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2 years ago

How is a planet different from a moon? (2 points)

Chemistry

1 answer:

earnstyle [38]2 years ago

7 0

Answer:

Explanation:

I just know.

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Which of the following pairs of elements is most likely to form an ionic compound?

Scrat [10]

Explanation:

magnesium and sodium because they are minerals and part of iron

5 0

3 years ago

A balloon contains 269.7 L of helium at 6.12ºC and 1.00 atm. What is the temperature (in ºC) of the gas if the volume has increa

MatroZZZ [7]

Answer:

T2= 7.3°C

Explanation:

To solve this problem we will use Charles law equation i.e,

V1/T1 = V2/T2

Given data

V1 = 269.7 L

T1 = 6.12 °C

V2= 320.4 L

T2=?

Solution:

Now we will put the values in equation

269.7 L / 6.12°C = 320.4 L / T2

T2= 320.4 L × 6.12°C/ 269.7 L

T2= 1960.85 °C. L /269.7 L

T2= 7.3°C

3 0

2 years ago

What is the density of an object that has a mass of 6.5 g and when placed in water displaces the volume from 4.5mL to 11.8mL? ro

KengaRu [80]

Answer:

$\rho =0.9g/mL$

Explanation:

Hello,

In this case, due to the volume displacement caused the by the object's submersion, it's volume is:

$V=11.8mL-4.5mL=7.3 mL$

In such a way, considering the mathematical definition of density, it turns out:

$\rho =\frac{m}{V}=\frac{6.5g}{7.3mL}\\ \\\rho =0.89g/mL$

Rounding to the nearest tenth we finally obtain:

$\rho =0.9g/mL$

Regards.

3 0

3 years ago

2567 milliliters to liters

andrew-mc [135]

Answer:

2.567 litres

Explanation:

I dont have a proper explanation sorry

6 0

2 years ago

A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f

a_sh-v [17]

Answer:

$m_{H_2O}=39.0g$

Explanation:

Hello,

In this case, is possible to infer that the thermal equilibrium is governed by the following relationship:

$\Delta H_{iron}=-\Delta H_{H_2O}\\m_{iron}Cp_{iron}(T_{eq}-T_{iron})=-m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})$

Thus, both iron's and water's heat capacities are: 0.444 and 4.18 J/g°C respectively, so one solves for the mass of water as shown below:

$m_{H_2O}=\frac{m_{iron}Cp_{iron}(T_{eq}-T_{iron})}{-Cp_{H_2O}(T_{eq}-T_{H_2O}} \\\\m_{H_2O}=\frac{32.5g*0.444\frac{J}{g^0C}*(59.7-22.4)^0C}{-4.18\frac{J}{g^0C}*(59.7-63.0)^0C} \\\\m_{H_2O}=39.0g$

Best regards.

8 0

2 years ago