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diamong [38]
3 years ago
11

A ship is sailing at a speed of 5 km/hr. What distance has it covered, 2 hours after it left the port?​

Chemistry
2 answers:
Fittoniya [83]3 years ago
6 0

Answer:

10km

Explanation:

xenn [34]3 years ago
3 0

Answer:

\boxed {\boxed {\sf 10 \ kilometers}}

Explanation:

Distance is the product of speed and time.

d=s \times t

The speed of the ship is 5 kilometers per hour. The time is 2 hours.

  • s= 5 km/hr
  • t= 2 hr

Substitute the values into the formula.

d= 5 \ km/hr * 2 \ hr

Multiply. Note that the hours will cancel each other out.

d= 5 \ km * 2 \\d=10 \ km

The ship covers a distance of <u>10 kilometers.</u>

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I’m not an expert and I haven’t taken chemistry in like a year but i believe you use stoichiometry to do this equation. It should look something like this.

8 0
2 years ago
Which part of the atom is involved in bonding?
bonufazy [111]

Answer:

Valence electrons (the electrons on the outermost shell of the atom) are responsible for bonding

Explanation:

8 0
3 years ago
what is the specific heat of a substance if 300 j are required to raise the temperature of a 267-g sample by 12 degrees c
slava [35]

Answer : The specific heat of the substance is 0.0936 J/g °C

Explanation :

The amount of heat Q can be calculated using following formula.

Q = m \times C \times \bigtriangleup T

Where Q is the amount of heat required = 300 J

m is the mass of the substance = 267 g

ΔT is the change in temperature = 12°C

C is the specific heat of the substance.

We want to solve for C, so the equation for Q is modified as follows.

C = \frac{Q}{m \times \bigtriangleup T}

Let us plug in the values in above equation.

C = \frac{300J}{267g \times 12 C}

C = \frac{300J}{3204 g C}

C = 0.0936 J/g °C

The specific heat of the substance is 0.0936 J/g°C

3 0
3 years ago
What temperature would 3.54 moles of xenon gas need to reach to exert a pressure of 1.57 atm at a volume of 34.6 l
ira [324]

Answer:

186.9Kelvin

Explanation:

The ideal gas law equation is PV = n R T

where

P   is the pressure of the gas

V   is the volume it occupies

n  is the number of moles of gas present in the sample

R  is the universal gas constant, equal to  0.0821 atm L /mol K

T  is the absolute temperature of the gas

Ensure units of the volume, pressure, and temperature of the gas correspond to R ( the universal gas constant, equal to  0.0821 atm L /mol K )

n = 3.54moles

P= 1.57

V= 34.6

T=?

PV = n R T

PV/nR = T

1.57 x 34.6/3.54 x 0.0821

54.322/0.290634= 186.908620464= T

186.9Kelvin ( approximately to 1 decimal place)

5 0
3 years ago
What would the products be for the reaction between Na3PO4 + MgSO4?
ivann1987 [24]

MgSO4 + Na3PO4 = Na2SO4 + Mg3(PO4)2

Answer: The products of Na3PO4 + MgSO4 are Na2SO4 + Mg3(PO4)2

Explanation:

7 0
2 years ago
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