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myrzilka [38]
2 years ago
13

All the objects in the force diagrams shown below have the same mass.

Chemistry
1 answer:
Stolb23 [73]2 years ago
3 0

Answer:

do u know what the answer is yet

Explanation:

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A student wants to make a 5.00%solution of rubidium chloride using 0.377g of the substance. what mass of water will be needed to
Alenkasestr [34]
 The mass of water that will  be needed to make the solution  is calculated as below

%  solution  =  mass of the solute/mass of the solvent(water) x100

%  solution = 5% = 5/100
mass of the solute =0.377 g
mass of the  solvent = ?

let the mass  of the solvent be  represented by Y

= 5/100 =0.377/y

by cross multiplication

5y=  37.7
divide both side by 5

y =7.54  grams
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3 years ago
Enough of a monoprotic acid is dissolved in water to produce a 0.0121 M solution. The pH of the resulting solution is 6.49. Calc
Hitman42 [59]
Calculate the H positive from the pH equation: pH equals -log (H positive). This would be 10 to the -6.49. Let's call the acid HA. To calculate Ka in this equation, Ka equals H positive times A- over HA. HA is going to be the 0 0121. So, Ka=(10^-6.49)^2/0.0121. This equals 1.05*10^-13/0.0121. Ka then equals 8.65*10^-12.
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3 years ago
This lab question is: How can you distinguish a physical change from a chemical change?
geniusboy [140]

Answer:b,and c

Explanation:

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Chemical equations must be balanced to satisfy ...?
murzikaleks [220]

Answer:

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Read 2 more answers
Be sure to answer all parts. The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calcu
pochemuha

Answer : The volume of CO_2 will be, 514.11 ml

Explanation :

The balanced chemical reaction will be,

HCO_3^-+HCl\rightarrow Cl^-+H_2O+CO_2

First we have to calculate the  mass of HCO_3^- in tablet.

\text{Mass of }HCO_3^-\text{ in tablet}=32.5\% \times 3.79g=\frac{32.5}{100}\times 3.79g=1.23175g

Now we have to calculate the moles of HCO_3^-.

Molar mass of HCO_3^- = 1 + 12 + 3(16) = 61 g/mole

\text{Moles of }HCO_3^-=\frac{\text{Mass of }HCO_3^-}{\text{Molar mass of }HCO_3^-}=\frac{1.23175g}{61g/mole}=0.0202moles

Now we have to calculate the moles of CO_2.

From the balanced chemical reaction, we conclude that

As, 1 mole of HCO_3^- react to give 1 mole of CO_2

So, 0.0202 mole of HCO_3^- react to give 0.0202 mole of CO_2

The moles of CO_2 = 0.0202 mole

Now we have to calculate the volume of CO_2 by using ideal gas equation.

PV=nRT

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = 37^oC=273+37=310K

n = number of moles of gas = 0.0202 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get :

(1.00atm)\times V=0.0202 mole\times (0.0821L.atm/mole.K)\times (310K)

V=0.51411L=514.11ml

Therefore, the volume of CO_2 will be, 514.11 ml

6 0
2 years ago
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