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Zigmanuir [339]

2 years ago

14

WILLL GIVE BRAINLIEST IF YOU ANSWERRR PLEASEEEE IM LITERALLY BEGGING YOU I PUT THIS QUESTION IN SOO MANY TIMES I HAVEN'T GOTTEN

AN ANSWER PLEASEEE HELPPP

1 answer:

tino4ka555 [31]2 years ago

6 0

Answer:

The answer is 375.54 g of AgBr

Explanation:

Mass (g) = Concentration (mol/L) x volume (L) x Molecular Weight of AgBr (g/mol)

Mass = 2M x 1L x 187.77 g/mol

Mass = 375.54g

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How many sigma bonds in a single bond

Katarina [22]

Answer:

single bond= 1 sigma bond

so basically 1

Explanation:

Usually, all bonds between atoms in most organic compounds contain one sigma bond each. If it is a single bond, it contains only sigma bond. Multiple bonds (double and triple), however, contains sigma and pi bonds. Double bonds have one each, and triple bonds have one sigma bond and two pi bonds

8 0

3 years ago

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What is #6 ,7,8 need help

weeeeeb [17]

#6 should be the independent variable because that's the one you can control

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3 years ago

Suppose 215 g of NO3- flows into a swamp each day. What volume of CO2 would be produced each day at 17.0°C and 1.00 atm?

charle [14.2K]

Answer:

The answer is " $41.23 \ L\ N_2$ "

Explanation:

$2 NO_3^{-} + 10 e^{-} + 12 H^{+} \longrightarrow N_2 + 6 H_2O\\\\= \frac{( 215 \ g \ NO_3^{-})}{(62.0049 \frac{\ g NO_3^{-}}{mol})} \times \frac{(1 \ mol \ N_2}{ 2 \ mol \ NO_3^{-})}\\\\$

$=3.46746789 \times 0.5\\\\= 1.733 \ mol \ N_2 \\\\\to V = \frac{nRT}{P} \\\\= (1.733 \ mol) \times (0.08205746 \frac{L\ atm}{Kmol}) \times \frac{ (17 + 273) K}{(1.00 atm)}\\\\= 41.23$

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3 years ago

1. which is not a component of biodiversity

Vera_Pavlovna [14]

The answer is C, i just took the k-12 test~

8 0

3 years ago

Be sure to answer all parts.The equilibrium constant, Kc, for the formation of nitrosyl chloride from nitric oxide and chlorine,

djverab [1.8K]

<u>Answer:</u> The reaction proceeds in the forward direction

<u>Explanation:</u>

For the given chemical equation:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

Relation of $K_p\text{ with }K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta n_g}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = ?

$K_c$ = equilibrium constant in terms of concentration = $6.5\times 10^4$

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = $35^oC=[35+273]K=308K$

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$K_p=6.5\times 10^4\times (0.0821\times 500)^{-1}\\\\K_p=1583.43$

$K_p$ is the constant of a certain reaction at equilibrium while $Q_p$ is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

The expression of $Q_p$ for above equation follows:

$Q_p=\frac{(p_{NOCl})^2}{p_{Cl_2}\times (p_{NO})^2}$

We are given:

$p_{NOCl}=1.76atm$

$p_{NO}=1.01atm$

$p_{Cl_2}=0.42atm$

Putting values in above equation, we get:

$Q_p=\frac{(1.76)^2}{0.42\times (1.01)^2}=7.23$

We are given:

$K_p=1583.43$

There are 3 conditions:

When $K_{p}>Q_p$ ; the reaction is product favored.
When $K_{p}$ ; the reaction is reactant favored.
When $K_{p}=Q_p$ ; the reaction is in equilibrium

As, $K_p>Q_p$ , the reaction will be favoring product side.

Hence, the reaction proceeds in the forward direction

4 0

3 years ago