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4 years ago

The formula of strontium

Chemistry

1 answer:

mrs_skeptik [129]4 years ago

6 0

Sr, because they are just basically asking for it so it is just Sr

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How much garbage does a average American generate?

ale4655 [162]

Well the average American makes about 29 pounds a week so, 52x29= 1508. the average American generates about 1508 pounds a year.

5 0

3 years ago

Solution A has a mass of 70 g. Solution B has a mass of 35 g. When they are mixed, a chemical reaction occurs in which a gas is

Nuetrik [128]

Mass of gas produced : 15 g

<h3>Further explanation </h3>

In general, the reaction takes place in the open air. If the reaction results in the form of gas as in combustion, the mass of the reaction results will be smaller than the original mass.

Conservation of mass applies to a closed system, where the masses before and after the reaction are the same

So In a closed/isolated system, the total mass of the substance before the reaction will be equal to the total mass of the reaction product.

Mass of reactants :

Mass of solution A = 70 g

Mass of solution B = 35 g

Mass of products :

Mass of mixture : 90 g

Mass of gas : X

Mass of gas X :

$\tt mass~reactants(mass~A+mass~B)-mass~products(mass~mixture+mass~gas)\\\\mass~gas=(70+35)-90\\\\mass~gas=15~g$

6 0

3 years ago

Compared to the atoms of nonmetals in period 3, the atoms of metals in period 3 have

siniylev [52]

A. Fewer valence electrons

Valence electrons are electrons that are held most loosely in the outermost shell.

4 0

3 years ago

Read 2 more answers

Consider the following reaction, equilibrium concentrations, and equilibrium constant at

REY [17]

Answer:

Equilibrium concentration of $H_{2}O$ is 12.5 M

Explanation:

Given reaction: $C_{2}H_{4}+H_{2}O\rightleftharpoons C_{2}H_{5}OH$

Here, $K_{c}=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}][H_{2}O]}$

where $K_{c}$ represents equilibrium constant in terms of concentration and species inside third bracket represent equilibrium concentrations

Here, $[C_{2}H_{4}]=0.015M$ , $[C_{2}H_{5}OH]=1.69M$ and $K_{c}=9.0$

So, $[H_{2}O]=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}]\times K_{c}}=\frac{1.69}{0.015\times 9.0}=12.5M$

Hence equilibrium concentration of $H_{2}O$ is 12.5 M

5 0

3 years ago

Consider the chemical reaction represented by the equation below:

svlad2 [7]

3 MgCl2 + 2 Al = 3 Mg + 2 AlCl3

Since we know that we have 12 moles of MgCl2 and enough Al, we don't need to figure out the exact amount of Al.

3 MgCl2 / 2 AlCl3 = 12 mol MgCl2 / x mol AlCl2

cross-multiply

3 MgCl2(x mol AlCl2) = 12 mol MgCl2(2 AlCl3)

3x = 24

x=8

8 moles AlCl3

Alternatively, we know that we have 12 moles of MgCl2, and a coefficient of 3. This indicates a multiplier of 4 for the entire equation, meaning 2 AlCl3 multiplied by (4) = 8 moles

6 0

3 years ago