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charle [14.2K]
2 years ago
12

4Hg(OH)2+ H3PO4 —> Hg3(PO4)2 + 5H20

Chemistry
1 answer:
kotegsom [21]2 years ago
7 0

Answer:

Hi. im an online tutor and i ca help you with all your assignments . check out our wesite https://toplivewriters.com

Explanation:

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What unit is mass generally measured in
Veronika [31]

Answer:

kilograms

Explanation:

hope this helps, pls mark brainliest :D

6 0
3 years ago
Read 2 more answers
How many moles of nitrogen trifluoride (NF3) can be produced from 9.65 mole of Fluorine gas (F2)
user100 [1]

Answer:

6.43 moles of NF₃.

Explanation:

The balanced equation for the reaction is given below:

N₂ + 3F₂ —> 2NF₃

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of NF₃.

Finally, we shall determine the number of mole of nitrogen trifluoride (NF₃) produced by the reaction of 9.65 moles of Fluorine gas (F₂). This can be obtained as follow:

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of NF₃.

Therefore, 9.65 moles of F₂ will react to to produce = (9.65 × 2)/3 = 6.43 moles of NF₃.

Thus, 6.43 moles of NF₃ were obtained from the reaction.

4 0
3 years ago
The half life of radon-222 is 3.8 days. How Much of a 100g sample is left after 15.2 days
professor190 [17]

Answer:  

6.2 g  

Explanation:  

In a first-order decay, the formula for the amount remaining after <em>n</em> half-lives is  

N = \frac{N_{0}}{2^{n}}  

where  

<em>N</em>₀ and <em>N</em> are the initial and final amounts of the substance  

1. Calculate the <em>number of half-lives</em>.  

If t_{\frac{1}{2}} = \text{3.8 da}  

n = \frac{t}{t_{\frac{1}{2}}} = \frac{\text{15.2 da}}{\text{3.8 da}}= \text{4.0}

2. Calculate the <em>final mass</em> of the substance.  

\text{N} = \frac{\text{100 g}}{2^{4.0}} = \frac{\text{100 g}}{16} = \text{6.2 g}

4 0
3 years ago
How much heat must be transferred to 100.0g of water to change its temperature by 35° C?
nekit [7.7K]

Answer:

14700J

Explanation:

From the question given, the following were obtained:

M = 100g

ΔT = 35° C

C = 4.2J/g °C

Q=?

The heat transferred can calculated for by using the following equation

Q = MCΔT

Q = 100 x 4.2 x 35

Q= 14700J

7 0
3 years ago
What is the pressure inside a 750 mL can of deodorant that starts at 15 degrees Celsius and 1.0 atm if the temperature is raised
Sonbull [250]

The answer is: the pressure inside a can of deodorant is 1.28 atm.

Gay-Lussac's Law: the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.

p₁/T₁ = p₂/T₂.  

p₁ = 1.0 atm.; initial pressure

T₁ = 15°C = 288.15 K; initial temperature.

T₂ = 95°C = 368.15 K, final temperature

p₂ = ?; final presure.

1.0 atm/288.15 K = p₂/368.15 K.  

1.0 atm · 368.15 K = 288.15 K · p₂.  

p₂ = 368.15 atm·K ÷ 288.15 K.  

p₂ = 1.28 atm.  

As the temperature goes up, the pressure also goes up and vice-versa.  

6 0
3 years ago
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