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3 years ago

What would the products of a double-replacement reaction between AgNO3

Chemistry

1 answer:

Radda [10]3 years ago

8 0

Answer:

D. AgCl and KNO₃

Explanation:

anions switch from the original reaction.

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What is the basic building block of matter to a chemist

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The basic building block of matter is the atom.

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3 years ago

5. __NH3 + __O2 >>>>>__ NO +__ H20

inn [45]

2 NH3+ 2 O2 —> 2 NO+ 3 H2O

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3 years ago

A chemist determined by measurements that 0.030 moles of barium participated in a chemical reaction. Calculate the mass of bariu

schepotkina [342]

Answer: 4.1 g of barium precipitated.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar Mass}}$

Given : moles of barium = 0.030

Molar mass of barium = 137 g/mol

$0.030=\frac{x}{137}$

x= 4.1 g

Thus there are 4.1 g of barium that precipitated.

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3 years ago

Cane sugar is being manufactured at the rate of 500 kg/hr from sugar canes that have the following composition (% by weight):

My name is Ann [436]

Answer:

a. 4166,67 b. 74553.0

Explanation:

6 0

2 years ago

If the sample has a total mass of 5.76 g and contains 1.79 g k, what are the percentages of kbr and ki in the sample by mass

Luden [163]

The percentages of KBr and KI in the sample by mass is 80.68 and 19.32 % respectively.

<h3>What is Molar Mass ?</h3>

Molar mass is defined as the mass contained in 1 mole of sample.

It is given that

the sample has a total mass of 5.76 g contains KBr and KI

and contains 1.79 gm of K is present

what is the percentage of KBr and KI in the sample

Molecular weight of K = 39

Molecular weight of Br = 79.9

Molecular weight of I = 126.9

In KBr the mass percentage of K is 39/(39+79.9) = 32.89%

In KI the mass percentage of K is 39/(39+126.9) = 23.5%

Let the mass of KBr present in the sample is x

K will be 0.3289 x

and let the mass of KI present be y

K will be 0.235y

x +y =5.76

0.3289x+0.235y = 1.79

0.0939y = 0.1045

y = 1.1125 gm

x = 5.76-1.1125

x = 4.6475 gm

% of KBr = (4.6475/5.76 )*100 = 80.68 %

% of KI = (1.1125/5.76) *100 = 19.32%

To know more about Molar Mass

brainly.com/question/12127540

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8 0

2 years ago