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3 years ago

What part of speech is heat

Chemistry

2 answers:

IRINA_888 [86]3 years ago

7 0

The part of speech that heat is categorized in is a noun

son4ous [18]3 years ago

3 0

Noun!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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Look at the picture

Trava [24]

C is the answer did y get it right

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3 years ago

Which one of the following does not represent 1.000 mol of the indicated

Katen [24]

Answer:

B. 26.00 g Fe

Atomic mass of Fe (Iron) is 55.845g

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3 years ago

Read 2 more answers

a sample of gold required 2.1200j of heat to melt it from room temperature, 22.0 degrees celsius to its melting point, 1064.4 de

ElenaW [278]

Mass of Gold = 267.165 × 0.01552494829

⇒ 4.1477228099

The amount of heat(q) required to raise m grams of a substance-specific C from T1 to T2 is given by

q=m C (T2-T1) ........1

Given : q= 2.1200 J

the initial temperature of gold, T1 = 22.0Celcius

the final temperature of gold, T2 = 1064.4Celcius

specific heat of gold = 0.131

putting values in eq 1:

⇒ 2.1200 = m × 0.131 × (1064.4-22)

⇒ 2.1200 = m × 0.131 × 1042.4

⇒ 2.1200 / 136.5544

⇒ 0.01552494829

Since 1g= 0.01552494829 Pounds

Mass of Gold = 267.165 × 0.01552494829

⇒ 4.1477228099

Learn more about temperature here: brainly.com/question/11464844

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7 0

1 year ago

A sample in the laboratory is found to contain 3.36 grams of hydrogen, 20.00 grams of carbon, and 26.64 grams of oxygen. The mol

KonstantinChe [14]

Answer:

Empirical formula is CH₂O.

Molecular formula = C₆H₁₂O₆

Explanation:

Given data:

Mass of hydrogen = 3.36 g

Mass of carbon = 20.00 g

Mass of oxygen = 26.64 g

Molar mass of compound = 180.156 g/mol

Empirical formula = ?

Molecular formula = ?

Solution:

Empirical formula:

It is the simplest formula gives the ratio of atoms of different elements in small whole number

Number of gram atoms of H = 3.36 / 1.01 = 3.3

Number of gram atoms of O = 26.64 / 16 = 1.7

Number of gram atoms of C = 20 / 12 = 1.7

Atomic ratio:

C : H : O

1.7/1.7 : 3.3/1.7 : 1.7/1.7

1 : 2 : 1

C : H : O = 1 : 2 : 1

Empirical formula is CH₂O.

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

Empirical formula mass = CH₂O = 12×1 + 2× + 16

Empirical formula mass = 30

n = 180.156 / 30

n = 6

Molecular formula = n (empirical formula)

Molecular formula = 6 (CH₂O)

Molecular formula = C₆H₁₂O₆

5 0

3 years ago

URGENT CHEMISTRY EXPERT!

vovangra [49]

Answer:

Part 1: 7.42 mL; Part 2: 3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ 2Cu₃(PO₄)₂(s)

Explanation:

Part 1. Volume of reactant

(a) Balanced chemical equation.

$\rm 2Na_{3}PO_{4} + 3CuCl_{2} \longrightarrow Cu_{3}(PO_{4})_{2} + 6NaCl$

(b) Moles of CuCl₂

$\text{Moles of CuCl}_{2} =\text{ 16.7 mL CuCl}_{2} \times \dfrac{\text{0.200 mmol CCl}_{2}}{\text{1 mL CuCl}_{2}} = \text{3.340 mmol CuCl}_{2}$

(c) Moles of Na₃PO₄

The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂

$\text{Moles of Na$_{3}$PO}_{4} = \text{3.340 mmol CuCl}_{2} \times \dfrac{\text{2 mmol Na$_{3}$PO}_{4}}{\text{3 mmol CuCl}_{2}} =\text{2.227 mmol Na$_{3}$PO}_{4}$

(d) Volume of Na₃PO₄

$V = \text{2.227 mmol Na$_{3}$PO}_{4}\times \dfrac{\text{1 mL Na$_{3}$PO}_{4}}{\text{0.300 mmol Na$_{3}$PO}_{4}} = \text{7.42 mL Na$_{3}$PO}_{4} \\\\\text{The reaction requires $\large \boxed{\textbf{7.42 mL Na$_{3}$PO}_{4}}$}$

Part 2. Net ionic equation

(a) Molecular equation

$\rm 2Na_{3}PO_{4}(\text{aq}) + 3CuCl_{2}(\text{aq}) \longrightarrow Cu_{3}(PO_{4})_{2}(\text{s}) + 6NaCl(\text{aq})$

(b) Ionic equation

You write molecular formulas for the solids, and you write the soluble ionic substances as ions.

According to the solubility rules, metal phosphates are insoluble.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)

(c) Net ionic equation

To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)

The net ionic equation is

3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)

7 0

3 years ago