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nataly862011 [7]
2 years ago
14

Which of the following steps will slow down the reaction between coal and oxygen?

Chemistry
2 answers:
Kay [80]2 years ago
7 0

Answer: I believe it it decrease temp of oxygen

Explanation:

irakobra [83]2 years ago
5 0

Answer:

decrease temperature of the oxygen

Explanation:

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Predict whether each of the following substances would be more soluble in water (polar solvent) or in a hydrocarbon such as hept
vlada-n [284]

Answer:

Vegetable oil will dissolve in heptane

Isopropyl alcohol will dissolve in water

Potassium Bromide will dissolve in water

Explanation:

In chemistry, like dissolves like. This implies that polar substances will dissolve in polar solvents and nonpolar substances will dissolve in nonpolar solvents. This is so because, dissolution of a solute in a solvent involves adequate intermolecular interaction between solute and solvent which isn't possible between a polar and a nonpolar substance.

5 0
3 years ago
You are trying to determine the specific heat of a metal. You heat the 97 g piece of metal to 100 °C and place it in a calorimet
Brut [27]

Answer : The specific heat of the metal is, 0.658J/g^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of metal = ?

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of metal = 97 g

m_2 = mass of water  = 122 g

T_f = final temperature of mixture = 28.9^oC

T_1 = initial temperature of metal = 100^oC

T_2 = initial temperature of water = 20.0^oC

Now put all the given values in the above formula, we get

97g\times c_1\times (28.9-100)^oC=-122g\times 4.18J/g^oC\times (28.9-20.0)^oC

c_1=0.658J/g^oC

Therefore, the specific heat of the metal is, 0.658J/g^oC

8 0
3 years ago
90 POINTSSSS!!!
Katarina [22]

Answer:

\large \boxed{\text{-2043.96 kJ/mol}}

Explanation:

Assume the reaction is the combustion of propane.

Word equation: propane plus oxygen produces carbon dioxide and water

Chemical eqn:    C₃H₈(g) +   O₂(g) ⟶   CO₂(g) +   H₂O(g)

Balanced eqn:    C₃H₈(g) + 5O₂(g) ⟶ 3CO₂(g) + 4H₂O(g)

(a) Table of enthalpies of formation of reactants and products

\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{C$_{3}$H$_{8}$(g)} & -103.85 \\\text{O}_{2}\text{(g)} & 0 \\\text{CO}_{2}\text{(g)} & -393.51 \\\text{H$_{2}$O(g)} & -241.82\\\end{array}

(b)Total enthalpies of reactants and products

\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)\\= \text{-2147.81 kJ/mol - (-103.85 kJ/mol)}\\=  \text{-2147.81 kJ/mol + 103.85 kJ/mol}\\= \textbf{-2043.96 kJ/mol}\\\text{The enthalpy change is $\large \boxed{\textbf{-2043.96 kJ/mol}}$}

ΔᵣH° is negative, so the reaction is exothermic.

5 0
2 years ago
An intravenous infusion is to contain 15 mEq of potassium ion and 20 mEq of sodium ion in 500 mL of 5% dextrose injection. Using
Studentka2010 [4]

Answer:

To supply the required ions it is necessary to inject 5,6mL of 6g/30mL solution and 131,1 mL of 0,9% solution.

Explanation:

1mEq of sodium are 59mg of NaCl and 1mEq of potassium are 75mg KCl

in intravenous infusion 15 mEq of K are:

15x75mg KCl = 1,125g of KCl

And 20 mEq of Na are:

20x59mg NaCl = 1,18g of NaCl

To supply the potassium ion it is necessary to inject:

1,125g of KCl×\frac{30mL}{6g} =<em> 5,6mL  of 6g/30mL solution</em>

And, to supply the sodium ion it is necessary to inject:

1,18g of NaCl×\frac{100mL}{0,9g} = <em>131,1 mL of 0,9% solution</em>

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I hope it helps!

6 0
3 years ago
What is the process of using one or more of your five senses to measure or collect data?
sveta [45]

Answer:

B no observation pls mark me branilest

7 0
2 years ago
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