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2 years ago

What is the answers? Is my answer right?

Chemistry

1 answer:

V125BC [204]2 years ago

5 0

Answer:

I say the correct answers are primary and secondary and teriary.

Explanation:

I say you are right!!

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What are two ways that scientists can report results of their experiments

tiny-mole [99]

Answer: Scientists often communicate their research results in three general ways. One is to publish their results in peer-reviewed journals that can be ready by other scientists. Two is to present their results at national and international conferences where other scientists can listen to presentations.

Explanation:

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3 years ago

How does thermal energy

BartSMP [9]

Objects would be like a lap, stove, & microwave. There’s many options.

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2 years ago

The addition of 250.0 J to 30.0 g of copper initially at 22.0°C will change its temperature to what final value? (Specific heat

WINSTONCH [101]

Answer:

Final temperature = $43.53^{\circ} C$

Explanation:

Given that,

Heat added, Q = 250 J

Mass, m = 30 g

Initial temperature, T₁ = 22°C

The Specific heat of Cu= 0.387 J/g °C

We know that, heat added due to the change in temperature is given by :

$Q=mc\Delta T\\\\Q=mc(T_2-T_1)\\\\\dfrac{Q}{mc}=(T_2-T_1)\\\\T_2=\dfrac{Q}{mc}+T_1$

Put all the values,

$T_2=\dfrac{250}{30\times 0.387}+22\\\\=43.53^{\circ} C$

So, the final temperature is equal to $43.53^{\circ} C$ .

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3 years ago

Changes in ________ move matter from one state to another.

kipiarov [429]

Temperature. Water is an example. When water is at room temp. its liquid. When water is at boiling temp. It is a gas. And when water is at freezing temp. Its a solid.

7 0

3 years ago

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The normal freezing point of water is 0.00 ⁰C. What is the freezing point of a solution containing450.0 mg of ethylene glycol (M

anyanavicka [17]

Answer:

Freezing T° of solution = - 8.98°C

Explanation:

We apply Freezing point depression to solve this problem, the colligative property that has this formula:

Freezing T° of pure solvent - Freezing T° of solution = Kf . m

Kf = 1.86°C/m, this is a constant which is unique for each solvent. In this case, we are using water

m = molality (moles of solute / 1kg of solvent)

We convert the mass of solvent from g to kg

1.5 g . 1kg/1000g = 0.0015 kg

We convert the mass of solute, to moles. Firstly we make this conversion, from mg to g → 450mg . 1g/1000mg = 0.450 g

0.450 g. 1mol / 62.07g = 7.25×10⁻³ moles

Molality → 7.25×10⁻³ mol / 0.0015 kg = 4.83 m

- Freezing T° of solution = 1.86°C /m . 4.83 m - Freezing T° of pure solvent

-Freezing T° of solution = 1.86°C /m . 4.83 m - 0°C

Freezing T° of solution = - 8.98°C

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2 years ago

Read 2 more answers