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3 years ago

Which two terms? i will mark brainlist

Chemistry

2 answers:

OLga [1]3 years ago

6 0

Answer:

I believe it's qualitative and continuous.

Explanation:

Liono4ka [1.6K]3 years ago

3 0

Answer:

Continuous, and Quantitative

Hope this helps :)

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The heat of fusion for ice is 334 joules per gram. Adding 334 joules of heat to one gram of ice at STP will

yulyashka [42]

C) change to water at the same temperature

Explanation:

Adding 334Joules of heat to one gram of ice at STP will cause ice to change to water at the same temperature.

The heat of fusion is the amount of energy needed to melt a given mass of a solid
It is also conversely the amount of energy removed from a substance to freeze it.
The addition of this energy does not cause a decrease or increase in temperature.
Only a phase change occurs.

Learn more:

Heat of fusion brainly.com/question/4050938

#learnwithBrainly

3 0

3 years ago

Given 1 cm3 = 1 mL<br> A box has dimensions 2.0 cm x 4.0 cm x 8.0 cm.

svp [43]

Answer: 64

Explanation:

you just multiple the 3 numbers to get the answer (i’m in chemistry and just did this question lol)

5 0

3 years ago

What group is the element barium(Ba) In?<br>A. 1<br>B. 6<br>C. 56<br>D. 4

LenaWriter [7]

Ummmmmmmmmmm 56 yes 56

7 0

2 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

For a lake in Michigan, researchers have determined that largemouth bass feed on smaller fish, which in turn feed on zooplankton

allsm [11]

Answer:

the expected decline in largemouth bass is 3,000 kg.

Option d) 3,000 kg is the correct answer.

Explanation:

Given the data in the question;

In 2016, there was 600,000 kg of zooplankton in the lake.

In 2017, an accidental runoff of insecticide near the lake caused a 50 percent decline of the zooplankton population in the lake.

Now,

The remaining mass of zooplankton after the 50% decline will be;

⇒ 600,000 kg zooplankton × 50%

⇒ 600,000 × 50/100

⇒ 300,000 kg of zooplankton

Now, with 10 percent trophic efficiency; smaller fish directly feed on zooplankton; decline in smaller fish mass will be;

⇒ 300,000 kg × 10%

⇒ 300,000 × 10/100

⇒ 30000 kg

Finally, with 10 percent trophic efficiency, largemouth bash directly feed on smaller fish, so the expected decline in mass of largemouth bash will be;

⇒ 30000 kg × 10%

⇒ 30000 kg × 10/100

⇒ 3,000 kg

Therefore, the expected decline in largemouth bass is 3,000 kg.

Option d) 3,000 kg is the correct answer.

7 0

3 years ago