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Chemistry

1 answer:

andrew11 [14]3 years ago

8 0

hellow! how r you? ｡◕‿◕｡

Explanation:

nice to know you then:)

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For the hypothetical compound AX2, which of the following statements is true?. a.) A is a nonmetal from Group 3A, and X is a met

Ierofanga [76]

The cations has positive charges that are metals while the anions have negative charges that are non-metals. Upon reaction, there is an exchange in charges that are reflected in the subscripts of the atoms. In this case, compound AX2 must have a cation, A belonging to group 2 A with +2 charge and anion, X belonging to Group 7A with -1 charge. Answer is D.

4 0

3 years ago

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How is cloning different from hybridization?

kvasek [131]

1. Hybridization is a method of sexual reproduction whereas cloning is a method of asexual reproduction.

2. Hybrid animals are sterile, but cloned animals are fertile.

3. Hybrid organisms contain DNA from both male and female parents but cloned organisms contain DNA from only one type of parent.

4. Hybrid has superior character over its parents (improved hybrid vigor) but clones are 100% identical to their parent.

5. Hybridization gives only one hybrid progeny, whereas through cloning unlimited identical organisms can be produced.

6 0

3 years ago

What volume of ammonia in liters, at STP, can be produced from 2.0 kg of hydrogen gas, H2, and an excess of nitrogen gas, N2? N2

Sonbull [250]

Ok so rewrite the equation for yourself :).

N2 + 3H2 -> 2NH3 (Always check if it is balanced).

Ok so we have 2kg of hydrogen conver this to grams and then to moles.
2kg -> 2000g
2000g divide by R.M.M of H2
R.M.M
2H - 2 * 1 -> 2
2000/2 - 1000 moles.
Look at the ratio of H2 to 2NH3.
We see that we get 2NH3 for every H2 therefore the ratio is 1:2.
This means that we HAVE 2000 moles of 2NH3.
At stp. we know that 1 moles = 22.4 Litres
We have 2000 moles that means that we produce 44800 litres of NH3 at stp.
2000 * 22.4 - 44800
Hope this helps :).

4 0

3 years ago

If a solution containing 51.429 g of mercury(ii) perchlorate is allowed to react completely with a solution containing 16.642 g

OLga [1]

Hg(No3)2 +NaSO4 --->2NaNO3 + HgSO4(s)
calculate the moles of each reactant
moles=mass/molar mass

moles of Hg(NO3)2= 51.429g/ 324.6 g/mol(molar mass of Hg(NO3)2)=0.158 moles

moles Na2SO4 16.642g/142g/mol= 0.117 moles of Na2SO4

Na2SO4 is the limiting reagent in the equation and by use mole ratio Na2So4 to HgSO4 is 1:1 therefore the moles of HgSO4 =0.117 moles

mass of HgSO4=moles x molar mass of HgSo4= 0.117 g x 303.6g/mol= 35.5212 grams

7 0

4 years ago

You have 363 mL of a 1.25M potassium chloride solution, but you need to make a 0.50M potassium chloride solution. How many milli

maxonik [38]

Answer:- 544.5 mL of water need to be added.

Solution:- It is a dilution problem. The equation used for solving this type of problems is:

$M_1V_1=M_2V_2$

where, $M_1$ is initial molarity and $M_2$ is the molarity after dilution. Similarly, $V_1$ is the volume before dilution and $V_2$ is the volume after dilution.