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Semenov [28]
3 years ago
5

An empty beaker weighs 20g and when filled with kerosene weighs 60g. If the volume of the kerosene is 15cm3, calculate the densi

ty of the kerosene
Chemistry
1 answer:
oee [108]3 years ago
6 0

Answer:

2.7 g/cm³

Explanation:

Step 1: Calculate the mass of kerosene

The mass of the full beaker (mFB) is equal to the sum of the masses of the empty beaker (mEB) and the mass of the kerosene (mK).

mFB = mEB + mK

mK = mFB - mEB

mK = 60 g - 20 g = 40 g

Step 2: Calculate the density of kerosene

Density (ρ) is an intrinsic property of matter. It can be calculated as the quotient between the mass of kerosene and its volume.

ρ = m/V

ρ = 40 g/15 cm³ = 2.7 g/cm³

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Answer:

The stomach, gallbladder, and pancreas work together as a team to perform the majority of the digestion of food. Food entering the stomach from the esophagus has been minimally processed — it has been physically digested by chewing and moistened by saliva, but is chemically almost identical to unchewed food

8 0
3 years ago
Starting with 0.657 g of lead(II) nitrate, a student collects 0.925 g of precipitate. If the calculated mass of precipitate is 0
tatuchka [14]

Answer:

101.2%

Explanation:

Given:

Theoretical yield of the precipitate = 0.914 g

Actual yield of the precipitate = 0.925 g

Now, the percent yield is given as a ratio of actual yield by theoretical yield expressed as a percentage.

Framing in equation form, we have:

\%\ yield=\frac{Actual\ yield}{Theoretical\ yield}\times 100

Now, plug in 0.925 g for actual yield, 0.914 g for theoretical yield and solve for % yield. This gives,

\%\ yield=\frac{0.925\ g}{0.914\ g}\times 100\\\\\%\ yield=1.012\times 100\\\\\%\ yield=101.2\%

Therefore, the percent yield is 101.2%.

5 0
4 years ago
H2SO4 is a strong acid because the first proton ionizes 100%. The Ka of the second proton is 1.1x10-2. What would be the pH of a
forsale [732]

Question options:

a) 2.05

b) 0.963

c) 0.955

d) 1.00

Answer:

b) 0.963

Explanation:

H2SO4→ HSO4- + H3O+

HSO4- + H2O ⇌ SO42- + H3O+

Construct ICE table:

        HSO4- (aq)    +    H2O        ⇌      SO42- (aq)     +     H3O+ (aq)

I          0.1                  solid &                   0                          0.1

C         -x                     liquid                 + x                            + x

E         0.1 - x          are ignored              x                          0.1 + x

Calculate x

Ka = products/reactants

  = \frac{[SO42-] [H3O+]}{[HSO4-]}

0.011 = \frac{x (0.1 + x)}{0.1 - x}

0.011 x (0.1 -x) = o.1x + x^2

0.0011 - 0.011 x - o.1x - x^2 = 0

0.0011 - 0.011 x - x^2 = 0

Use formula to solve for quadratic equation

x = { -b +,-\sqrt{b^2 - 4ac / 2a

a = -1, b = -0.111, c = 0.001

Solve for x

x = \sqrt[-(-o.111)]{(-0.111)^2 - 4(-1) (0.0011) }  / 2(-1)

x = 0.111 +,- \sqrt{0.012321 + 0.0044} / -2

x = 0.111 +,- \sqrt{0.016721} / -2

x = \frac{0.111 +, - 0.1293}{-2}

x = \frac{0.111 + 0.1293}{-2}   , x = \frac{0.111  - 0.1293}{-2}

x = \frac{0.2403}{-2}    , x = \frac{0.0183}{-2}

x = - 0.12015  , x = 0.00915

x cannot be negative, so

x = 0.00915 M

Calculate [H3O+]

[H3O+] = 0.1 M + x

[H3O+] = 0.1 M + 0.00915 M

[H3O+] = 0.10915 M

Clculate pH

pH = - log [ H3O+]

pH = - log [ 0.10915]

pH = 0.963

8 0
3 years ago
The recommended single dose for acetaminophen is 10.0 to 15.0 mg/kg of body weight for adults. Using this guideline, calculate t
marshall27 [118]

Answer:

1.041g is the maximum sigle dose for the person

Explanation:

The maximum dose for acetaminophen is 15.0mg / kg of body weight for adults

To find the maximum single dosage for a person who weighs 153lb we must convert the lb to kg (1lb = 0.4536kg):

153lb * (0.4536kg / 1lb) =<em> 69.4kg is the mass of the person</em>

<em />

As the maximum dose is 15.0mg / kg, the dose of the person is:

69.4kg * (15.0mg acetaminophen / kg) =

<h3>1041mg = 1.041g is the maximum sigle dose for the person</h3>
7 0
3 years ago
Select all that apply.<br><br> The stability of atomic nuclei is related to the _____.
ohaa [14]
Neutrons and protons
8 0
3 years ago
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