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3 years ago

Please help it’s a test due in five minutes! I’ll give 10 points

Chemistry

1 answer:

barxatty [35]3 years ago

3 0

Answer: negative acellaration or mass.

Explanation:

the first reason why is that i got that quistion right. and when objects are unbalanced it gives negative acellaration

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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

3 years ago

Potassium cyanide is a toxic substance, and the median lethal dose depends on the mass of the person or animal that ingests it.

kakasveta [241]

Answer:

The volume will be 89.6875 ml

Explanation:

So to count this we will use a single proportion.

0.0640 mol - 1000 ml

5.74×10−3 mol - x ml

x ml=5.74×10−3 mol*1000 ml/0.0640 mol=89.6875 ml

6 0

3 years ago

When 1 mol of sodium nitrate, is dissolved in 1 cubic decimeter of water, 40 kJ of heat energy is absorbed. What is the drop in

vesna_86 [32]

Answer:

1.0 ° C

Explanation:

The molar mass for Sodium Nitrate NaNO₃ = (23+14+(16×3)) = 85

Number of moles of NaNO₃ = mass of NaNO₃ /molar mass of NaNO₃

⇒ 17/85 = 1.38 moles

Since 1 mole of NaNO₃ dissolved in 1 cubic decimeter of water, 40 kJ of heat energy is absorbed.

when 1.38 mole of NaNO₃ dissolved in 1 cubic decimeter of water, x kJ of heat energy is absorbed..

Then; x kJ of 1.38 mole of NaNo₃ = 1.38 × 40 kJ =55.2 kJ of heat absorbed.

Using the relation : Q = mcΔT to determine the temperature drop ; we get:

55.2 = 17 × 4 (ΔT)

55.2 = 68 ΔT

ΔT= 0.8 ° C

ΔT ≅ 1.0 ° C

Therefore, the drop in temperature when 17.0g of sodium nitrate is dissolved in 1 cubic decimeter of water is 1.0 ° C

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mote1985 [20]

1- Base substitutions.

2- Deletions.

3- Insertions.

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