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4 years ago

What is the name of H2S? A. sulfuric acid B. sulfurous acid C. hydrosulfurous acid D. hydrosulfuric acid

Chemistry

2 answers:

kondor19780726 [428]4 years ago

7 0

Correct answer is D! Good luck!

damaskus [11]4 years ago

3 0

<h3>correct answer is D. Hydrosulfuric acid</h3><h3>100% sure.. mark as Brainliest please</h3>

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Which missing item would complete this alpha decay reaction?<br><br> 257/100 Fm= ________+4/2He

maria [59]

Answer: ₉₈²⁵³Cf

253 is a superscript to the left of the symbol, Cf, which represents the mass number, and 98 is a subscript to the left of the same symbol, which represents the atomic number.

Explanation:

1) The alpha decay equation shows that the isotope Fm - 257, whose nucleus has 100 protons and 157 neutrons, emitted an alpha particle (a nucleus with 2 protons and 2 neutrons).

2) Therefore:

i) the mass number decreased in 4, from 257 to 257 - 4 = 253.

2) the atomic number decreased in 2, from 100 to 100 - 2 = 98.

3) Hence the formed atom has atomic number 98, which is californium, Cf, and the isotope is californium - 253.

4) The item that completes the given alpha decay reaction is:

₉₈²⁵³ Cf.

5) The complete alfpha decay reaction is:

₁₀₀²⁵⁷ Fm → ₉₈²⁵³Cf + ₂⁴He

You can verify the mass balance:

257 = 253 + 4, and

100 = 98 + 2

4 0

3 years ago

Read 2 more answers

Magnesium sulfate heptahydrate is heated until all the water is driven off. The sample loses 11.80 grams upon heating. What was

iren [92.7K]

Hello!

The initial mass of Magnesium Sulfate Heptahydrate (MgSO₄·7H₂O) is 23,08 g

The chemical reaction for the dehydrating of Magnesium Sulfate Heptahydrate (MgSO₄·7H₂O) is the following:

MgSO₄·7H₂O(s) + Δ → MgSO₄(s) + 7H₂O(g)

We know that the sample loses 11,80 g upon heating. That mass is the mass of Water that is released as vapor. Knowing that piece of information, we can apply the following conversion factor to go from the mass of water to the moles of water and back to the mass of the original compound (mi).

$mi=11,80gH_2O* \frac{1 mol H_2O}{18gH_2O}* \frac{1molMgSO_4.7H_2O}{7molH_2O}* \frac{246,47 g MgSO_4.7H_2O}{1 mol MgSO_4.7H_2O} \\ \\ mi=23,08gMgSO_4.7H_2O$

Have a nice day!

6 0

3 years ago

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

3 years ago

A solution has a poh of 7. 1 at 10∘c. what is the ph of the solution given that kw=2. 93×10−15 at this temperature? remember to

Gnom [1K]

A solution has a pOH of 7. 1 at 10∘c. Then the pH of the solution given that kw=2. 93×10−15 at this temperature is 7.4 .

It is given that,

pOH of solution = 7.1

Kw =2.93×10^(-15)

Firstly, we will calculate the value of pKw

The expression which we used to calculate the pKw is,

pKw=-log [Kw]

Now by putting the value of Kw in this expression,

pKw =−log{2.93×10^(-15)}

pKw =15log(2.93)

pKw=14.5

Now we have to calculate the pH of the solution.

As we know that,

pH+pOH=pKw

Now put all the given values in this formula,

pH+7.1=14.5

pH=7.4

Therefore, we find the value of pH of the solution is, 7.4.

learn more about pH value:

brainly.com/question/12942138

#SPJ4

7 0

1 year ago

Try to rationalize the sign of ΔS∘rxn in each case.

bonufazy [111]

Answer:

a) decrease, gas

b) increase, gas

c) liquid

d) increase, solid

Explanation:

Entropy refers to the degree of disorderliness of a system. If the number of moles of gas increases from left to right in a reaction, the entropy of the system increases positively.

Similarly, when the number of liquid molecules remain constant, there could only be a very little increase in entropy.

However, solids have the least entropy and the entropy of a system decreases when a system yields solid products.

6 0

3 years ago