Ask question

Ask question

All categories

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

You might be interested in

In Valence Shell Electron Pair Repulsion Theory (VSEPR),

VashaNatasha [74]

Answer: look at the close because that is the answer

Explanation:

7 0

2 years ago

Read 2 more answers

1. What are commonly called as "shooting stars"?

nydimaria [60]

Answer:

B

B

A

C

D

Explanation:

I think dont take my word for though

6 0

3 years ago

Read 2 more answers

In the Bohr model, ____ electrons fill the first energy level, ____ electrons fill the second energy level, and ___ electrons fi

Setler [38]

The answer is d Thus, the first energy level holds 2 * 1^2 = 2 electrons, while the second holds 2 * 2^2 = 8 electrons. Each orbital. The third energy level can hold up to 18 electrons, meaning that it is not full when it has only electrons.

8 0

2 years ago

Given the standard heats of reaction

ANTONII [103]

Answer:

Explanation:

M(s) → M (g ) + 20.1 kJ --- ( 1 )

X₂ ( g ) → 2X (g ) + 327.3 kJ ---- ( 2 )

M( s) + 2 X₂(g) → M X₄ (g ) - 98.7 kJ ----- ( 3 )

( 3 ) - 2 x ( 2 ) - ( 1 )

M( s) + 2 X₂(g) - 2 X₂ ( g ) - M(s) → M X₄ (g ) - 98.7 kJ - 2 [ 2X (g ) + 327.3 kJ ] - M (g ) - 20.1 kJ

0 = M X₄ (g ) - 4 X (g ) - M (g ) - 773.4 kJ

4 X (g ) + M (g ) = M X₄ (g ) - 773.4kJ

heat of formation of M X₄ (g ) is - 773.4 kJ

Bond energy of one M - X bond = 773.4 / 4 = 193.4 kJ / mole

6 0

3 years ago

Oh, no! You just spilled 85.00 mL of 1.500 M sulfuric acid on your lab bench and need to clean it up immediately! Right next to

vredina [299]

Explanation:

We will balance equation which describes the reaction between sulfuric acid and sodium bicarbonate: as follows.

$H_2SO_4(aq) + 2NaHCO_3(s) \rightarrow Na2SO_4(aq) + 2H_2O(l) + 2CO_2(g)$

Next we will calculate how many moles of $H_2SO_4$ are present in 85.00 mL of 1.500 M sulfuric acid.

As, Molarity = $\frac{\text{moles of solute}}{\text{liters of solution
}}$

1.500 M = $\frac{n}{0.08500 L
}$

n = 0.1275 mol $H_2SO_4$

Now set up and solve a stoichiometric conversion from moles of $H_2SO_4$ to grams of $NaHCO_3$ . As, the molar mass of $NaHCO_3$ is 84.01 g/mol.

$0.1275 mol H_2SO_4 \times (\frac{2 mol NaHCO_3}{1 mol H_2SO_4}) \times (\frac{84.01 g NaHCO_3}{1 mol NaHCO_3})$

= 21.42 g $NaHCO_3$

So unfortunately, 15.00 grams of sodium bicarbonate will "not" be sufficient to completely neutralize the acid. You would need an additional 6.42 grams to complete the task.

4 0

2 years ago