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saveliy_v [14]
3 years ago
6

Which of the following are examples of conservable quantities? a. potential energy and length c. mechanical energy and mass b. m

echanical energy and length d. kinetic energy and mass
Chemistry
1 answer:
Blababa [14]3 years ago
7 0

Answer:

b

Explanation:

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If a reaction ends with 5.5 grams of products, how many grams of reactants were required to start the reaction?
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5.5 grams of reactants. According to the Law of Conservation of Mass, mass isn’t created or lost through any chemical changes, so the total mass should remain constant from the initial reactants to the final products.
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3 years ago
A car traveling with constant speed travels 150km in 7200 s. What is the speed of the car
neonofarm [45]
75km/hr

A car traveling with constant speed travels 150km in 7200 s. What is the speed of the car
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3 years ago
A 34.0 g piece of metal is heated to 92.0°C then placed in a beaker of water containing 22.0 g of water at 19.0°C. The temperatu
lapo4ka [179]

Answer:

0.1988 J/g°C

Explanation:

-Qmetal = Qwater

Q = mc∆T

Where;

Q = amount of heat

m = mass of substance

c = specific heat of substance

∆T = change in temperature

Hence;

-{mc∆T} of metal = {mc∆T} of water

From the information provided in this question, For water; m= 22.0g, ∆T = (24°C-19°C), c = 4.18J/g°C.

For metal; m= 34.0g, ∆T = (24°C-92°C), c = ?

Note that, the final temperature of water and the metal = 24°C

-{34 × c × (24°C-92°C)} = 22 × 4.18 × (24°C-19°C)

-{34 × c × (-68°C)} = 459.8

-{34 × c × -68} = 459.8

-{-2312c} = 459.8

+2312c = 459.8

c = 459.8/2312

c = 0.1988

The specific heat capacity of the metal is 0.1988 J/g°C

6 0
2 years ago
Combustion of 9.511 grams of c4h10 will yield ____ grams of CO2
Flauer [41]

Answer:

\boxed{28.81}

Explanation:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:      58.12                   44.01

           2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

m/g:     9.511

1. Moles of C₄H₁₀

\text{Moles of C$_{4}$H$_{10} $} = \text{ 9.511 g C$_{4}$H$_{10} $} \times \dfrac{\text{1 mol C$_{4}$H$_{10} $}}{\text{ 58.12 g C$_{4}$H$_{10} $}} = \text{0.1636 mol C$_{4}$H$_{10}$}

2. Moles of CO₂

The molar ratio is 8 mol CO₂:2 mol C₄H₁₀

\text{Moles of CO}_{2} =\text{0.1636 mol C$_{4}$H$_{10} $} \times \dfrac{\text{8 mol CO}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \text{0.6546 mol CO}_{2}

3. Mass of CO₂

\text{Mass of CO}_{2} = \text{0.6546 mol CO}_{2} \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO}_{2}} = \textbf{28.81 g CO}_{2}\\\\\text{The combustion will form $\boxed{\textbf{28.81 g CO}_{2}}$}

8 0
3 years ago
Unequal heating of Earth's surface over a SMALL area causes ..??? (please answer fast)
ra1l [238]
Global win I think brace it say around small areas
5 0
2 years ago
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