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2 years ago

Relative formula mass of ammonium sulphate?

Chemistry

1 answer:

MissTica2 years ago

6 0

Relative atomic mass?
actually molar mass?
Mr= 14(N)+4•1(H)+28(S)+3•16(O)
Mr=94

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Spell out the full name of the compound.

Natalka [10]

Explanation:

Step one look for the longest chain of carbon atoms

Longest chain is 7 C atoms

Step 2 look for double bonds or others functional groups

it is present in 3rd carbon

Therefore IUPAC name is 3-heptene

From point of stereochemistry it can also be written as trans-3-heptene as the hydrogens are placed in opposite side of the C=C bond.

Hope this helps...

3 0

3 years ago

Balanced Equation: C3H,OH,) + __02 - __CO2 + H20

gavmur [86]

2C3H7OH + 9O2 → 6CO2 + 8H2O

8 0

2 years ago

Hydrochloric acid reacts with calcium to form hydrogen and calcium chloride. If 100 grams of hydrochloric acid reacts with 100 g

LenaWriter [7]

Answer is hydrochloric acid

Hope this helps!

5 0

2 years ago

Read 2 more answers

The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 ×

Anuta_ua [19.1K]

Answer:

$Kc=~1.49x10^3^4}$

Explanation:

We have the reactions:

A: $N_2_(_g_) + O_2_(_g_) 2NO_(_g_)~~~~~~Kc = 4.3x10^-^2^5$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~~~Kc = 6.4x10^9$

Our target reaction is:

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)$

We have $NO_(_g_)$ as a reactive in the target reaction and $NO_(_g_)$ is present in A reaction but in the products side. So we have to flip reaction A.

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

Then if we add reactions A and B we can obtain the target reaction, so:

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~Kc=6.4x10^9$

For the final Kc value, we have to keep in mind that when we have to add chemical reactions the total Kc value would be the multiplication of the Kc values in the previous reactions.

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)~~~Kc=\frac{6.4x10^9}{4.3x10^-^2^5}$

$Kc=~1.49x10^+^3^4}$

3 0

2 years ago

A cylinder with a moving piston expands from an initial volume of 0.350 L against an external pressure of 1.90 atm. The expansio

myrzilka [38]

Answer:

1.84 L

Explanation:

Using the equation for reversible work:

$W = -P*(V_{2} - V_{1})$

Where:

W is the work done (J) = -287 J.

Since the gas did work, therefore W is negative.

P is the pressure in atm = 1.90 atm.

However, work done is in joules and pressure is in atm. We can use the values of universal gas constant as a convenient conversion unit. R = 8.314 J/(mol*K); R = 0.0821 (L*atm)/(mol*K)

Therefore, the conversion unit is 0.0821/8.314 = 0.00987 (L*atm)/J

$V_{1}$ is the initial volume = 0.350 L

$V_{2}$ is the final volume = ?

Thus:

(-287 J)*0.00987 (L*atm)/J = -1.9 atm*( $V_{2}$ - 0.350) L

$V_{2}$ = [(287*0.00987)+(1.9*0.350)]/1.9 = (2.833+0.665)/1.9 =1.84 L

7 0

3 years ago