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Len [333]
3 years ago
13

A scientist has 8.5 g of silver nitrate and needs to prepare 1.5 L of a 0.020 M solution. Will there be enough silver nitrate? I

f so, how much silver nitrate will be leftover?
A) Nothing leftover
B) 3.4 g
C) 0.040 g
D) 2.3 g
Chemistry
1 answer:
andrey2020 [161]3 years ago
4 0

Answer:

is

Explanation:

the fest president is

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Where are the alkali metals, the alkaline earth metals, the halogens, and the noble gases located on the periodic table?
ivolga24 [154]
The Alkali Metals- (Group 1A)can be found (excluding hydrogen) in the first column of the periodic table. They all have 1 valence electron and tend to form +1 cations when forming Ionic bonds with non metals. 

The Alkali Earth Metals: (Group 2A) can be found in the second column/group on the periodic table. These elements form +2 positively charged cations when forming Ionic bonds with non metals. They also all have 2 valence electrons. 

The Halogens: are the elements that make up the second to last group on the periodic table. These elements include Chlorine, Fluorine, Bromine, Iodine, and Astatine. These elements have 7 valence electrons. The usually form a -1 charged anion when forming ionic bonds with metals. 

The Noble gases: the very last group on the periodic table. All these elements have 8 valance electrons. Due to the octet rule these atoms are already very stable and rarely react with other elements. 

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3 years ago
All systems have. A. Output and feedback. B. Process and function. C. input,process,and sensors.
zvonat [6]
I think the Answer is C
7 0
4 years ago
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Where are intrusive igneous rocks formed? What size of crystals will they have?
vladimir2022 [97]

Answer:

It is formed when magma is trapped deep inside the Earth.

Explanation:

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medium grained rocks ( crystal 2mm to 5mm) coarse grained rocks with crystal over 5mm in size.

6 0
3 years ago
If 15.5 mL of 0.225 M aqueous magnesium chloride is added to 37.5 mL of 0.250 M aqueous lead(II) nitrate, then what mass of lead
Evgen [1.6K]

Answer:

B. 0.971 g

Explanation:

When MgCl₂(aq) reacts with Pb(NO₃)₂(aq), PbCl₂(s) and Mg(NO₃)₂(aq) are produced:

MgCl₂(aq) + Pb(NO₃)₂(aq) →, PbCl₂(s) + Mg(NO₃)₂(aq)

Thus, we need to find imiting reactant finding moles of each reactant:

<em>Moles MgCl₂:</em>

15.5mL = 0.0155L * (0.225 mol / L) = 3.49x10⁻³ moles

<em>Moles Pb(NO₃)₂:</em>

37.5mL = 0.0375L * (0.250mol / L) = 9.38x10⁻³ moles

As the ratio of the reactants is 1:1, the moles of PbCl₂ are 3.48x10⁻³ moles.

We need to convert thes moles to mass using molar mass of PbCl₂ (278.1g/mol), thus:

3.48x10⁻³ moles * (278.1g/mol) =

0.968g of PbCl₂ are precipitate

Thus, right answer is:

<h3>B. 0.971 g</h3>
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3 years ago
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