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2 years ago

Atoms like sodium and lithium often lose electrons, which makes them

Chemistry

2 answers:

zvonat [6]2 years ago

7 0

More reactive than others

Savatey [412]2 years ago

4 0

Answer:

Good reducing agents

Explanation:

Atoms that have a trend to lose electrons are less electronegative than others and also this makes them good reducing agents. This means that they can reduce other species by oxidating themselves

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Will Maynez burns a 0.6-g peanut beneath 50 g of water, which increases in temperature from 22°C to 50°C. (The specific heat cap

Neko [114]

Answer:

40% of the energy release by the peanut is 3500 calories

Explanation:

One calorie is defined as the amount of energy required to increase the temperature of one gram of water for one degree Celsius (or one Kelvin)

Equation for energy gain by water is

Q = mcΔT

where, m is the mass of the object

c is the specific heat capacity

ΔT is the change in temperature

c = 1.0 cal/g?°C.

m = 50 g

ΔT = 50°C - 22°C

= 28°C

Q = (50)× (1)× (28)

= 1400calories

The peanut contain 1400calories of energy .

amount that 40% of energy is released to water ,

so,

Q = 1400 calories / 0.4

= 3500 calories

Therefore, 40% of the energy release by the peanut is 3500 calories

7 0

2 years ago

Part c how long does it take to raise the temperature of the air in a good-sized living room (3.00m×5.00m×8.00m) by 10.0∘c? note

postnew [5]

Do you have any multiple choices

4 0

3 years ago

The solubility table helps predict the products of what kind of reactions?

Ganezh [65]

Answer:

B. double replacement

Explanation:

Rate, or give brainliest if you can :)

3 0

2 years ago

The powder mixture (Cu, Al and Fe) = 10g was oxidized from sufficient chloride acid. 1) What are the possible reactions to the p

Mandarinka [93]

Answer:

1) 2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g

Explanation:

1) Possible reactions

2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Mass of each metal

a) Mass of Cu

The waste was the unreacted copper.

Mass of Cu = 2.5 g

b) Masses of Al and Fe

We have two relations :

Mass of Al + mass of Fe = 10 g - 2.5 g = 7.5 g

H₂ from Al + H₂ from Fe = 6.38 L at NTP

i) Calculate the moles of H₂

NTP is 20 °C and 1 atm.

$\begin{array}{rcl}pV & = & n RT\\\text{1 atm} \times \text{6.38 L} & = & n \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{293.15 K}\\6.38 & = & 24.06n \text{ mol}^{-1} \\n & = & \dfrac{6.38}{24.06 \text{ mol}^{-1} }\\\\ & = & \text{0.2652 mol}\\\end{array}$

(ii) Solve the relationship

Let x = mass of Al. Then

7.5 - x = mass of Fe

Moles of Al = x/27

Moles of Fe = (7.5 - x)/56

Moles of H₂ from Al = (3/2) × Moles of Al = (3/2) × (x/27) = x /18

Moles of H₂ from Fe = (1/1) × Moles of Fe = (7.5 - x)/56

∴ x/18 + (7.5 - x)/56 = 0.2652

56x + 18(7.5 - x) = 267.3

56x + 135 - 18x = 267.3

38x = 132.3

x = 3.5 g

Mass of Al = 3.5 g

Mass of Fe = 7.5 g - 3.5 g = 4.0 g

The masses of the metals are Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g

4 0

3 years ago

A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by O 3 ( g ) + NO ( g )

Aleksandr [31]

Answer:

(a) 7.11x10⁻⁴ M/s

(b) 2.56 mol.L⁻¹.h⁻¹

Explanation:

(a) The reaction is:

O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)

The reaction rate of equation (1) is given by:

$rate = k*[O_{3}][NO]$ (2)

k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹

[O₃]₀ = 2.35x10⁻⁶ M

[NO]₀ = 7.74x10⁻⁵ M

Hence, to find the inital reacion rate we will use equation (2):

$rate = k*[O_{3}]_{0}[NO]_{0} = 3.91 \cdot 10^{6} M^{-1}s^{-1}*2.35\cdot 10^{-6} M*7.74 \cdot 10^{-5} M = 7.11 \cdot 10^{-4} M/s$

Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s

(b) The number of moles of NO₂(g) produced per hour per liter of air is:

t = 1 h

V = 1 L

$\frac{\Delta[NO_{2}]}{\Delta t} = rate$

$\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}$

Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹

I hope it helps you!

5 0

2 years ago