Help On Chemistry Questions

Help with chemistry questions?

Can someone please explain how i would answer these questions thanks!

1. Determine the amount concentration, in mol/L, of 3.6×10-2 g of magnesium chloride dissolved in 65 mL of solution\

2. What mass, in grams, of ammonium phosphate is present in 1.40 L of a 125 mmol/L solution?

3.20 mL of 0.095 mol/L sulfuric acid solution is diluted up to 50 mL. What is the new concentration of the acid, in mol/L?

4.Question 8 options:
0.400 L of distilled water is added to 100 mL of a 2.10 mol/L solution of nitric acid. What is the new concentration of the acid, in mol/L?

Help with chemistry questions?

1) Given the reaction:
131 I ----------->131 Xe+ X
53 54

What particle is represented by X?
a) alpha
b) beta
c) neutron
d) proton

2) Given the nuclear equation:
9 Be + X----->6 Li + 4 He
4 3 2
What is the identity of particle X?
a) 1 H
1

b) 2 H
1

c) 0 e
-1

d) 1 n
0
-1

3) In a nuclear fusion reaction, the mass of the products is
a) less than the mass of the reactants because some of the mass has been converted to energy
b) less than the mass of the reactants because some of the energy has been converted to mass
c) more than the mass of the reactants because some of the mass has been converted to energy
d) more than the mass of the reactants because some of the energy has been converted to mass

Thanks!

Help with chemistry questions!!?

Chemistry question.....

1. This is an example of a question with poorly constructed answer choices since only one choice contains the right answer and the other part is ambiguous. Weight is a force and is proportional to acceleration (F=ma, Newton's second law). Even if flying level, and only experiencing the acceleration due to gravity at 10 km, the weight of a person will be essentially indistinguishable from the weight at the ground. The force of gravity is still quite strong even at 250 km above the surface (orbit of ISS), and even the orbit of the moon, which is of course what helps keep the Earth and Moon in orbit around their center of mass.

Questions 2, 3 and 4 are "d", "d" and "b", respectively. The gram is NOT the SI unit for mass, the kilogram is.

As for the 5th question, the indicator of precision is the closeness to other measurements of the same thing. The more agreement between measurements, the more precision in the measurements. Here's another reference to reinforce my point. https://labwrite.ncsu.edu/Experimental%2...

The number of significant digits is related to precision, usually, the more precise the measurement, the greater the number of significant digits, but the number of significant digits is not a definition of precision.

Chemistry Questions HELP?

Part 1)
(1.16 g Zn) / (65.380 g Zn/mol) = 0.0177 mol Zn
(29.2 g Ar) / (39.9480 g Ar/mol) = 0.731 mol Ar
(70.5 g Ta) / (180.94788 g Ta/mol) = 0.390 mol Ta
(2.28 × 10^−2 g Li) / (6.9410 g Li/mol) = 0.00328 mol Li

Part 2)
(6.67 mol W) × (183.840 g W/mol) = 1226 g = 1.23 kg W
(0.582 mol Ba) × (137.3270 g Ba/mol) = 79.9 g Ba
(68.3 mol Xe) × (131.2930 g Xe/mol) = 8967 g = 8.97 kg Xe
(1.56 mol S) × (32.0655 g S/mol) = 50.0 g S

Part 3)
(16.5 g Sr) / (87.620 g Sr/mol) × (6.022 × 10^23 atoms/mol) = 1.13 × 10^23 atoms Sr
(26.8 g Fe) / (55.8452 g Fe/mol) × (6.022 × 10^23 atoms/mol) = 2.89 × 10^23 atoms Fe
(8.19 g Bi) / (208.9804 g Bi/mol) × (6.022 × 10^23 atoms/mol) = 2.36 × 10^22 atoms Bi
(33.9 g P) / (30.97376 g P/mol) × (6.022 × 10^23 atoms/mol) = 6.59 × 10^23 atoms P

Help Chemistry Question?

if 0.240 of an unknown triprotic acid (H3X) requires 18.00 mL of a 0.500 M NaOH solution for neutralization, what is the molecular weight of the acid? If it is a common phosphorous containing ox-acid, what is the formula and name?

I assume you mean 0.240 GRAMS of H3X...
3NaOH + H3X ==> 3H2O + Na3X
moles NaOH used = 0.018L x 0.500 mol/L = 0.009 moles
moles H3X present = 1/3 x 0.009 moles = 0.003 moles H3X
molar mass of H3X = 0.240 g/0.003 moles = 80 g/mole
Since there are 3 hydrogens (3.03) and P (30.97), that accounts for 34, leaving 80 - 34 = 46 for the oxygens.
Each O = 16, therefore there would be 46/16 = 2.88 O's, or 3 within experimental error.
H3PO3 would be likely. It is called phosphorous aicd. It has a molar mass of 82, close to the calculated 80.

HELP! CHEMISTRY QUESTION?

Ether, (C2H5)2O, which was originally used as an anesthetic but has been replaced by safer and more effective medications, is prepared by the reaction of ethanol with sulfuric acid.

2C2H5OH + H2SO4 ⟶ (C2H5)2 + H2SO4∙H2O

What is the percent yield of ether if 1.17 L (d = 0.7134 g/mL) is isolated from the reaction of 1.500 L of C2H5OH (d = 0.7894 g/mL)?

Help with Chemistry questions?

1. formula: m=mol/kg
18.02 g H2O x(kg/1000 g)= .01802 kg solvent
m=2.1 x 10^-4 mol/.01802 kg= .012 m

2. formula: M=mol/L
20+10=30 mL convert to L=30 x10^-3 L
find mol H=20 x10^-3 L x(.100 mol/L)= .002 mol H
find mol H=10 x10^-3 L x(.500 mol/L)= .005 mol H
now add up mol to get total H=.007 mol
plug into eqn and solve for M
M=.007 mol/30 x10^-3 L= .233 M
ANS D is correct

3. given: 250 g sln, 5.26% solute (assume % is g)
want: mol solute
know: MM=180.18 g/mol solute

step 1: g sln, go to g, go to mol
step 2: check ans

250 g sln x(5.26 g/100 g sln) x(mol/180.18 g)= .073 mol

HELP, chemistry questions?

What is the molarity of the solution?

First, balance the equation, It's a redox reaction so break it into the 2 half eqaution

reduction half equation
I2 + 2e ------> 2I-

oxidation half equation
H3AsO3(aq) ----------------> H3AsO4(aq )
balance the O by adding H2O on the side with the least O
H3AsO3(aq) + H2O --------> H3AsO4(aq )
balance the H you added in the H2O by adding H+ to the other side
H3AsO3(aq) + H2O --------> H3AsO4(aq ) + 2H+

work out how many electrons are transferred. To do this tally up the charge on each side of the arrow
RHS = 2 x H+ = 2+
LHS = 0
Now add enough electrons to the most positive side to make each side equal, that is, add 2 e to the RHS

balanced oxidation half equation
H3AsO3(aq) + H2O --------> H3AsO4(aq ) + 2H+ + 2e


Now add the two half equations together.
I2 + 2e ----------> 2I-
H3AsO3(aq) + H2O --------> H3AsO4(aq ) + 2H+ + 2e
---------------------------------------...
H3AsO3 + I2 + H2O ---------> H3AsO4 + 2I- + 2H+

Now, you can see that 1 mole I2 reacts with 1 mole of H3AsO3

moles H3AsO3 = molarity x litres
= 0.2317 M x 0.04500 L
= 0.0104265 mol

Since 1 mole I2 reacts with 1 mole H3AsO3 then this is how many moles of I2 there must have been

mass I2 = molar mass x moles
= 253.8 g/mol x 0.0104265 mol
= 2.646 g


Approach the second problem the same way, ,first get your balanced equation.
Then determine moles of reagent used and calculate how many moles of CH3CH2OH must have been present to react with this many moles of reagent.
Then convert this to the % ethanol there must have been present in the blood.