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3 years ago

A piece of metal has a volume of 30.0cm3 and a mass of 252g. What is its density? what metal do you think this is?

Chemistry

1 answer:

Alborosie3 years ago

3 0

Answer:

Explanation:

get density = D = m / V = 0.252 / 0.00003 = 8400
metal will be Cu => bronze

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What is 7% as a decimal?

Colt1911 [192]

.07! you divide 7 by 100%

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3 years ago

A gas sample occupies 2.1 L at a pressure of 101 kPa.

Verizon [17]

Answer:

D) 0.9 L

Explanation:

At constant temperature,

PV = Constant

so,

P1.V1 = P2. V2

101 × 2.1 = 235 × V2

V2 = 0.9 L

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2 years ago

Read 2 more answers

calculate the mols of alt gas if the volume is 0.97 liters at a temperature of 12 C and the pressure is 152 Kpa’s

katrin2010 [14]

Answer:

0.062mol

Explanation:

Using ideal gas law as follows;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821Latm/molK)

T = temperature (K)

Based on the information provided;

P = 152 Kpa = 152/101 = 1.50atm

V = 0.97L

n = ?

T = 12°C = 12 + 273 = 285K

Using PV = nRT

n = PV/RT

n = (1.5 × 0.97) ÷ (0.0821 × 285)

n = 1.455 ÷ 23.39

n = 0.062mol

4 0

3 years ago

Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t

Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)$

Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

$Q_1=m\times \Delta H_{fusion}$

where,

$Q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

$\Delta H_{fusion}$ = enthalpy change for fusion = $3.35\times 10^5J/Kg$

Now put all the given values in $Q_1$ , we get:

$Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J$

<u>For process 2 :</u>

$Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})$

where,

$Q_2$ = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

$c_{p,l}$ = specific heat of liquid water = $4186J/Kg^oC$

$T_1$ = initial temperature = $0^oC$

$T_2$ = final temperature = $100^oC$

Now put all the given values in $Q_2$ , we get:

$Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC$

$Q_2=4.186\times 10^5J$

From this we conclude that, $Q_1$ that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0

3 years ago

Could a mixture be made up of only elements and no compounds. explain

lbvjy [14]

Yes, because a mixture is 2 or more substances that are mixed together (not chemically). A mixture could be two different elements physically combined in a set ratio.

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3 years ago