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3 years ago

Which parts of sedimentary rock formation include the breakdown and carrying away of existing rock? Check all that apply.

Chemistry

1 answer:

stiks02 [169]3 years ago

3 0

Answer

Weathering and erosion are two processes of which sedimentary rocks are broken down and carry away existing rocks.

Explanation:

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Does the bowling ball have more potential energy or kinetic energy as it is halfway through its fall? Why?

ahrayia [7]

Answer:

more kinetic

Explanation:

I think the kinetic energy is 75 percent while the potential energy is 25 percent

5 0

3 years ago

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

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3 years ago

Drug testing of athletes and other individuals is big business.

ziro4ka [17]

Answer:

bioanalytical chemistry

Explanation:

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3 years ago

Which describes the path of energy that a gas powered car uses to drive along a road?

harkovskaia [24]

Answer:

mechanical energy to electrical energy to light energy

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3 years ago

A volume of 105 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If

NemiM [27]

Answer:

<h3>25.0 grams is the mass of the steel bar.</h3>

Explanation:

Heat gained by steel bar will be equal to heat lost by the water

$Q_1=-Q_2$

Mass of steel= $m_1$

Specific heat capacity of steel = $c_1=0.452 J/g^oC$

Initial temperature of the steel = $T_1=2.00^oC$

Final temperature of the steel = $T_2=T=21.50^oC$

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2= 105 g$

Specific heat capacity of water= $c_2=4.18 J/g^oC$

Initial temperature of the water = $T_3=22.00^oC$

Final temperature of water = $T_2=T=21.50^oC$

$Q_2=m_2c_2\times (T-T_3)-Q_1=Q_2(m_1c_1\times (T-T_1))=-(m_2c_2\times (T-T_3))$

On substituting all values:

$(m_1\times 0.452 J/g^oC\times (21.50^o-2.00^oC))=-(105 g\times 4.18 J/g^oC\times (21.50^o-22.00^o))\\\\m_1*8.7914=241.395\\\\m_1=\frac{219.45}{8.7914} \\\\m_1=24.9\\\\ \approx25 \texttt {grams}$

<h3>25.0 grams is the mass of the steel bar.</h3>

3 0

3 years ago

Read 2 more answers