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mojhsa [17]
4 years ago
10

if a plot weight (in g) vs. volume (in ml) for a metal gave the equation y= 13.41x and r^2=0.9981 what is the density of the met

al?
Chemistry
2 answers:
Bumek [7]4 years ago
8 0

ANS: density = 13.41 g/ml

Density (d) of a substance is the mass (m) occupied by it in a given volume (v).

Density = mass/volume

i.e. d = m/v

m = (d) v -----(1)

The given equation from the plot of weight vs volume is :

y = 13.41 x ----(2)

Based on equations (1) and (2) we can deduce that the density of the metal is 13.41 g/ml

Phoenix [80]4 years ago
7 0

Answer:

13.41 g/mL is the density of the metal.

Explanation:

y= 13.41\times x

Slope intercept form of line:

y = mx +c

y = y, x= x m = 13.41 and c = 0

y= 13.41x + 0

Where y and x parameter as are the weights in grams and volume in milliliter respectively.

\frac{Mass}{Voluime}=\frac{y}{x}=14.41 g/mL

13.41 g/mL is the density of the metal.

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A sample of oxygen gas at a pressure of 1.14 atm and a temperature of 21.5 degrees Celsius, occupies a volume of 815 ml. If the
maria [59]

Answer:

The volume of the gas will be 573.52 mL.

Explanation:

Boyle’s law states that <u>the  pressure of a fixed amount of gas at a constant temperature is inversely proportional  to the volume of the gas</u>.

The equation for Boyle's law is:

P₁V₁ = P₂V₂

We want to know V₂, that is, the volume of the container after the compression. We rearrange the equation and calculate:

V₂ = P₁V₁ ÷ P₂

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7 0
4 years ago
An unknown piece of metal absorbs 1350 J of heat as 55.0 g of the metal heats up from 20.0 oC to 47.0 oC
mash [69]

Answer:

C=0.91\frac{J}{g\°C}

Explanation:

Hello there!

In this case, according to the given information, it is possible to recall the equation to calculate the heat, Q, in these calorimetry problems as shown below:

Q=mC(T_f-T_i)

Thus, given the absorbed heat, mass and temperatures, we can easily calculate the specific heat of the metal as shown below:

C=\frac{Q}{m(T_f-T_i)}

Then, by plugging in we obtain:

C=\frac{1350J}{55.0g(47.0\°C-20.0\°C)} \\\\C=0.91\frac{J}{g\°C}

Best regards!

7 0
3 years ago
Perform each of the following unit conversions using the conversion factors given below: 1 atm = 760 mmHg = 101.325 kPa
Alenkasestr [34]
  unit  coversation
1.429  atm  - 1086mmhg

9361 pa-9.36 KPa  -  70.21  mmhg

725 mmhg -0.95 atm-  96.26  kpa

calculation

(a)     1 atm =  760  mmhg
    1.429 atm = ?
1.429  x760/1 = 1086.34  mm hg

(B)   1  mmhg  =  101.325  kpa
              ?      =9361 KPa
9361   x1 /101.25  =70.21  mmhg

760  mm hg= 101.325 KPa
70.21  mm hg=?

70.21  x101.325/760  = 9.36 Kpa

(C ) 1 atm = 760 mmhg
       ?   =  725
= 725 x1/ 760=0.95  atm


1 atm = 101.325 kpa
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0.95 x101.325/1 = 96.26 KPa
5 0
4 years ago
Read 2 more answers
A spirit burner used 1.00 g methanol to raise the temperature of 100.0 g water in a metal can from 28.00C to 58.0C. Calculate th
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Complete question:

A spirit burner used 1.00 g methanol to raise the temperature of 100.0 g water in a metal can from 28.00C to 58.0C. Calculate the heat of combustion of methanol in kJ/mol.

Answer:

the heat of combustion of the methanol is 402.31 kJ/mol

Explanation:

Given;

mass of water, m_w = 100 g

initial temperature of water, t₁ = 28 ⁰C

final temperature of water, t₂ = 58 ⁰C

specific heat capacity of water = 4.184 J/g⁰C

reacting mass of the methanol, m = 1.00 g

molecular mass of methanol = 32.04 g/mol

number of moles = 1 / 32.04

                             = 0.0312 mol

Apply the principle of conservation of energy;

n\Delta H_{methanol} = Q_{water}\\\\n\Delta H_{methanol} =  mc\Delta t\\\\n\Delta H_{methanol} =  100 \times 4.184\times (58-28)\\\\n\Delta H_{methanol} =  12,552 \ J\\\\n\Delta H_{methanol} =  12.552 \ kJ\\\\\Delta H_{methanol} = \frac{12.552}{n} \\\\H_{methanol} = \frac{12.552 \ kJ}{0.0312 \ mol} \\\\\Delta H_{methanol} = 402.31 \ kJ/mol

Therefore, the heat of combustion of the methanol is 402.31 kJ/mol

                       

5 0
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