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Shkiper50 [21]
2 years ago
14

1. All of the following energy resources are fossil fuels except

Chemistry
1 answer:
solong [7]2 years ago
4 0
C. Biomass is the correct answer
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Pascal difference has a value of 1000 millipascals difference

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Why is it logical to assume that they hydrogen ion concentration in an aqueous solution of a strong monoproctic acid equals the
natka813 [3]
It is logical to assume that they hydrogen ion concentration in an aqueous solution of a strong monoproctic acid equals the molarity of the acid because ions which are charged,for example ammonium ion (NH4+), which can be derived by the addition of a proton to a molecular base.
3 0
3 years ago
The zinc in a copper-plated penny will dissolve in hydrochloric acid if the copper coating is filed down in several spots (so th
LekaFEV [45]

Answer:

A penny dissolves in hydrochloric acid if the copper coating is filed down in several spot... ... When The Zinc In A Certain Penny Dissolves, The Total Volume Of Gas ... is filed down in several spots (so that the hydrochloric acid can reach the zinc). The reaction between the acid and the zinc 2H+(aq)+Zn(s)→H2(g)+Zn2+(aq) .

6 0
3 years ago
A sample of calcium phosphate was found to have a mass of 125.3 g. How many molecules were contained in the sample?
Viktor [21]

The answer for the following problem is mentioned below.

  • <u><em>Therefore number of molecules(N) present in the calcium phosphate sample are  19.3 × 10^23 molecules.</em></u>

Explanation:

Given:

mass of calcium phosphate (Ca_{3}(PO_{4} )_{2} ) = 125.3 grams

We know;

molar mass of calcium phosphate  (Ca_{3}(PO_{4} )_{2} ) = (40×3) + 3 (31 +(4×16))

molar mass of calcium phosphate  (Ca_{3}(PO_{4} )_{2} ) = 120 + 3(95)

molar mass of calcium phosphate  (Ca_{3}(PO_{4} )_{2} )  = 120 +285 = 405 grams

<em>We also know;</em>

No of molecules at STP conditions(N_{A}) = 6.023 × 10^23 molecules

To solve:

no of molecules present in the sample(N)

We know;

\frac{m}{M} =\frac{N} }{}N÷N_{A}

\frac{405}{125.3} =\frac{N}{6.023*10^23}

N =(405×6.023 × 10^23) ÷ 125.3

N = 19.3 × 10^23 molecules

<u><em>Therefore number of molecules(N) present in the calcium phosphate sample are  19.3 × 10^23 molecules</em></u>

3 0
3 years ago
29.5 g of mercury is heated from 32°C to 161°C, and absorbs 499.2 joules of heat in the process. Calculate the specific heat cap
Finger [1]

Answer:

c = 0.13 j/ g.°C

Explanation:

Given data:

Mass of mercury = 29.5 g

Initial temperature = 32°C

Final temperature = 161°C

Heat absorbed = 499.2 j

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Q = m.c. ΔT

ΔT  = T2 - T1

ΔT  = 161°C - 32°C

ΔT  = 129 °C

Q = m.c. ΔT

c = Q / m. ΔT

c = 499.2 j / 29.5 g. 129 °C

c =  499.2 j / 3805.5 g. °C

c = 0.13 j/ g.°C

5 0
3 years ago
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