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3 years ago

How did Robert Goddard contribute to space exploration?

Chemistry

1 answer:

stealth61 [152]3 years ago

6 0

Answer:

He was the first person to use liquid fuel for rockets.

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Which word equation shows lithium oxide being formed from the reaction between oxygen and lithium? oxygen lithium oxide right ar

Oliga [24]

The equation of the reaction for the formation lithium oxide from lithium and oxygen is given as:

$2Li + O_2 \rightarrow Li_2O$

<h3>What is the equation for the formation of Lithium oxide?</h3>

Lithium oxide is a compound formed from the reaction between lithium and oxygen.

The equation for the chemical reaction is given below:

$2Li + O_2 \rightarrow Li_2O$

Therefore, the reaction between lithium and oxygen yields lithium oxide.

Learn more about reactions at: brainly.com/question/26018275

#SPJ4

5 0

2 years ago

Why do cells divide?

Zielflug [23.3K]

Cells divide for so living things can grow

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3 years ago

The science of using tree rings to determine absolute age is called

leva [86]

Option c. dendrochronology

tree rings or dendrochronology they allow to use it in calibration for carbon-14 on temporal placements of fragments of wood (from long dead trees).

Example Bristle cone pines (1957) 4723 years old

4 0

4 years ago

Read 2 more answers

What is not produced during glycosis

a_sh-v [17]

The energy containing electron transporters of FADH2 are not produced during glycolysis.

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4 years ago

What is the OH- of {H+} = 4.0 x 10 to the power of -8

True [87]

Answer:

At standard room temperature, $[{\rm OH^{-}] \approx 2.5 \times 10^{-7}\; \rm M$ when $[{\rm H^{+}] = 4.0 \times 10^{-8}\; \rm M$ .

Explanation:

The following equilibrium goes on in water:

${\rm H_{2}O}\, (l) \rightleftharpoons {\rm H^{+}}\, (aq) + {\rm OH^{-}}\, (aq)$ .

The forward reaction is known as the self-ionization of water. The ionization constant of water, $K_{\rm w}$ , gives the equilibrium position of this reaction:

$K_{\rm w} = [{\rm H^{+}] \cdot [{\rm OH^{-}}]$ .

At standard room temperature ( $25\; {\rm ^{\circ}C}$ ), $K_{\rm w} \approx 10^{-14}$ . Also, $[{\rm H^{+}}] = 4.0 \times 10^{-8}\; \rm mol \cdot L^{-1}$ . Substitute both values into the equation and solve for $[{\rm OH^{-}}]$ .

$\begin{aligned} {[}{\rm OH^{-}}{]} &= \frac{K_{\rm w}}{[{\rm H^{+}}]} \\ &\approx \frac{10^{-14}}{4.0 \times 10^{-8}} = 2.5 \times 10^{-7}\end{aligned}$ .

In other words, in an aqueous solution at standard room temperature, $[{\rm OH^{-}] \approx 2.5 \times 10^{-7}\; \rm M$ when $[{\rm H^{+}] = 4.0 \times 10^{-8}\; \rm M$ .

5 0

3 years ago