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3 years ago

What is 108 g +10+5+3

Chemistry

1 answer:

natali 33 [55]3 years ago

7 0

ANSWER : 108 + 10 = 118

118 + ( 5 + 3 )

118 + 8 = 126

= 126 g

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Na2CO3(aq) + CaCl2(aq) — 2 NaCl(aq) + CaCO3(s)

attashe74 [19]

Answer:

100 mL

Explanation:

Given data:

Mass of CaCO₃ produced = 2.00 g

Molarity of CaCl₂ = 0.200 M

Volume of CaCl₂ needed = ?

Solution:

Chemical equation:

Na₂CO₃ + CaCl₂ → 2NaCl + CaCO₃

First of all we will calculate the number of moles of CaCO₃.

Number of moles = mass/molar mass

Number of moles = 2.00 g / 100.09 g/mol

Number of moles = 0.02 mol

Now we will compare the moles of CaCO₃ and CaCl₂.

CaCO₃ : CaCl₂

1 : 1

0.02 : 0.02

Thus, 0.02 moles of CaCl₂ react,

Volume of CaCl₂ reacted:

Molarity = number of moles / volume in L

0.200 M = 0.02 mol / volume in L

Volume in L = 0.02 mol / 0.200 M

Volume in L = 0.1 L

Volume in mL:

0.1 L × 1000 mL/1L

100 mL

5 0

2 years ago

What is the molar mass of Ca(NO3)2?

gregori [183]

Answer:

C.164.1g/mol

Explanation:

40+14×2+16×6

164

7 0

2 years ago

What is instantaneous acceleration?A. how fast a speed is changing at a specific instant B. how fast a velocity is changing at a

joja [24]

If i had to answer i would choose a hope this helps

8 0

2 years ago

Which of these statements is correct?

Crazy boy [7]

Answer:

Explanation:

3 0

2 years ago

The equilibrium constant for the reaction AgBr(s) Picture Ag+(aq) + Br− (aq) is the solubility product constant, Ksp = 7.7 × 10−

barxatty [35]

Answer:

The reaction will be non spontaneous at these concentrations.

Explanation:

$AgBr(s)\rightarrow Ag^+(aq) + Br^- (aq)$

Expression for an equilibrium constant $K_c$ :

$K_c=\frac{[Ag^+][Br^-]}{[AgCl]}=\frac{[Ag^+][Br^-]}{1}=[Ag^+][Br^-]$

Solubility product of the reaction:

$K_{sp}=[Ag^+][Br^-]=K_c=7.7\times 10^{-13}$

Reaction between Gibb's free energy and equilibrium constant if given as:

$\Delta G^o=-2.303\times R\times T\times \log K_c$

$\Delta G^o=-2.303\times R\times T\times \log K_{sp}$

$\Delta G^o=-2.303\times 8.314 J/K mol\times 298 K\times \log[7.7\times 10^{-13}]$

$\Delta G^o=69,117.84 J/mol=69.117 kJ/mol$

Gibb's free energy when concentration $[Ag^+] = 1.0\times 10^{-2} M$ and $[Br^-] = 1.0\times 10^{-3} M$

Reaction quotient of an equilibrium = Q

$Q=[Ag^+][Br^-]=1.0\times 10^{-2} M\times 1.0\times 10^{-3} M=1.0\times 10^{-5}$

$\Delta G=\Delta G^o+(2.303\times R\times T\times \log Q)$

$\Delta G=69.117 kJ/mol+(2.303\times 8.314 Joule/mol K\times 298 K\times \log[1.0\times 10^{-5}])$

$\Delta G=40.588 kJ/mol$

For reaction to spontaneous reaction: $\Delta G$ .
For reaction to non spontaneous reaction: $\Delta G>0$ .

Since ,the value of Gibbs free energy is greater than zero which means reaction will be non spontaneous at these concentrations

5 0

2 years ago