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Ulleksa [173]
1 year ago
14

A student takes a measured volume of 3.00 M HCl to prepare a 50.0 mL sample of 1.80 M HCl. What volume of 3.00 M HCl did the stu

dent use to make the sample? Use M subscript i V subscript i equals M subscript f V subscript f.. 3.70 mL 16.7 mL 30.0 mL 83.3 mL
Chemistry
2 answers:
Pani-rosa [81]1 year ago
8 0

83.3 mL

Explanation:

M_{i} = 3M

V_{i} = 50mL

M_{f} = 1.80M

Unknown:

V_{f} = ?

Solution:

To solve this problem we must understand that the initial and final number of moles remains the same.

  Number of moles = molarity x volume

Molarity x volume of initial solution = Molarity x volume of final solution

 Therefore:

   M_{i}  x  V_{i}     = M_{f} x   V_{f}

M is the molarity

V is the volume

i = initial state

f = final state

    input the parameters;

  3 x 50 = 1.8 x V_{f}

      V_{f} = 83.3 mL

learn more:

Molarity brainly.com/question/9324116

#learnwithBrainly

forsale [732]1 year ago
3 0

Answer:

83.3

Explanation:

I’m smart

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4.82 g of an unknown metal is heated to 115.0∘C and then placed in 35 mL of water at 28.7∘C, which then heats up to 34.5∘C. What
nikitadnepr [17]

<u>Answer:</u> The specific heat of metal is 2.34 J/g°C

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of water = 1 g/mL

Volume of water = 35 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{35mL}\\\\\text{Mass of water}=(1g/mL\times 35mL)=35g

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]      ......(1)

where,

q = heat absorbed or released

m_1 = mass of metal = 4.82 g

m_2 = mass of water = 35 g

T_{final} = final temperature = 34.5°C

T_1 = initial temperature of metal = 115°C

T_2 = initial temperature of water = 28.7°C

c_1 = specific heat of metal = ?

c_2 = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

4.82\times c_1\times (34.5-110)=-[35\times 4.186\times (34.5-28.7)]

c_1=2.34J/g^oC

Hence, the specific heat of metal is 2.34 J/g°C

4 0
1 year ago
2.92 A 50.0-g silver object and a 50.0-g gold object are both added
Trava [24]

Answer:

82.9 mL  

Explanation:

1. Volume of silver

\begin{array}{rcl}\text{Density}&=& \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho&=& \dfrac{m}{V}\\\\V &=& \dfrac{m}{\rho}\\\\& = & \dfrac{\text{50.0 g}}{\text{10.49 g$\cdot$mL}^{-1}}\\\\& = & \text{4.766 mL}\\\end{array}\\\text{The volume of the silver is $\large \boxed{\textbf{4.766 mL}}$}

2. Volume of gold

\begin{array}{rcl}V& = & \dfrac{\text{50.0 g}}{\text{19.30 g$\cdot$mL}^{-1}}\\\\& = & \text{2.591 mL}\\\end{array}\\\text{The volume of the gold is $\large \boxed{\textbf{2.591 mL}}$}

3. Total volume of silver and gold

V = 4.766 mL + 2.591 mL = 7.36 mL

4 New reading of water level

V = 75.5 mL + 7.36 mL = 82.9 mL

3 0
2 years ago
n the diagram shown, when an object ‘X’ is brought near the ring shaped magnet, the magnet moves away from it. Four friends are
kati45 [8]

Answer:

Akash

Explanation:

it could be a magnet with the same poles facing eachoher

6 0
1 year ago
Salts and acids are examples of inorganic compounds called _____, which dissociate in water to release ions.
Blizzard [7]

Answer:

Salts and acids are examples of inorganic compounds called <u><em>electrolytes</em></u>.

Explanation:

Electrolytes are the substances which dissociates into ions when dissolved in water and due to this they are able to conduct electric current through them.  These compounds in solid form does not conduct electricity due to the absence of free ions.

For example: Sodium chloride , sulfuric acid etc.

NaCl(aq)\righarrow = Na^+(aq)+Cl^-(aq)

H_2SO_4(aq)\rightarrow SO_4^{2-}(aq)+2H^+(aq)

4 0
2 years ago
Read 2 more answers
En una determinación cuantitativa se utilizan 17.1 mL de Na2S2O3 0.1N para que reaccione todo el yodo que se encuentra en una mu
lozanna [386]

Answer:

La cantidad de yodo en la muestra es 0.217 g

Explanation:

Los parámetros dados son;

Normalidad de la solución de Na₂S₂O₃ = 0.1 N

Volumen de la solución de Na₂S₂O₃ = 17.1 mL

Masa de muestra = 0.376 g

La ecuación de reacción química se da de la siguiente manera;

I₂ + 2Na₂S₂O₃ → 2 · NaI + Na₂S₄O₆

Por lo tanto, el número de moles de sodio por 1 mol de Na₂S₂O₃ en la reacción = 1 mol

Por lo tanto, la normalidad por mol = 1 M × 1 átomo de Na = 1 N

Por lo tanto, 0.1 N = 0.1 M

El número de moles de Na₂S₂O₃ en 17,1 ml de solución 0,1 M de Na₂S₂O₃ se da de la siguiente manera;

Número de moles de Na₂S₂O₃ = 17.1 / 1000 × 0.1 = 0.00171 moles

Lo que da;

Un mol de yodo, I₂, reacciona con dos moles de Na₂S₂O₃

Por lo tanto;

0,000855 moles de yodo, I₂, reaccionan con 0,00171 moles de Na₂S₂O₃

La masa molar de yodo = 253.8089 g / mol

La masa de yodo en la muestra = 253.8089 × 0.000855 = 0.217 g.

5 0
2 years ago
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