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raketka [301]
3 years ago
7

How much heat is required to increase the temperature of 10.0 grams of water 6.0oC? (The specific heat of water is 4.18 J/g x oC

) 250 J
Chemistry
1 answer:
scoray [572]3 years ago
8 0

Heat required in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:<span>

Heat = mC (T2-T1)
Heat = 10.0 g (4.18 J/g-C ) ( 6.0 C )
<span>Heat = 250.8 J</span></span>

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A gravel can be broken down into different kinds of substances by physical processes. No chemical reactions are needed to separa
Vlada [557]

Answer:

MIXTURE

Explanation:

A mixture is a substance composed of a combination of other different substances. These component(s) of a mixture are physically combined, meaning that there is no chemical linkage between the individual components/constituents of a mixture.

This is the case of the gravel described in this question. The components of gravel can be separated using physical means because they are not chemically bonded to one another, hence, no chemical reactions are needed to separate different parts of gravel into pure substances. This makes gravel a MIXTURE.

6 0
3 years ago
Read 2 more answers
How many liters of 4.0 M NaOH solution will react with 0.60 liters 3.0 M H2SO4?
slava [35]

Answer:

A. 0.90 L.

Explanation:

  • NaOH solution will react with H₂SO₄ according to the balanced reaction:

<em>H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.</em>

<em>1.0 mole of H₂SO₄ reacts with 2.0 moles of NaOH.</em>

  • For NaOH to react completely with H₂SO₄, the no. of millimoles should be equal.

<em>∴ (MV) NaOH = (xMV) H₂SO₄.</em>

x for H₂SO₄ = 2, due to having to reproducible H⁺ ions.

<em>∴ V of NaOH = (xMV) H₂SO₄/ M of NaOH</em> = 2(0.6 L)(3.0 M)/(4.0 M) = <em>0.90 L.</em>

4 0
3 years ago
6. How will you obtain ? (a) Magnesium oxide from magnesium. (b) Silver chloride from silver nitrate. (c) Nitrogen dioxide from
Reil [10]

Answer:

a) reaction with oxygen

2mg +o2---------2mgo

b) Agno3+NaCl ----------AgCl+NaNo3

8 0
2 years ago
The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W
Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹  

Explanation:

Assume the rate law is  

rate = k[A][B]²

If you are comparing two rates,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}

You are cutting each concentration in half, so

\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}

Then,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}

8 0
2 years ago
Explain specifically how an electron gives off light in an atom.
exis [7]

Answer:

Then, at some point, these higher energy electrons give up their "extra" energy in the form of a photon of light, and fall back down to their original energy level.

Explanation:

When properly stimulated, electrons in these materials move from a lower level of energy up to a higher level of energy and occupy a different orbital.

3 0
2 years ago
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