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3 years ago

Which changes would cause an increase in the resistance of a wire? Check all that apply.

Chemistry

2 answers:

chubhunter [2.5K]3 years ago

9 0

Answer:

Increasing its temperature.

Longer

and thinner

Explanation:

e2021

OLEGan [10]3 years ago

6 0

Answer:

increasing its temperature

using a longer wire

using a thinner wire

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How many mols are in 100.g of Fe?

PolarNik [594]

Answer:

1.79 moles

Explanation:

1.79 moles

3 0

3 years ago

Read 2 more answers

The combustion of palmitic acid is represented by the chemical equation: C16H32O2(s) + 23O2(g) → 16 CO2(g) + 16 H2O(l) The magni

solniwko [45]

Answer:

The correct option is C

Explanation:

From the question we are told that

The reaction is

$C_{16}H_{32}O_2(g) + 23O_2(g) \to 16 CO_2(g) + 16 H_2O(l)$

Generally $\Delta H = \Delta U + \Delta N* RT$

Here $\Delta H$ is the change in enthalpy

$\Delta U$ is the change in the internal energy

$\Delta N$ is the difference between that number of moles of product and the number of moles of reactant

Looking at the reaction we can discover that the elements that was consumed and the element that was formed is $O_2$ and $CO_2$ and this are both gases so the change would occur in the number of moles

$\Delta H = \Delta U + [16 -23]* RT$

$\Delta H = \Delta U -7RT$

The negative sign in the equation tell us that the enthalpy $\Delta_r H$ would be less than the Internal energy $\Delta_r U$

4 0

3 years ago

A colorless liquid has a molar mass of 60.01 g/mol. When the liquid was analyzed, it was 46.7% nitrogen and 53.3% oxygen. What i

MAXImum [283]

first we have to find the empirical formula of the compound

empirical formula is the simplest ratio of whole numbers of components making up a compound

for 100 g of the compound

N O

mass 46.7 g 53.3 g

number of 46.7 g/ 14 g/mol 53.3 g/ 16 g/mol

moles = 3.34 mol = 3.33 mol

divide by the least number of moles

3.34/3.33 = 1.00 3.33/ 3,33 = 1.00

therefore number of atoms are

N - 1

O - 1

empirical formula is - NO

mass of empirical unit - 14 g/mol + 16 g/mol = 30 g

molecular formula is actual composition of elements in the compound

molecular mass - 60.01 g/mol

number of empirical units = molecular mass / empirical unit mass

= 60.01 g/mol / 30 g = 2

there are 2 empirical units

2(NO)

molecular formula = N₂O₂

6 0

3 years ago

Calculate the ph of a 0.021 m nacn solution. [ka(hcn) = 4.9  10–10]

ohaa [14]

Answer is: pH of solution of sodium cyanide is 11.3.
Chemical reaction 1: NaCN(aq) → CN⁻(aq) + Na⁺(aq).
Chemical reaction 2: CN⁻ + H₂O(l) ⇄ HCN(aq) + OH⁻(aq).
c(NaCN) = c(CN⁻) = 0.021 M.
Ka(HCN) = 4.9·10⁻¹⁰.
Kb(CN⁻) = 10⁻¹⁴ ÷ 4.9·10⁻¹⁰ = 2.04·10⁻⁵.
Kb = [HCN] · [OH⁻] / [CN⁻].
[HCN] · [OH⁻] = x.
[CN⁻] = 0.021 M - x..
2.04·10⁻⁵ = x² / (0.021 M - x).
Solve quadratic equation: x = [OH⁻] = 0.00198 M.
pOH = -log(0.00198 M) = 2.70.
pH = 14 - 2.70 = 11.3.

4 0

3 years ago

Britney added 0.05 moles of copper(II) nitrate solution to 0.1 moles of sodium hydroxide solution and

Rama09 [41]

The percent yield of copper hydroxide is 84%

<h3>Stoichiometry</h3>

From the question, we are to determine the percent yield of copper hydroxide

First, we will determine the theoretical mass

From the given balanced chemical equation, we have

Cu(NO₃)₂ + 2NaOH -- Cu(OH)₂ + 2NaNO₃

This means,

1 mole of copper(II) nitrate reacts with 2 moles of sodium hydroxide to produce 1 mole of copper hydroxide

Therefore,
0.05 mole of copper(II) nitrate reacts with 0.1 mole of sodium hydroxide to produce 0.05 mole of copper hydroxide

The theoretical number of moles of copper hydroxide that is produced is 0.05 mole

Now, for the theoretical mass

Using the formula,

Mass = Number of moles × Molar mass

Molar mass of copper hydroxide = 97.56 g/mol

Then,

Theoretical mass = 0.05 × 97.56

Theoretical mass of copper of hydroxide produced is = 4.878 g

Now, for the percent yield of copper hydroxide

Percent yield is given by the formula,

$Percent\ yield = \frac{Actual\ yield}{Theoretical\ yield} \times 100\%$

Then,

$Percent\ yield\ of\ copper\ hydroxide= \frac{4.1}{4.878}\times 100\%$

$Percent\ yield\ of\ copper\ hydroxide= 84\%$

Hence, the percent yield of copper hydroxide is 84%.

Learn more on Stoichiometry here: brainly.com/question/9372758

7 0

3 years ago