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Aneli [31]
3 years ago
14

Can sum1 please help me? if u can

Chemistry
1 answer:
kiruha [24]3 years ago
6 0

Answer:

donde estan las preguntas

Explanation:

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How many grams of ammonium chloride (NH4Cl) would be required to make a saturated solution in 1000 grams of water at 50 degrees
Pavel [41]
<span>Since,
1000 grams of water = 1000 mL of water</span><span>
So, 
At any of the given temperature:
</span>1000 mL = 10 x 100 mL
<span>
moles of NH4Cl = 53.5/53.49
                          = 1.0 m
                          = 1.0 mol/Kg
Delta T = 2 x 1.86 x 1.0
             = 3.72 c
             = - 3.72 °C</span>
3 0
3 years ago
1. How many grams would 8.1 x 1021 molecules of sucrose (C12H22011)<br> weigh?
Nimfa-mama [501]

Answer:

Mass = 4.6 g

Explanation:

Given data:

Number of molecules of sucrose = 8.1 ×10²¹ molecules

Mass of sucrose = ?

Solution:

First of all we will calculate the number of moles by using Avogadro number.

1 mole × 8.1 ×10²¹ molecules / 6.022×10²³ molecules

1.35 × 10⁻² mol

Mass of sucrose:

Mass = number of moles × molar mass

Molar mass = 342.3 g/mol

Mass = 1.35 × 10⁻² mol ×342.3 g/mol

Mass = 462.1  × 10⁻² g

Mass = 4.6 g

7 0
4 years ago
After undergoing alpha decay, an atom of radium-226 becomes
Zanzabum
The answer is b. radon-222. The alpha decay means that it will emit an alpha particle when decays. The alpha particle has two protons and two neutrons. So Radium(88) minus two protons will become Radon(86). And the atomic mass will become 226-4=222.
8 0
3 years ago
The number of nitrogen atoms in one mole of nitrogen gas are...
9966 [12]

Explanation:

The number of nitrogen atoms in one mole of nitrogen gas are <em><u>6.02214179×1023 nitrogen </u></em><em><u>atoms</u></em><em><u>.</u></em><em><u> </u></em>

<em>Hope this helps... </em>

3 0
3 years ago
580 grams of boiling water (temperature 100°C, specific heat capacity 4.2 J/K/gram) are poured into an aluminum pan whose mass i
fenix001 [56]

Answer:

81.04°C

Explanation:

Heat loss by water = Heat gained by Aluminum

Heat loss by water;

H = MCΔT

ΔT = 100 -  T2

M = 580g

c = 4.2

H = 580 * 4.2 (100 - T2)

H =  243600  - 2436T2

Heat ganed by Aluminium

H = MCΔT

ΔT = T2 - 24

M = 900g

c = 0.9

H = 900 * 0.9 (T2 - 24)

H = 810 T2 - 19440

243600  - 2436T2 = 810 T2 - 19440

243600 + 19440 =  810 T2 + 2436T2

263040 = 3246 T2

T2 = 81.04°C

Assumption;

Assume that energy diffuses throughout the pan and water so that all parts reach the same final temperature.

3 0
4 years ago
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