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3 years ago

BRAINLIEST IF CORRECT (put them in order)

Chemistry

2 answers:

Nostrana [21]3 years ago

6 0

Answer:

what grade are you in and i think it is D

Explanation:

Butoxors [25]3 years ago

6 0

Answer:

1 a person sings into a microphone

2 the sound is converted into an electrical signal

3 the analog signal is converted into a digital signal made of many samples of the digital sound

4 the samples of the sound is converted into 1s and 0s

5 the digital pattern is converted to a medium such as a hard disk or flash drive

Explanation:

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When 0.50 L of a 12 M solution is diluted to 1.0 L, what is the resulting molarity?

Lyrx [107]

Answer:

The resulting molarity is 6M.

Explanation:

A dilution consists of the decrease of concentration of a substance in a solution (the higher the volume of the solvent, the lower the concentration).

We use the formula for dilutions:

C1 x V1 = C2 x V2

12 M x 0,5L = C2 x 1,0 L

C2= (12 M x 0,5 L)/1,0 L

5 0

2 years ago

Connie was making sodium chloride by adding an acid to an alkali. She followed the progress of the reaction with a pH sensor. At

Misha Larkins [42]

⇒Answer:

When the pH sensor hits pH=7.

⇒Explanation:

Because pH=7 is the indicator that the acid and alkali have been neutralized.

5 0

2 years ago

Can someone help please I will give Brainlyiest

kotykmax [81]

Answer:

See Explanation

Explanation:

moles of NH₃ = 11.9g/17.03 g/mol = 0.699 mole

moles of CN₂OH₄ = 1/2(0.699) mole =0.349 mole

Theoretical yield of CN₂OH₄ = (0.349 mole)(60 g/mole) = 20.963 grams

%Yield = Actual Yield/Theoretical Yield x 100%

= 18.5g/20.963g x 100% = 88.25%

7 0

3 years ago

Read 2 more answers

Please help I need to correct I don't understand it

victus00 [196]

Answer is d. in hetrogeneous you can separate things from each other

4 0

2 years ago

What is the total energy change for the following reaction:CO+H2O-CO2+H2

Alekssandra [29.7K]

Answer:

$\large \boxed{\text{-41.2 kJ/mol}}$

Explanation:

Balanced equation: CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)

We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

$\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)$

(a) Enthalpies of formation of reactants and products

$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}$

(b) Total enthalpies of reactants and products

$\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}$

(c) Enthalpy of reaction $\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}$

4 0

2 years ago