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2 years ago

Calculate the mass of 5.22 mol of He.

Chemistry

1 answer:

andrew-mc [135]2 years ago

4 0

Answer:

The mass is 20, 89 grams of He.

Explanation:

We calculate the weight of 1 mol of He, from the atomic weight of each element obtained from the periodic table: Weight 1 mol He= 4, 0020602 g. 1 mol He-----4, 0020602 g. 5,22 mol He---x= (5,22 mol He x4, 0020602 g)/ 1 mol He= 20, 89 g

hope this helps also can i get brainliest

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Explanation:

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3 years ago

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2 years ago

The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that

kakasveta [241]

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is $1.1\times 10^6$ M.

$HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)$

This value indicates that

A. $CN^-$ is a stronger base than $ONO^-$

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is $ONO^-$

D. The conjugate acid of CN- is HCN

Answer: A. $CN^-$ is a stronger base than $ONO^-$

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When $K_{p}>1$ ; the reaction is product favoured.

When $K_{p};$ ; the reaction is reactant favored.

$When K_{p}=1$ ; the reaction is in equilibrium.

As, $K_p>>1$ , the reaction will be product favoured and as it is a acid base reaction where $HONO$ acts as acid by donating $H^+$ ions and $CN^-$ acts as base by accepting $H^+$

Thus $HONO$ is a strong acid thus $ONO^-$ will be a weak conjugate base and $CN^-$ is a strong base which has weak $HCN$ conjugate acid.

Thus the high value of K indicates that $CN^-$ is a stronger base than $ONO^-$

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2 years ago

Question

Black_prince [1.1K]

Answer:

efef2w

Explanation:

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2 years ago

Write the balanced nuclear equation for the following. (Use the lowest possible coefficients. Omit states-of-matter in your answ

olga55 [171]

Answer:

$_{93}^{232}\text{Np} + _{-1}^{0}\text{e} \longrightarrow _{92}^{232}\text{U}$

Explanation:

The unbalanced nuclear equation is

$_{93}^{232}\text{Np} + _{1}^{0}\text{e} \longrightarrow ?$

It is convenient to replace the question by an atomic symbol, $_{x}^{y}\text{Z}$ , where x = the atomic number, y = the mass number, and Z = the symbol of the element.

$_{93}^{232}\text{Np} + _{1}^{0}\text{e} \longrightarrow _{x}^{y}\text{Z}$

Then your equation becomes

$_{93}^{232}\text{Np} + _{1}^{0}\text{e} \longrightarrow _{x}^{y}\text{Z}$

The main point to remember in balancing nuclear equations is that the sums of the superscripts and of the subscripts must be the same on each side of the equation.

Then

93 – 1 = x, so x = 92

232 + 0 = y, so y = 232

Element 92 is uranium, so the nuclear equation becomes

$_{93}^{232}\text{Np} + _{-1}^{0}\text{e} \longrightarrow _{92}^{232}\text{U}$

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2 years ago