All categories

3 years ago

How much heat is needed to raise the temperature of 50.0 g of water from 4.5 ⁰C to 83.0 ⁰C?

Chemistry

1 answer:

Alex_Xolod [135]3 years ago

6 0

The heat needed to raise the temperature of 50.0 g of water from 4.5 ⁰C

to 83.0 ⁰C is 16.46KJ

Explanation:

To solve the problem,

we need the value of the water and its given by,

C=4.18 × J/ g ∘ C

To increase the temperature we need m grams of water by n ∘ C

And the formula for the heat is given by, heat= m × n ×specific

heat

q = m ⋅ c . Δ T

where q denotes the heat absorbed

m denotes the mass sample

c denotes the specific heat

Δ T denotes the change in the temperature

Based on the data it is written as,

q= 50.0 g . 4.18 J / g ∘ C (83.0-4.5) ∘ C

Simplifying the equation we get,

q=16406.5 J

Express it in the form of significant figure

16406.5 J . 1 KJ / 10^ 3 J

Solving the equation we get

16.46KJ

The heat needed to raise the temperature of 50.0 g of water from 4.5 ⁰C

to 83.0 ⁰C is 16.46 KJ

You might be interested in

Select the correct answer. Elements Electronegativity aluminum (Al) 1.61 calcium (Ca) 1 carbon (C) 2.55 chlorine (Cl) 3.16 fluor

marishachu [46]

Answer: Fe - O

Explanation:

1) Start by showing the chart information in a more understandable way:

Elements ---------- Electronegativity

aluminum (Al) ----- 1.61

calcium (Ca) ------- 1

carbon (C) ---------- 2.55

chlorine (Cl) -------- 3.16

fluorine (F) --------- 4

hydrogen (H) ------ 2.2

iron (Fe) ------------- 1.83

nitrogen (N) --------- 3.04

oxygen (O) ----------- 3.44

phosphorus (P) ----- 2.19

potassium (K) -------- 0.82

silicon (Si) ------------- 1.9

sulfur (S) --------------- 2.58

2) Find the electronegativity differences for the atoms in each compound:

i) Ca – Cl: 3.16 - 1 = 2.16

ii) Fe – O: 3.44 - 1.83 = 1.61

iii) H – S: 2.58 - 2.2 = 0.38

iii= F – F = 4 - 4 = 0

3) Analysis

i) The higher the electronegativity difference between two non-metal elements, the the more the polar the covalent bond. Therefore, the bond H - S is the most polar among the non-metal - non-metal bonds listed. But 0.38 is a small difference, so this is not very polar.

ii) For metal - non-metal bonds, when the difference in electronegativities is too high (close to or greater than 2.0) the ionic character is dominant, and so you cannot classify the bond as polar but as ionic. Therefore, Ca - Cl with 2.16 electronegativity difference is a ionic bond.

iii) The bond in Fe – O has electronegativty difference of 1.61, so it is yet covalent, and highly polar.

Read more on Brainly.com - brainly.com/question/1309389#readmore

8 0

3 years ago

Read 2 more answers

3.8 liters of sulfur vapor, S8(g), at 921.4°C and 5.87 atm is burned in excess pure oxygen gas to give sulfur dioxide gas measur

lilavasa [31]

Answer:

116.5 g of SO₂ are formed

Explanation:

The reaction is:

S₈(g) + 8O₂(g) → 8SO₂ (g)

Let's identify the moles of sulfur vapor, by the Ideal Gases Law

We convert the 921.4°C to Absolute T° → 921.4°C + 273 = 1194.4 K

5.87 atm . 3.8L = n . 0.082 L.atm/mol.K . 1194.4K

(5.87 atm . 3.8L) / (0.082 L.atm/mol.K . 1194.4K) = n → 0.228 moles of S₈

Ratio is 1:8, 1 mol of sulfur vapor can produce 8 moles of dioxide

Then, 0.228 moles of S₈ must produce (0.228 . 8) /1 = 1.82 moles

We convert the moles to g → 1.82 moles . 64.06 g /1mol = 116.5 g

3 0

3 years ago

Read 2 more answers

Which of the following is not found in the nucleus of an atom?

Ulleksa [173]

A. electron. The nucleus has protons and neutrons, quark is the particle which forms protons and neutrons.

8 0

4 years ago

Government should use informed science to help make policies protect all citizens True or False

Sloan [31]

Answer:

yes

Explanation:

8 0

3 years ago

Complete and balance the chemical equations for the precipitation reactions, if any, between the following pairs of reactants, a

Crank

Explanation:

a. $Pb(NO_3)_2(aq) + Na_2SO_4(aq)$ → ?

$Pb(NO_3)_2(aq) + Na_2SO_4(aq)\rightarrow PbSO_4(s)+2NaNO_3(aq)$

$Pb(NO_3)_2(aq)\rightarrow Pb^{2+}(aq)+2NO_3^{-}(aq)$

$Na_2SO_4(aq)\rightarrow 2Na^++SO_4^{2-}(aq)$

$Pb^{2+}(aq)+2NO_3^{-}(aq)+2Na^++SO_4^{2-}(aq)\rightarrow PbSO_4(s)+2Na^++2NO_3^{-}(aq)$

Removing common ions from both sides, we get the net ionic equation:

$Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)$

b. $NiCl_2(aq) + NH_4NO_3(aq)$ →

$NiCl_2(aq) + NH4NO_3(aq) \rightarrow Ni(NO_3)_2+NH_4Cl(aq)$

No precipitation is occuring.

c. $Fe_Cl2(aq) + Na_2S(aq)$ →

$FeCl_2(aq) + Na_2S(aq)\rightarrow FeS(s)+2NaCl(aq)$

$FeCl_2(aq)\rightarrow Fe^{2+}(aq)+2Cl^{-}(aq)$

$Na_2S(aq)\rightarrow 2Na^++S{2-}(aq)$

$Fe^{2+}(aq)+2Cl^{-}(aq)+2Na^++S^{2-}(aq)\rightarrow FeS(s)+2Na^++2Cl^{-}(aq)$

Removing common ions from both sides, we get the net ionic equation:

$Fe^{2+}(aq)+S^{2-}(aq)\rightarrow FeS(s)$

d. $MgSO_4(aq) + BaCl_2(aq)$ →

$MgSO4(aq) + BaCl2(aq)\rightarrow BaSO_4(s)+MgCl_2$

$MgSO_4(aq)\rightarrow Mg^{2+}(aq)+SO_4^{2-}(aq)$

$BaCl_2(aq)\rightarrow Ba^{2+}+2Cl^{-}(aq)$

$Mg^{2+}(aq)+SO_4^{2-}(aq)+Ba^{2+}+2Cl^{-}(aq)(aq)\rightarrow BaSO_4(s)+Mg^{2+}(aq)+2Cl^{-}(aq)$

Removing common ions from both sides, we get the net ionic equation:

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

5 0

3 years ago