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Doss [256]
3 years ago
8

The specific silver is How many joules of energy are needed to warm 4.37 g of silver from 25.0 degrees * C to 27.5 degrees * C ?

Chemistry
1 answer:
Alexxandr [17]3 years ago
6 0

0.24J/g*degC * 4.37g * 2.5degC = 2.622J

The 2.5 degC is the difference between 25 and 27.5 deg C.

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A scuba tank contains a mixture of oxygen (O2) and nitrogen (N2) gas. The oxygen has a partial pressure of PO2=5.62MPa. The tota
dmitriy555 [2]

Answer:

21.16 MPa

Explanation:

Partial pressure of oxygen = 5.62 MPa

Total gas pressure = 26.78 MPa

But

Total pressure of the gas= sum of partial pressures of all the constituent gases in the system.

This implies that;

Total pressure of the system = partial pressure of nitrogen + partial pressure of oxygen

Hence partial pressure of nitrogen=

Total pressure of the system - partial pressure of oxygen

Therefore;

Partial pressure of nitrogen= 26.78 - 5.62

Partial pressure of nitrogen = 21.16 MPa

7 0
2 years ago
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Which celestial object is found in kuiper belt
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The Kuiper belt is home to three officially recognized dwarf planets: Pluto, Haumea, and Makemake. Some of the Solar System's moons, such as Neptune's Triton and Saturn's Phoebe, are also thought to have originated in the region.

4 0
3 years ago
1. Displacement vectors of 10 m west and 14 m west make a resultant vector that is
Iteru [2.4K]

Answer:

B 24 m west I belive that it is the answer

3 0
3 years ago
What is the mass of 3.25 moles of sodium hydroxide (NaOH)?
Dimas [21]

Answer:

<em>1 mole is equal to 1 moles NaOH, or 39.99711 grams.</em>

Explanation:

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5 0
3 years ago
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A star is estimated to have a mass of 2.0 x 10 ^36kg. Assuming it to be a sphere of average radius of 7.0 x 10 ^5 km. Calculate
Montano1993 [528]

Answer:

<em>a)</em> <em>1.392 x 10^6 g/cm^3</em>

<em>b) 8.69 x 10^7 lb/ft^3</em>

<em></em>

Explanation:

mass of the star m =  2.0 x 10^36 kg

radius of the star (assumed to be spherical) r = 7.0 x 10^5 km = 7.0 x 10^8 m

The density of substance ρ = mass/volume

The volume of the star = volume of a sphere = \frac{4}{3}\pi  r^{3}

==> V = \frac{4}{3}*3.142*(7.0*10^8)^{3} = 1.437 x 10^27 m^3

density of the star ρ = (2.0 x 10^36)/(1.437 x 10^27) = 1.392 x 10^9 kg/m^3

in g/cm^3 = (1.392 x 10^9)/1000 = <em>1.392 x 10^6 g/cm^3</em>

in lb/ft^3 =  (1.392 x 10^9)/16.018 = <em>8.69 x 10^7 lb/ft^3</em>

6 0
2 years ago
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