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MAXImum [283]
3 years ago
9

Addition of 1 M HBr to 0.1 M solutions of which

Chemistry
1 answer:
fgiga [73]3 years ago
8 0

Addition of 1 M HBr to 0.1 M solutions of the compound (A) NaHSO3 results in evolution of a colorless gas.

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How many grams of sodium chloride are present in a 0.75 M solution with a volume of 500.0 milliliters?
saveliy_v [14]
The molarity of a solution is a type of expression of concentration equal to the number of moles solute per liter solution. In this problem, we are given the molarity equal to 0.75 M and a volume equal to 500 milliliters. <span>500 milliliters is equal to 0.5 liters. we multiply M and L to get the number of moles then multiply by the molar mass of NaCl. The answer is 21.92 grams.</span>
6 0
4 years ago
B
aev [14]

Answer:

mgh is the formula for potential energy

8 0
3 years ago
Which is heavier einsteinium (Es) or europium (Eu)
Maksim231197 [3]

Answer:

Einsteinium is heavier

Explanation:

Atomic weight is what we are referring to when figuring out how heavy an element is and is measured in amu (atomic mass units)

(This is also the same as the molar mass how heavy 6.022 x 10^23 atoms of a substance or element is which is measured in grams)

The atomic weight of an element can be found on a periodic table.

Europium has an atomic mass of approximately 152 amu (atomic mass units)

Einsteinium has an atomic mass of approximately 252 amu.

Einsteinium has a higher atomic weight so it's heavier.

There's another easier way of going about it where you don't even have to think about the actual number and basically just look at the element's atomic number and position on the periodic table.

Eu is 63

Es is 99

As the atomic number increases on the periodic table this means it's number of protons is greater.

Protons lie in the nucleus which has most of the atom's weight so the more protons the heavier it is.

Einsteinium has a higher atomic number meaning it has more protons so we can assume it's the heavier one.

Hope this helps!

3 0
3 years ago
If the toxic quantity is 1.5 gg of ethylene glycol per 1000 gg of body mass, what percentage of ethylene glycol is fatal
Elina [12.6K]

Answer:

The percentage of ethylene glycol that is fatal is 0.15 %

Explanation:

If the toxic quantity of ethylene glycol in a 1kg or 1000 g body weight is 1.5 g then the percentage of ethylene glycol that is fatal is

\frac{1.5}{1000} ˣ 100 = 0.15%

Hence, the percentage of ethylene glycol that is toxic for any body weight is 0.15%. This percentage is very important in various aspects of science including drug discovery and food production/processing

4 0
3 years ago
How much CuSO4*5H2O and Na2CO3 is needed to obtain 5g of CuCO3?
Svetllana [295]
I think the reaction that is occurring is CuSO₄(aq)+Na₂CO₃(aq)⇒CuCO₃(s)+Na₂SO₄(aq).  Water in the hydrate will just become part of water in solution so it does not really matter for this question.  
SInce we now know the balanced chemical equation, you can use stoichiometry to find the amount of reactants needed to produce 5g CuCO₃.  
The first step to any stoichiometry question is to convert the mass given into moles.  To do this you have to divide the mass by the molar mass of the compound the mass is referring to.  The molar mass of CuCO₃ is 123.5g/mol so you have to divide 5g by 123.5g/mol to get 0.04047 moles of CuCO₃.
The next step in this qustoin is to find the number of moles of each reactant using the moles of product we found in the first step.  To do this, we need go back and look at the equation to find the molar ratios which is shown through the coefficients.  Since all of the coeficients in the chemical equation is 1, we know the molar ratios are all `1 to 1 and therefore the number of moels of each reactant is equal to the number of moles of the product found in step one.  This means that you started off with 0.04047 moles of both CuSO₄ and Na₂CO₃.
The final step is to multiply the number moles of each reactant by its molar mass.  the molar mass of CuSO₄ is 159.6g/mol and the molar mass of Na₂CO₃ is 106g/mol.  When you multiply 0.04047mol by 159.6g/mol you get 6.4g and when you multiply 0.04047mol by 106g/mol you get 4.29g.
Therefore you started with 6.4g of CuSO₄ and 4.29g of Na₂CO₃.
I hope this helps.
5 0
3 years ago
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