How many grams are in 23.0 moles of carbon (C)?

Please help...awarding brainliest if correct and lots of points.

abruzzese [7]

Silver chloride produced : = 46.149 g

Limiting reagent : CuCl2

Excess remains := 3.74 g

<h3>Further explanation</h3>

Reaction

silver nitrate + copper(II) chloride ⇒ silver chloride + copper(II) nitrate

Required

silver chloride produced

limiting reagent

excess remains

Solution

Balanced equation

2AgNO3 (aq) + CuCl2 (s) → 2AgCl(s) + Cu(NO3)2(aq)

mol AgNO3 :

= 58.5 : 169,87 g/mol

= 0.344

mol CuCl2 :

=21.7 : 134,45 g/mol

= 0.161

mol ratio : coefficient of AgNO3 : CuCl2 :

= 0.344/2 : 0.161/1

= 0.172 : 0.161

CuCl2 as a limiting reagent

mol AgCl :

= 2/1 x 0.161

= 0.322

Mass AgCl :

= 0.322 x 143,32 g/mol

= 46.149 g

mol remains(unreacted) for AgNO3 :

= 0.344-(2/1 x 0.161)

= 0.022

mass AgNO3 remains :

= 0.022 x 169,87 g/mol

= 3.74 g

8 0

3 years ago

A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions

Mandarinka [93]

Answer:

$\boxed{\text{-55.8 kJ/mol NaOH}}$

Explanation:

NaOH + HNO₃ ⟶ NaNO₃ + H₂O

There are two energy flows in this reaction.

$\begin{array}{cccl}\text{Heat from neutralization} & + &\text{Heat absorbed by water} & = 0\\q_{1} & + & q_{2} & =0\\n\DeltaH & + & mC\Delta T & =0\\\end{array}$

Data:

V(base) = 100.0 mL; c(base) = 0.300 mol·L⁻¹

V(acid) = 100.0 mL; c (acid) = 0.300 mol·L⁻¹

T₁ = 35.00 °C; T₂ = 37.00 °C

Calculations:

(a) q₁

$n_{\text{NaOH}} = \text{0.1000 L } \times \dfrac{\text{0.300 mol}}{\text{1 L}} = \text{0.0300 mol}\\\\n_{\text{HNO}_{3}} = \text{0.1000 L } \times \dfrac{\text{0.300 mol}}{\text{1 L}} = \text{0.0300 mol}$

We have equimolar amounts of NaOH and HNO₃

n = 0.0300 mol

q₁ = 0.0300ΔH

(b) q₂

V = 100.0 mL + 100.0 mL = 200.0 mL

m = 200.0 g

ΔT = T₂ - T₁ = 37.00 °C – 35.00 °C = 2.00 °C

q₂ = 200.0 × 4.184 × 2.00 = 1674 J

(c) ΔH

0.0300ΔH + 1674 = 0

0.0300ΔH = -1674

ΔH = -1674/0.0300

ΔH = -55 800 J/mol

ΔH = -55.8 kJ/mol

$\Delta_{r}H^{\circ} = \boxed{\textbf{-55.8 kJ/mol NaOH}}$

6 0

3 years ago

A 25.0 sample of liquid has a mass of 22.1 grams what's the liquids density

lyudmila [28]

The density of the liquid is 0.884 g/mL.

<em>Density = mass/volume</em> = 22.1 g/25.0 mL = 0.884 g/mL

7 0

3 years ago

What is transmutation?

Tanzania [10]

Answer:

D. the conversion of one element into another

Explanation:

Transmutation is the action of changing or the state of being changed into another form. Therefore, D is the correct answer choice.

4 0

3 years ago

Read 2 more answers

Compound has a molar mass of and the following composition: elementmass % carbon47.09% hydrogen6.59% chlorine46.33% Write the mo

snow_lady [41]

The given question is incomplete. The complete question is:

Compound X has a molar mass of 153.05 g/mol and the following composition:

element mass %

carbon 47.09%

hydrogen 6.59%

chlorine 46.33%

Write the molecular formula of X.

Answer: The molecular formula of X is $C_6H_{10}Cl_2$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 47.09 g

Mass of H = 6.59 g

Mass of Cl = 46.33 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{47.09g}{12g/mole}=3.92moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.59g}{1g/mole}=6.59moles$

Moles of Cl = $\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{46.33g}{35.5g/mole}=1.30moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.92}{1.30}=3$

For H = $\frac{6.59}{1.30}=5$

For Cl = $\frac{1.30}{1.30}=1$

The ratio of C : H: Cl= 3: 5 :1

Hence the empirical formula is $C_3H_5Cl$

The empirical weight of $C_3H_5Cl$ = 3(12)+5(1)+1(35.5)= 76.5g.

The molecular weight = 153.05 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{153.05}{76.5}=2$

The molecular formula will be= $2\times C_3H_5Cl=C_6H_{10}Cl_2$

7 0

3 years ago