structure. So, the temperature remains constant until the ice has melted. Hope this helped. Good luck on your project!

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12.5 g of water are warmed from 12C to 72C. How much energy in J were absorbed? (specific heat of water=4.18 J/gC)

Marta_Voda [28]

Answer

Energy = 3135 J

Explanation

Given:

Mass of water = 12.5 g

Final temperature = 72 C

Initial temperature = 12 C

Specific heat of water = 4.18 J/g C

Required: Energy

Solution

Q = m x Cp x ΔT

where Q is heat absorbed or released

m is the mass

Cp is the specific heat capacity

ΔT = final temperature - initial temperature

Q = 12.5 g x 4.18 J/gC x (72 - 12 C)

Q = 3135 J

7 0

1 year ago

5. How many grams of cobalt are in 1.52 x 1024 atoms of cobalt?

klasskru [66]

149 g of cobalt are present in 1.52 x $10^{24}$ atoms of cobalt.

Explanation:

One mole of an atom is equal to the Avagadro number of the atom.

$1 \ mol \ of \ Co = 6.023\times10^{23} atoms\ of \ Co$

The number of atoms per mol is divided by the given atoms of Cobalt.

= $(1.52 \times 10^{24}) \times$ $\frac{1 \ mol\ Co}{6.023 \times10^{23}\ atoms\ Co}$

= 2.52 mol of Co

To covert, the moles into grams multiply the mole value by its molar mass

The Molar mass of Cobalt is 58.93

= 2.52 $\times$ 58.93

= 149 g.

6 0

3 years ago

If 25.16 g of chlorine react with 12.99 g of manganese metal, what is the empirical formula of the compound?

WARRIOR [948]

We use the given masses of the reactants to calculate the moles of Mn and Cl. Empirical formula represents the simplest mole ratio of atoms present in a compound.

Moles of Mn = $12.99 g Mn * \frac{1 mol Mn}{54.94 g Mn} = 0.236 mol Mn$

Moles of Cl = $25.16 g Cl_{2} *\frac{1 mol Cl_{2}}{70.91 g Cl_{2}} * \frac{2 mol Cl}{1 mol Cl_{2}}$ = 0.710 mol Cl

Simplest mole ratio: $Mn_{\frac{0.236}{0.236}}Cl_{\frac{0.710}{0.236}}$

So the empirical formula is $MnCl_{3}$

6 0

3 years ago

I neeeeeeeeeeeddddddddd help!!!

Elza [17]

Answer:

Explanation:

Butane I think, if right pls mark brainliest.

6 0

4 years ago

Read 2 more answers

2al(s)+3cl2(g)→2alcl3(s) δh∘ = -1408.4 kj. how much heat (in kilojoules) is released on reaction of 4.70 g of al?

Annette [7]

Answer:
- 122.58 kJ

Solution:
According to following equation,

2 Al + 3 Cl₂ → 2 AlCl₃ δH° = -1408.4 kJ

When,
54 g (2 mole) Al on reaction releases = 1408.4 kJ heat
So,
4.70 g of Al will release = X kJ of Heat

Solving for X,
X = (4.70 g × 1408.4 kJ) ÷ 54 g

X = - 122.58 kJ

7 0

4 years ago