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2 years ago

Someone help me please i’ll give brainly

Chemistry

1 answer:

drek231 [11]2 years ago

7 0

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2>

$\huge\boxed{Sublimation}$

<h2>_____________________________________</h2><h2>Sublimation:</h2>

Sublimation is a process in which the solid molecules gain energy from their surrounding and they now have enough energy to overcome the force of attraction with the neighboring molecules, thus they escape into the surrounding without going through the liquid phase, that means they directly jump to the vapor phase.

<h2>_____________________________________</h2><h2>Endothermic</h2>

Those process which absorb energy from their surrounding to form the product are called endothermic process. Since the SUBLIMATION requires additional energy, it is an endothermic change.

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2><h2 />

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If 34.7 g of AgNO₃ react with 28.6 g of H₂SO₄ according to this UNBALANCED equation below, how many grams of Ag₂SO₄ could be for

Luba_88 [7]

The number of grams of Ag2SO4 that could be formed is 31.8 grams

<u><em> calculation</em></u>

Balanced equation is as below

2 AgNO3 (aq) + H2SO4(aq) → Ag2SO4 (s) +2 HNO3 (aq)

Find the moles of each reactant by use of mole= mass/molar mass formula

that is moles of AgNO3= 34.7 g / 169.87 g/mol= 0.204 moles

moles of H2SO4 = 28.6 g/98 g/mol =0.292 moles

use the mole ratio to determine the moles of Ag2SO4

that is;

the mole ratio of AgNo3 : Ag2SO4 is 2:1 therefore the moles of Ag2SO4= 0.204 x1/2=0.102 moles

The moles ratio of H2SO4 : Ag2SO4 is 1:1 therefore the moles of Ag2SO4 = 0.292 moles

AgNO3 is the limiting reagent therefore the moles of Ag2SO4 = 0.102 moles

<h3> finally find the mass of Ag2SO4 by use of mass=mole x molar mass formula</h3>

that is 0.102 moles x 311.8 g/mol= 31.8 grams

3 0

3 years ago

What volume of 15.6M NH4OH is needed to make 500. mL of 3.00M solution?

Mariana [72]

Answer:

V₁ = 96.2 mL

Explanation:

Given data:

Initial volume of NH₄OH required = ?

Initial molarity = 15.6 M

Final molarity = 3.00 M

Final volume = 500.0 mL

Solution:

Formula:

M₁V₁ = M₂V₂

M₁ = Initial molarity

V₁ = Initial volume of NH₄OH

M₂ =Final molarity

V₂ = Final volume

Now we will put the values.

15.6 M ×V₁ = 3.00 M×500.0 mL

15.6 M ×V₁ = 1500 M.mL

V₁ = 1500 M.mL /15.6 M

V₁ = 96.2 mL

6 0

3 years ago

There are four hundred participants in the local Thanksgiving 5K race. If 51% of the participants are female, what is the ratio

Paha777 [63]

Answer:

51:49 is the ratio of female participants to male participants.

Explanation:

Percentage of females in the race = 51 %

Percentage of males in the race = 100% - 51 % = 49%

Total participants in race = 400

Number of women participants = $51% of 400=\frac{51}{100}\times 400=204$

Number of men participants = $49% of 400=\frac{49}{100}\times 400=196$

Ratio of female participants to male participants :

$\frac{204}{196}=\frac{51}{49}=51:49$

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3 years ago

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Answer:

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3 years ago

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Alexandra [31]

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<em>shell and an overall positive charge..therefore it is an</em> ionic bond

7 0

3 years ago