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umka2103 [35]
3 years ago
5

What is true about past volcanic eruptions?

Chemistry
1 answer:
Nimfa-mama [501]3 years ago
4 0

Answer:

It is either

Past volcanic eruptions leave marks on rocks near the volcano.

Past volcanic eruptions leave marks in the atmosphere.

My guess is C.

Explanation:

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Iron oxide reacts with aluminum to give aluminum oxide and iron. What kind of chemical reaction is this?
Dima020 [189]
FeO + Al -> AlO + Fe 

As you can see, this type of reaction would lead to be a Single Placement (or Replacement) reaction!

Hope this helps! Best wishes!
3 0
3 years ago
Three people made the
mojhsa [17]

Answer:

the answer is b hope it helped

Explanation:

8 0
3 years ago
What does unsaturated zone mean?
Nadya [2.5K]
The unsaturated zone is the portion of the subsurface above the groundwater table. The soil and rock in this zone contains air as well as water in its pores. ... Unlike the aquifers of the saturated zone below, the unsaturated zone is not a source of readily available water for human consumption
6 0
3 years ago
Read 2 more answers
I want to know the steps.
Artyom0805 [142]

The answer for the following problem is described below.

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

Explanation:

Given:

enthalpy of combustion of glucose(ΔH_{f} of C_{6}H_{12} O_{6}) =-1275.0

enthalpy of combustion of oxygen(ΔH_{f} of O_{2}) = zero

enthalpy of combustion of carbon dioxide(ΔH_{f} of CO_{2}) = -393.5

enthalpy of combustion of water(ΔH_{f} of H_{2} O) = -285.8

To solve :

standard enthalpy of combustion

We know;

ΔH_{f}  = ∈ΔH_{f} (products) - ∈ΔH_{f} (reactants)

C_{6}H_{12} O_{6} (s) +6 O_{2}(g) → 6 CO_{2} (g)+ 6 H_{2} O(l)

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [0 - 1275]

ΔH_{f} = 6 (-393.5) + 6(-285.8)  - 0 + 1275

ΔH_{f} = -2361 - 1714 - 0 + 1275

ΔH_{f} =-2800 kJ

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

7 0
3 years ago
How much heat, in joules, must be added to a 75.0-g iron block with a specific heat of 0.449 J/g °C to increase its temperature
svp [43]

Answer:

50,849.25 Joules

Explanation:

The amount of heat, Q, required to raise the temperature of a body with mass, m, and specific heat capacity, c is given by:

Q = mcΔT, where ΔT represents the change in temperature.

In the case of the iron block:

m = 75 g

c = 0.449 J/g °C

ΔT = 1535 - 25 = 1510 °C

Therefore,

Q = 75 g x 0.449 J/g °C x 1510 °C

     = 50,849.25 Joules

<em>Hence, </em><em>50,849.25 Joules </em><em> of heat must be added to  a 75.0-g iron block with a specific heat of 0.449 J/g °C to increase its temperature from 25 °C to its melting temperature of 1535 °C</em>

5 0
3 years ago
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